AP Physics 2
Table of Contents
Unit 1: Fluids……….…………….……..………..3
Unit 2: Thermal Physics..……………….……….15
Unit 3: Electrostatics…………………...………..30
Unit 4: Circuits……………………………......…50
Unit 5: Magnetism……….……………….……..67
Unit 6: Light Waves………………………......…86
Unit 7: Particle Physics…..…………………….112
2
AP Physics 2
Unit 1: Fluids
Section 1.1 – Pressure……………………………………………………………………………4
Section 1.2 – Buoyancy………………………………………………………………..…………8
Section 1.3 – Fluid Dynamics………………………………………………..............…………10
3
AP Physics 2
1.1 Pressure
Unit 1: Fluids
Focus Question: How does pressure change with depth?




Solid – Shape and volume do not readily change.
Liquid – Takes on the shape of its container, but volume does not readily change (liquids are
incompressible).
Gas – No fixed shape of fixed volume. Will expand to fill whatever container it’s in.
Fluid Does not maintain a fixed mass.
 Flows under pressure.
 Includes liquids, gasses, and plasmas.
Density & Specific Gravity

Density – The ratio of an objects mass to the volume it occupies:
𝜌=
*units of density = 𝑘𝑔/𝑚

𝑀
𝑉
Specific Gravity – Specific gravity is the ratio of an object’s density to the density of water:
𝜌
𝑆𝐺 =
𝜌
The value of density of water in Physics 2 can be taken as 1000 kg/m 3
Pressure
 Pressure – When a force is applied to a surface, the pressure on the surface is the force applied on the
surface divided by the surface area:
𝐹
𝑃=
𝐴
Units of Pressure: Pascals (Pa)
Pressure is sometimes measured in atmospheres, where 1 atmosphere is equal to the pressure exerted by the
earth’s atmosphere at sea level: 1 atm = 1.013x105 Pa

Pascal’s Principle - If pressure is applied to a confined fluid then the
pressure at all points in the fluid is greater than the hydrostatic
pressure. The pressure throughout the fluid is increased everywhere.
The pressure on each piston is equal:
𝑃 =𝑃
𝑭𝟏 𝑭𝟐
=
𝑨𝟏 𝑨𝟐
4
Example A: A car lift uses pistons to raise a car. Compressed air
exerts a force on a piston with a radius of 5.0 cm, and the pressure on
this piston is transmitted to a larger piston with a radius of 20 cm to
lift a car with a mass of 1200 kg.
a) What force does the compressed air exert?
b) How does the work done by the pistons compare?
a) Pascal’s Principle:
=
→
=
→𝐹 =
(𝐹 )
(5 𝑐𝑚)
𝑚
(1200 𝑘𝑔) 10
= 𝟕𝟓𝟎 𝑵
(20 𝑐𝑚)
𝑠
b) Since a constant volume is transmitted, the volume is equated (one piston pushes a volume of air to
the other, that volume is the same):
∆𝑥
𝜋𝑟
𝑉 = 𝑉 → 𝜋𝑟 ∆𝑥 = 𝜋𝑟 ∆𝑥 →
=
∆𝑥
𝜋𝑟
With the ratio of the distances moved by the pistons, the ratio of the work can be found:
𝑊
𝐹 ∆𝑥
=
𝑊
𝐹 ∆𝑥
→𝐹 =
Substitute in
∆
:
=
∆
𝑊
𝐹
=
𝑊
𝐹
𝑟
𝑟
=
(20 𝑐𝑚)
= 1 → 𝑾𝟏 = 𝑾𝟐
(1200 𝑘𝑔)(10 ) (5 𝑐𝑚)
(750 𝑁)
Example B: A swimming pool is a rectangular prism with a width of 8 m, a length of 14 m, and an average
depth of 1.5 m.
a) Calculate the weight of the water in the swimming pool.
b) Calculate the force the Earth’s atmosphere exerts on the pool’s surface.
c) A frisbee of radius 𝑟 = .1 𝑚 rests at a depth of 1 m below the surface of the pool. Calculate the
pressure exerted on the top surface of the frisbee from the water.
a) The density of water is 1000 kg/m3
(8 𝑚)(14 𝑚)(1.5 𝑚) = 168000 𝑘𝑔
𝜌 = → 𝑀 = 𝜌𝑉 = 1000
b) Atmosphere pressure is 1.013x105 Pa
𝑃 = → 𝐹 = 𝑃(𝐴) = 1.01𝑥10 𝑃𝑎(8 𝑚)(14 𝑚) = 1.1𝑥10 𝑁
c) The mass of the volume of water only directly above the frisbee
exerts pressure on it:
𝑃= =
𝑃=
=
=
= 𝜌𝑔ℎ
𝑃 = (1000
)(10 )(1 𝑚) = 10,000 𝑃𝑎
5

Hydrostatic Pressure
 A fluid exerts equal pressure in all directions: Pressure of a fluid is due to the weight of the fluid
above it.
 When a fluid is at rest, the force it exerts is always perpendicular to a surface in contact with the
fluid.
 A fluid exerts pressure on a submerged object in all directions.
 Hydrostatic Equation - Used to calculate the pressure at a point h meters below the surface of a
fluid:
Hydrostatic pressure is the force per unit area exerted on an object underwater by the column of water above it:
𝐹𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑤𝑎𝑡𝑒𝑟 𝑎𝑏𝑜𝑣𝑒
𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑏𝑜𝑣𝑒
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑏𝑜𝑣𝑒
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 𝑎𝑏𝑜𝑣𝑒 𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝑚𝑔 𝜌𝑉𝑔 𝜌(𝐴ℎ)𝑔
𝑃=
=
=
𝐴
𝐴
𝐴
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
Hydrostatic Pressure Formula: 𝑷 = 𝝆𝒈𝒉
*The pressure in a uniform liquid is the same at every point ℎ meters below the surface.

Gauge Pressure
A pressure gauge indicates pressure above atmospheric pressure. The pressure given by the gauge is
called gauge pressure. To find absolute pressure: Add the measured pressure to atmospheric pressure.
When submerged is water, the pressure on an object is due to the water above, and the atmospheric
above it.
𝑷𝑨𝑩𝑺 = 𝑷𝑨𝑻𝑴 + 𝑷𝑮𝑨𝑼𝑮𝑬
*Atmospheric pressure on Earth’s Surface varies with weather conditions and altitude *air compressible
*The weight of the atmosphere is large. We are not crushed because our cells are close to atmospheric pressure.
Example C: Determine an expression for the absolute pressure at each
of the labeled points.
A: 𝑃 = 𝑃
since A is open to the atmosphere.
B: 𝑃 = 𝑃
+ 𝜌𝑔ℎ
C: 𝑃 = 𝑃
+ 2𝜌𝑔ℎ
D: The pressure at D is equal to the pressure at B since they
have the same depth: 𝑃 = 𝑃
+ 𝜌𝑔ℎ
E: The pressure at E is equal to the pressure at C since they have the same depth:
𝑃 =𝑃
+ 2𝜌𝑔ℎ
6
Example D: A cube of wood with sides of 10 m floats at the interface of oil and water as shown. The cube’s
lower surface is 2.0 m below the interface. The density of oil is 600 kg/m 3.
a) What is the gauge pressure at the upper face of the block?
b) What is the gauge pressure at the lower face of the block?
c) What is the absolute pressure at the lower face of the block?
a) The top surface is under 2 m of oil, so the pressure is due to the force of the oil only:
(2 𝑚) = 12,000 𝑃𝑎 = 12 𝑘𝑃𝑎
𝑃 = 𝜌 𝑔ℎ = 600
10
b) The bottom surface is under all 10 m of oil. Additionally, it is under 2 m of water.
𝑃 = 𝜌 𝑔ℎ + 𝜌
𝑔ℎ
(10 𝑚) + 1000
(2 𝑚) = 80,000 𝑃𝑎
→ 𝑃 = 600
10
10
c) The absolute pressure is the gauge pressure added to atmosphere pressure:
𝑃 = 80,000 𝑃𝑎 + 10,130 𝑃𝑎 = 90,130 𝑃𝑎 = 90 𝑘𝑃𝑎
This is equal to about 9 atmospheres (9 atm)
Rate your understanding: Pressure
0
1
I’m certainly feeling
pressure.
I can understand fluids
with help.
2
3
4
I can solve problems
involving fluids with
minor errors.
I can solve problems
involving fluids with no
errors.
I can explain and teach
pressure.
7
AP Physics 2
1.2 Buoyancy
Unit 1: Fluids
Focus Question: What is Archimedes’s Principle

Buoyant Force – Upward force on an object in a fluid; the buoyant force acts against gravity.
*An object will float if: The buoyant force is greater than the weight.
*When an object floats, the upward buoyant force on the object is greater than the downward force due to
gravity. There is a buoyant force because pressure increases with depth.

Archimedes Principle- The upward buoyant force that is exerted on an
object submerged in a liquid is equal to the weight of the liquid
displaced by the object.
𝐹 = 𝑚𝑔
→ 𝑚𝑎𝑠𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒
𝑭𝑩 = 𝝆𝒈𝑽𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒅
*An object floats if its density is less than the liquid.
Example A: What fraction of a piece of aluminum (𝜌 = 2700 𝑘𝑔/𝑚 ) will be submerged when floating in
mercury (𝜌 = 13600 𝑘𝑔/𝑚 )?
Since the aluminum floats, the buoyant force is equal to its weight.
𝐹 =0=𝐹 −𝐹 →𝐹 =𝐹
𝐹 =𝜌
𝑔𝑉
𝐹 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑖𝑒𝑐𝑒 𝑜𝑓 𝑎𝑙𝑢𝑚𝑖𝑛𝑢𝑚 = 𝜌
𝑔𝑉
When calculating volumes, only the submerged height is used for the
displaced mercury, the total height is used for the volume of the aluminum
itself. Both volumes have the same cross-sectional area.
𝜌
𝑔(𝐴ℎ
)=𝜌
𝑔(𝐴ℎ
)
ℎ
𝜌
2700 𝑘𝑔/𝑚
=
=
= 𝟎. 𝟐
ℎ
𝜌
13600 𝑘𝑔/𝑚
Example B: The figure shows a floating block of density 640 kg/m3 floating in water.
Calculate the fraction h/L.
𝜌
𝑔𝑉
=𝜌
𝑔𝑉
𝜌
𝐴(𝐿 − ℎ) = 𝜌
𝐴(𝐿)
𝒉
1000𝐿 − 1000ℎ = 640𝐿 → −1000ℎ = −360𝐿 → =. 𝟑𝟔
𝑳

In general, the fraction of an object that is submerged when it floats equals that ratio of the object’s
density to that of the liquid.
In water, the fraction submerged is equal to the specific gravity of the material:
𝜌
𝑔𝑉
=𝜌
𝑔𝑉
→𝜌
𝐴ℎ
=𝜌
𝐴ℎ
ℎ
𝜌
→
=
= 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
𝜌
ℎ
8
Example C: Totally submerged object: An apple with is hung from a spring scale. When
the object is in the air, the spring scale reads 2.0 N. When the apple is fully submerged in
water, the scale reads 1.5 N.
a) Calculate the density of the apple.
b) The spring scale is removed while the apple is fully submerged. Will the apple
float up or sink?
a) Since the object is at rest, the sum of the forces on it are zero. Weight acts
downward, and the upward force of the spring scale holds it upwards. It is also
pushed up by the buoyant force. Because the spring scale reads 2 N when the apple is not in water,
that is the weight the apple.
𝐹 =0=𝐹 +𝐹 −𝐹 =0
Since the apple is fully submerged, all of its volume is used when applying Archimedes’ Principle. The volume
of the apple is its mass dived by its density. The mass the apple is its weight (which is known) divided by
acceleration due to gravity.
𝑚
→𝐹 +𝜌
𝑔𝑉
=𝑊
→𝐹 +𝜌
𝑔
=𝑊
𝜌
→𝜌
𝑔
𝜌
→𝜌
=𝑊
=
−𝐹 →𝜌
(1000
)(2 𝑁)
2 𝑁 − 1.5 𝑁
=
𝜌
𝑊
= 𝟒𝟎𝟎𝟎
𝑊
−𝐹
𝒌𝒈
𝒎𝟑
b) Since the apple in this problem is more dense than water, it will sink.
When an object is fully submerged in water , it will accelerate up if it is less dense than water. If it is
more dense than water, it will acceleration down.
Example D: A spherical air balloon has a radius of 9.5 m and is filled with helium. What is the maximum mass
that the ball can lift? (𝜌 = .18 , 𝜌
= 1.29 𝑘𝑔/𝑚 ). The mass of the balloon can be neglected, but not
the mass of the helium.
The buoyant force has to support the cargo and the weight of the helium in the balloon.
𝐹 =0=𝐹 −𝑊
𝑚
→ 𝜌 𝑔𝑉 = 𝜌 𝑔𝑉 + 𝑚
𝑔
𝑚
→ (𝜌 − 𝜌 )(𝑉)
𝑘𝑔
𝑘𝑔 4
= 1.29
− .18
𝜋(9.5 𝑚) = 4000 𝑘𝑔
𝑚
𝑚
3
Rate your understanding: Buoyancy
0
1
My grade will not stay
afloat.
−𝑊
I can understand fluids
with help.
2
3
4
I can solve problems
involving fluids with
minor errors.
I can solve problems
involving fluids with no
errors.
Eureka! I can explain
and teach fluids.
9
AP Physics 2
1.3 Fluid Dynamics
Unit 1: Fluids
Focus Question: What effects the velocity of flowing ideal fluid?
 Laminar Flow – Trajectory of each particle is smooth and trajectories
do not cross. Also called “streamline” flow.

Turbulent Flow- Flow where trajectories of water particles cross.
Occurs is the flow goes above some critical velocity.
*An eddy current is when current circles back on itself. Eddy currents disrupt the flow of a fluid.
The Ideal Fluid
For fluids in motion, fluids are assumed to be ideal fluids. An ideal fluid has the following properties:
o The fluid exhibits laminar flow.
o The fluid is non-viscous, so there is no drag force.
*Viscosity is analogous to friction for the flow of fluids.
o The fluid is incompressible; the density is the same everywhere in the fluid.
o The fluid is steady.
*velocity, density, and pressure are constant over time.
Equation of Continuity
Mass flow rate stays the same:
𝑚𝑎𝑠𝑠 𝑚𝑎𝑠𝑠 𝜌𝑉 𝜌𝑉
=
→
=
𝑡
𝑡
𝑡
𝑡
𝜌𝐴 𝑣 = 𝜌𝐴 𝑣
𝑨𝟏 𝒗𝟏 = 𝑨𝟐 𝒗𝟐
Bigger cross section = slower flow. Smaller cross section = faster flow.
*Fluid speeds up where flow is constricted. This is based on conservation of mass.
Example A: Water runs through a water main of with a circular cross section of radius 5 cm 2 with a velocity of
6 m/s.
a) Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-section radius of
1.0 cm2.
b) How long will it take the water out of the faucet to fill a 200 L tub?
a) Applying the continuity formula:
𝐴 𝑣 =𝐴 𝑣 →𝑣 =
=
𝑣
(
→𝑣 =( (
) )
) )
6
= 150 𝑚/𝑠
b) The volume flow rate can be calculated using either cross section:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 = = (𝐴) = 𝑣𝐴 = 150
There are 1000 L in a cubic meter:
.
(𝜋)(. 01 𝑚) ) = .047
= 4.3 𝑠
10
Bernoulli’s Equation
Bernoulli’s Equation is derived from applying conservation of
energy to the flow of an ideal fluid:
Going back to Physics 1, the change in energy is due to work
done by outside forces. The work done on the fluid from the
outside forces exerting pressure is equal to the total change in
energy:
𝑊
= ∆𝐾 + ∆𝑈
The work done on each end of the fluid flow is:
𝑊 = 𝐹∆𝑥 = (𝑃𝐴)∆𝑥 = 𝑃𝑉
The net work is then 𝑊
= 𝑃 𝑉 − 𝑃 𝑉 since both ends of the flow have forces in opposite directions.
The change in kinetic energy is: ∆𝐾 = 𝑚𝑣 − 𝑚𝑣
The change in potential energy is: ∆𝑈 = 𝑚𝑔𝑦 − 𝑚𝑔𝑦
The work energy relationship is then:
1
1
𝑃 𝑉 − 𝑃 𝑉 = 𝑚𝑣 − 𝑚𝑣 + 𝑚𝑔𝑦 − 𝑚𝑔𝑦
2
2
Since 𝜌 = 𝑚𝑉, dividing the entire equation by volume yields:
1
1
𝑃 − 𝑃 = 𝜌𝑣 − 𝜌𝑣 + 𝜌𝑔𝑦 − 𝜌𝑔𝑦
2
2
𝟏
𝟏
→ 𝑷𝟏 + 𝝆𝒗𝟐𝟏 + 𝝆𝒈𝒚𝟏 = 𝑷𝟐 + 𝝆𝒗𝟐𝟐 + 𝝆𝒈𝒚𝟐
𝟐
𝟐
*Fluid velocity is low where pressure is high and fluid velocity is high where pressure is low.
Example B: A sealed tank contains water filled to a height of h = 8.0
m. The air above the water is at 40 atm of pressure. Water flows out
of a hole at a height a = 5 m above the ground.
a) Find the speed at which the water leaves the whole.
b) Find the horizontal range travel by the water.
c) How does the height a, of the hole affect the horizontal
range of the water?
a) Bernoulli’s equation will be applied for the top of the water
and the opening. The water level at the top is assumed to be
constant (otherwise this is Calc II problem), and the opening is open to the atmosphere, so it is at
atmospheric pressure:
1
1
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
2
2
→𝑣 =
2
→𝑣 =
2(
(
.
)
(
) ( .
)
(
)
= 𝟔𝟑 𝒎/𝒔
11
*If the top of the container is also open to the atmosphere, the equations becomes:
1
1
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
2
2
𝑣 =
2𝑔 𝑦 − 𝑦
The result above is a statement of Torricelli's Theorem
b) Kinematics, WOW. OMG. This is projectile problem with an initial horizontal velocity of v=63 m/s. The
water accelerates down due to gravity. Such fun.
Find time in the air using the y-direction: 𝑣 = 0, 𝑎 = −10 , ∆𝑦 = −5 𝑚, 𝑡 =?
1
∆𝑦 = 𝑣 𝑡 + 𝑎𝑡 → 𝑡 =
2
2∆𝑦
=
𝑎
2(−5𝑚)
−10
=1𝑠
Find the horizontal displacement since the velocity in the x-direction doesn’t change:
𝑚
(1 𝑠) = 𝟔𝟑 𝒎
∆𝑥 = 𝑣 𝑡 = 63
𝑠
c) The low the opening is in the container of the water, the faster the water will travel, giving it a larger
horizontal velocity when it leaves the container. However, a lower opening leads to less time in the air.
Example C: In a hydroelectric power plant, water leaves a dam
from a point 50 m beneath the surface. It enters a pipe of radius
80 cm and is incident on a turbine through a pipe of radius 40 cm.
a) Find the speed of the water as it hits the turbine.
b) Calculate the pressure at point 2.
a) Both the top of the reservoir and the turbine are open to the
atmosphere. Also, the water level of the reservoir is assumed to
be constant, so the velocity at 1 is zero.
1
1
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
2
2
→ 𝑣 = 2𝑔(𝑦 − 𝑦 ) =
𝑚
𝑣 = 2(10 )(350 𝑚 − 0 𝑚) = 𝟖𝟒 𝒎/𝒔
𝑠
b) Before applying Bernoulli’s, the speed at 2 needs to be found with continuity:
𝐴
𝜋(. 40 𝑐𝑚)
𝑚
𝑚
𝐴 𝑣 =𝐴 𝑣 →𝑣 =
𝑣 =
84
= 21
𝐴
𝜋(. 80 𝑐𝑚)
𝑠
𝑠
Either 1 or 3 can be used as the other point when applying Bernoulli’s. 1 is simpler:
1
1
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
2
2
1
→ 𝑃 = 𝜌𝑔𝑦 − 𝜌𝑣 − 𝜌𝑔𝑦
2
𝑘𝑔
𝑚
1
𝑘𝑔
𝑚
𝑘𝑔
𝑚
(350 𝑚) − 1000
(50 𝑚)
→ 𝑃 = 1000
10
21
− 1000
10
𝑚
𝑠
2
𝑚
𝑠
𝑚
𝑠
→ 𝑃 = 𝟐. 𝟖𝒙𝟏𝟎𝟔 𝑷𝒂
12

Venturi Tube – A venturi tube measures pressure different between areas in a
pipe with different diameters in order to measure the flow rate of a fluid. A tube
called a manometer connects sections of the venturi with different known
diameters. The fluid flow speed is found by measuring the height of a fluid in the
manometer, such as mercury.
Example D: Venturi Tube: The section of pipe shown has a crosssectional area of 𝐴 =40 cm2 at its wide section and 𝐴 = 10 cm2 at the
narrow portion. Water flows out of the pipe at a rate of 3𝑥10 𝑚 /𝑠.
a) Find the maximum velocity of the water.
b) Find the pressure difference between the wide and narrow
sections.
c) Find the different in height between the sections of the
manometer. 𝜌
= 13.6𝑥10 𝑘𝑔/𝑚
a) The volume flow rate at 1, where 𝑉 = 𝐴𝑣
𝐴 𝑣 =𝐴 𝑣
→𝑣 =
𝐴 𝑣
𝐴
=
3𝑥10
10𝑥10
𝑚
= 3 𝑚/𝑠
b) In the pipe itself, there is no height difference.
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
→ 𝑃 − 𝑃 = 𝜌𝑣 − 𝜌𝑣
→𝑃 −𝑃 =
1000
3
−
= 𝟒𝟐𝟎𝟎 𝑷𝒂
c) The two columns of mercury have the same pressure difference as the wide and narrow sections of the
pipe, since the two ends of the columns are open to these sections of pipe.
𝑃 −𝑃 =𝑃 −𝑃
When applying Bernoulli’s, the velocities are canceled out since the mercury levels are constant.
𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦 = 𝑃 + 𝜌𝑣 + 𝜌𝑔𝑦
→ 𝑃 − 𝑃 = 𝜌𝑔𝑦 − 𝜌𝑔𝑦
→𝑦 −𝑦 =
=
=. 𝟎𝟑 𝒎
.

/
)(
/
)
Airplane Wings - The shape of an aircraft wing forces air flowing above the wing to travel faster than air
flowing below the wing. The higher pressure below the wing results in a net upward force on the wing.
13
Example F: Airplane Lift – An airplane of wings with an area of 5.0 m2. Air flows over the top of the wing at
250 m/s and under the wing at 210 m/s.
a) Find the mass of the airplane so that the lift on the wing supports the airplane.
b) Suppose an airplane flies where the air is thinner. How does thinner air affect the required size of the
wings?
a) The airplane keeps its altitude since the lift force on the wing cancels out the force of the airplane’s
weight:
𝐹=𝐹
− 𝑚𝑔 = 0
𝐹
→ 𝑚𝑔 = 𝐹
→𝑚=
𝑔
The lift force for one wing is 𝐹 = 𝐴∆𝑃. For the two airplane wings, 𝐹
= 2𝐴∆𝑃.
Bernoulli’s formula is applied to find the pressure difference between the top and bottom surfaces of the
wings. The heights of the wings surfaces are assumed to be assumed since the difference is negligible
compared to the difference in speeds:
1
1
𝑃
+ 𝜌𝑣
+ 𝜌𝑔𝑦
=𝑃
+ 𝜌𝑣
+ 𝜌𝑔𝑦
2
2
1
1
𝑃
−𝑃
= ∆𝑃 = 𝜌𝑣
− 𝜌𝑣
2
2
1
𝑘𝑔
𝑚
𝑚
∆𝑃 = 1.2
250
− 210
= 11040 𝑃𝑎
2
𝑚
𝑠
𝑠
𝐹
2𝐴∆𝑃 2(5 𝑚)(11040)
𝑚=
=
=
= 11040 𝒌𝒈
𝑔
𝑔
10
b) If the air is thinner, the pressure difference will be smaller, requiring more surface area.
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AP Physics 2
Unit 2: Thermal Physics
Section 2.1 – Temperature …………………………………….………………….……………16
Section 2.2 – Heat Transfer…………………………………………………………………..…19
Section 2.3 – Gas Laws…………………………………..………………………………...…...23
Section 2.4 – The First Law of Thermodynamics ……….………………………………….….26
Section 2.5 – The Second Law of Thermodynamics……….……………………….……….…30
15
AP Physics 2
2.1 Temperature
Unit 2: Thermal Physics
Focus Question: What determines the temperature of a substance?
Temperature Scales
 Celsius – Temperature scale based on the freezing and boiling point of water.
 Kelvin – Temperature scale with the same increments are Celsius, but with sets 0 degrees at absolute
zero (the universal minimum temperature), which is about 273 ℃.
𝑇 = 𝑇℃ + 273
 Fahrenheit – Used in the best country in the world to measure temperature in everyday life.
9
𝑇℉ = 𝑇℃ + 32°
5
*When a formula in thermal physics has temperature in it, the Kelvin temperature is to be used. Celsius can be
used if the formula involves a temperature difference, ∆𝑇.
Temperature and Kinetic Energy
 Temperature – A measure of the average kinetic energy of individual molecules in a substance.
*Kinetic Theory – Substances are made of many tiny particles that move erratically.
o In solids, particles are held together, but vibrate.
o In liquids, particles can move across each other.
o In gases, particles move all over across all available volume.

Temperature can be related to the average kinetic energy, of the molecules in a substance:
By considering the collisions of gas particles, the equation for average kinetic energy was derived:
𝑲𝒂𝒗𝒈 =
𝟑
𝒌 𝑻
𝟐 𝑩
KB : Boltzmann’s constant = 1.38𝑥10
𝑇: Temperature in Kelvin (all formulas in this unit use temperature in Kelvin)
The molecules in a gas move around at different speeds, 𝐾 , is the based on the average speed of all the
particles.
*At 0 K, molecules stop moving completely (since their kinetic energy is zero)

Root mean-square speed – Root mean square speed can be though of as somewhat of an average speed
of all molecules moving in a substance. By using the definition of kinetic energy and comparing to the
formula for average kinetic energy vs. temperature:
1
3
𝐾 = 𝑚𝑣
= 𝑘 𝑇
2
2
→ 𝒗𝒓𝒎𝒔 =
𝟑𝒌𝑩 𝑻
𝒎
𝑚:mass of one molecule of a substance.
16
Example A: The mass of an oxygen molecule is 5.3x10-26 kg. Oxygen molecules in the room have a root meansquare speed of 500 m/s. What is the temperature in the room?
𝑣
3𝑘 𝑇
𝑣 𝑚
→𝑇=
𝑚
3𝑘
=
𝑇=
500
(5.3𝑥10
𝑘𝑔)
3(1.38𝑥10
)
= 320 𝐾
*This is equivalent to 47 ℃, which is about 119 ℉. Not an ideal room to be in.
Example B: An ideal gas is held in a container at temperature T. The temperature is suddenly increased to 3T.
a) What happens to the average kinetic energy of molecules in the container?
b) What happens to root mean-square speed of molecules in the container?
a)
=
→
= 3 → 𝐾 = 3𝐾
Average kinetic energy also triples. Average kinetic and temperature are directly proportional.
b)
=
→𝑣
= √3𝑣
The root mean square speed increases by a factor of √3.
Thermal Expansion
Most substances expand when heated. In a solid, particles vibrate faster, leader to particles colliding and
increasing the distance between particles.

Linear Expansion
∆𝐿 = 𝐿 𝛼∆𝑇
𝐿 = 𝐿 + 𝐿 𝛼∆𝑇 → 𝐿 = 𝐿 (1 + 𝛼∆𝑇)
𝛼=
∆ /
∆
, coefficient of linear expansion. This is a
constant that varies by material and is equal to the fractional change in object’s length per degree. The
units of 𝛼 are 1/℃.
Example C: A steel railroad track has a length of 10 m in the winter when the temperature is 0 ℃. What is the
length of the track when the temperature is 40 ℃? (𝛼 for steel = 11𝑥10 ℃ )
∆𝐿 = 𝐿 𝛼∆𝑇 = (10 𝑚) 11𝑥10
1
(40 ℃ − 0℃) = .0044
℃
𝐿 = 𝐿 + ∆L = 10 m + .044 = 10.0044 m
Railroad tracks have gaps, called “expansion joints” to all linear thermal expansion without deforming the track.
17

Constructing a Temperature Scale – Linear thermal expansion is used to measure temperature. To
construct a Celsius temperature scale based on thermal expansion:
o Take a substance with known thermal expansion, such as mercury, and arrange it in a line in a
container that does not significantly expand thermally compared to the substance within.
o Measure the length of the substance at some lower temperature (the Celsius scale uses the
freezing point of water, Fahrenheit used a glass of ice water), make this 0 °.
o Measure the length of the substance at some higher temperature (Celsius uses water’s melting
point, Fahrenheit used human body temperature), make this 100 ° (or some high temperature).
o Write a linear equation using the two points and the slope of the segment between them,
temperature can be found by writing a linear equation of the form in the form y = mx:
𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 = 𝑠𝑙𝑜𝑝𝑒 𝑥 𝑙𝑒𝑛𝑔𝑡ℎ.

Area Expansion
Consider a square where both sides increase by linear thermal expansion:
𝐴 = 𝐿 = (𝐿 + 𝛼𝐿 ∆𝑇) = 𝐿 + 2𝛼𝐿 ∆𝑇 + 𝛼 𝐿 (∆𝑇)
→ 𝐴 = 𝐿 + 2𝛼𝐴 ∆𝑇 → 𝐴 = 𝐴 + 2𝛼𝐴 ∆𝑇
→ 𝑨 = 𝑨𝟎 + 𝜸𝑨𝟎 ∆𝑻
𝛾: coefficient of area expansion, 𝛾 = 2𝛼
*Thermal expansion of a 2D or 3D object occurs outward. If an object with a hole
thermally expands, the hole will not shrink.

Volume Expansion
∆𝑉 = 𝛽𝑉 ∆𝑇
*𝛽 is the coefficient of volume expansion where 𝛽 = 3𝛼
Example D: A 1 liter glass container of water is completed filled at room temperature (20 ℃). Calculate how
much water spills out of the glass when, if any, when the temperature is raised to 80 ℃.
(𝛽
= 2.1𝑥10 ℃ , 𝛽
= 2.7𝑥10 ℃ )
Expansion of the water: ∆𝑉 = 𝛽𝑉 ∆𝑇 = (2.1𝑥10−4 ℃−1 )(1 𝐿)(80℃ − 20℃) = 0.0126 𝐿
Expansion of the glass: ∆𝑉 = 𝛽𝑉 ∆𝑇 = (2.7𝑥10−5 ℃−1 )(1 𝐿)(80℃ − 20℃) = .0016 𝐿
Since the water expanded much more than the glass, water will spill out. The amount that spills out is
(.0126-.0016)=.011 L
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AP Physics 2
2.2 Temperature
Unit 2: Thermal Physics
Focus Question: What happens when two objects at different temperatures come into contact?
At high temperatures, particles vibrated faster than in substances at lower temperature.
When a fast particle hits a slower particle, energy is transferred, with the slow moving
object gaining energy and the fast on losing energy. When a hot substance comes into
contact with a cool substance, the average kinetic energy of their molecules become more
equal over time due to these collisions.

Heat – Heat is the transfer of thermal energy between two objects in thermal contact to a temperature
difference. Heat flows from high temperature to low temperature.
Variable for amount of heat: Q
*units of heat: Joules
(4.186 J = 1 cal)
Zeroth Law of Thermodynamics

Thermal Contact – Two objects are in thermal contact if thermal energy can be exchange between them.

Thermal Equilibrium – Thermal equilibrium occurs when two objects in thermal have no exchange in
energy between them.

Zeroth Law – Two objects in thermal equilibrium have the same temperature.
“If A and B are both in thermal equilibrium with C, then A and B are in equilibrium with each other.”
Methods of Heat Transfer

Conduction – Heat transferred when two objects in contact have a temperature difference. Heat flows
from hot to cold.
Conduction occurs is there is a temperature difference between two parts of a conducting medium. The rate of
heat transfer (power) through a conductor can be given by:
Q/t: rate of heat transfer
𝑸 𝒌𝑨(𝑻𝑯 − 𝑻𝑪 )
=
𝒕
𝑳
k – thermal conductivity, a constant that depends on the type of material
The rate of thermal energy transfer is quicker for shorter lengths and larger temperature differences.
*Materials that are good thermal conductors have a higher value of 𝑘. Good thermal conductors allow heat to
easily flow through them.
Example A: A brick has a thermal conductivity of k = 0.710 J / m·s·C° and the
dimensions shown. Two heat sources are placed in contact with the ends of the
brick, as shown. Calculate the amount of heat transfer though the brick one hour.
Rate of heat transfer: 𝑄 =
(
)
=
Heat transferred in one hour: 𝑄 = 22.2
.
(
℃
℃)
.
(1 ℎ𝑜𝑢𝑟) 3600
= 22.2 𝐽/𝑠
= 𝟕𝟗𝟗𝟐𝟎 𝑱
19

Convention – Heat transfer by movement by molecules in a fluid.
*Convention occurs because a substance’s density decreases when temperature increases.
As air gets hotter, it gets less dense and rises, colder air takes it place.

Radiation – Heat transfer by electromagnetic waves.
*Radiation is the only form of heat transfer than can occur in a vacuum since electromagnetic waves can
travel in a vacuum.
Phase Changes
As heat is added to a solid:
1) As heat is added, particles in the solid vibrate faster. Heat added increases the internal energy which
causes temperature to increase.
2) At the melting point, temperature will remain constant despite added heat. Instead, added heat works to
overcome the attractive force between particles. A mixture of solid and liquid exists.
3) When heat is added to liquid, the increase in internal energy increases temperature.
4) At the boiling point, temperature is constant with liquid and gas present. As with at the melting point,
added heat works increases electrical potential energy.
5) Adding heat to vapor increases the temperature.
During a phase change, heat added goes to change the state. Temperature does not change during a phase
change.
20
Specific Heat Capacity
 Specific Heat: The amount of heat needed to produce a given temperature change in a solid, liquid or
gas, assuming there is no change of state –
Q: amount of heat needed to increase the temperature of a mass m of a substance by ∆𝑇 when there is no
change of state.
𝑸 = 𝒎𝒄∆𝑻
c – specific heat capacity (units:
), depends on the substance being heat and its phase
Example B: The specific heat of iron is 450 J/kgC. A 5.0 kg block of iron is dropped off the roof a 200 m tall
building. Suppose all the mechanical energy of the block is converted into heat. If the block is 25 ℃ when it’s
dropped, what is its temperature after it strikes the ground?
(200 𝑚) = 10000 𝐽
Amount of energy: 𝑈 = 𝑚𝑔ℎ = (5 𝑘𝑔) 10
The 10,000 J are converted to kinetic energy, which all converts to heat when the block hits the ground:
𝑄 = 𝑚𝑐 𝑇 − 𝑇 → 𝑇 =
+𝑇
10000 𝐽
→𝑇 =(
+ 25℃ = 𝟐𝟗. 𝟒 ℃
(5 𝑘𝑔) 450
℃
Example C: 600 g of water at 10 ℃ is mixed with 200 g of water at 90 ℃. What is the final temperature of the
mixture?
0th Law: The hot water will transfer energy to the cool water until thermal equilibrium is reached (both will
have the same final temperature). *this is also an application of conservation of energy
𝑄 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 = 𝑄 𝑙𝑜𝑠𝑡 𝑏𝑦 ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟
𝑚
𝑐
𝑇 −𝑇
=𝑚 𝐶
(𝑇 − 𝑇 )
The specific heat capacity cancels out since both masses are water and have identical capacities. The increase of
temperature of the cold water equals in the decrease in the temperature of the hot water.
(. 6 𝑘𝑔) 𝑇 − 10℃ = (. 2 𝑘𝑔) 90℃ − 𝑇
𝑻𝒇 = 𝟑𝟎 ℃
Latent Heat
 Latent Heat - The amount of heat needed to melt or vaporize a substance:
o Amount of heat needed to melt mass m of a solid:
𝑸 = 𝒎𝑳𝒇
Lf: Latent heat of fusion; amount of heat needed to melt mass m of a solid.
o Amount of heat needed to vaporize mass m of a fluid:
𝑸 = 𝒎𝑳𝒗
Lv: Latent heat of vaporization; amount of heat needed to vaporize mass m of a liquid.
Both latent heat of fusion and latent heat of vaporization are properties of the substance being heated.
The units of latent heat are .
21
Example D: Heat Transfer and Phase Change - How much heat is needed to raise the temperature of ice at
-20℃, melt the ice and then the raise the temperature of the water to 30 ℃?
*the melting point ice is 0℃. The specific heat capacities are 2100
for ice and 4190
for water. The
latent heat of fusion of ice is 3.3x105 J/kg.
Total heat required = heat to warm ice + heat to melt ice + heat to warm water
𝑄 = 𝑚𝑐 ∆𝑇 + 𝑚𝐿 + 𝑚𝑐
∆𝑇
𝑄 = (2 𝑘𝑔) 2100
(0 − (−20℃)) + (2 𝑘𝑔) 3.3𝑥10
+ (2 𝑘𝑔) 4190
(30 ℃ − ℃)
→ 𝑸 = 𝟏𝒙𝟏𝟎𝟔 𝑱
*most the heat was used to melt the ice.
Example E: Incomplete Phase Change – A 5.0 kg block of 0 ℃ ice is added to a container filled with 10 kg of
15 ℃ water.
a) Calculate the final temperature.
b) Calculate the mass of the ice that was melted.
a) If the energy that can be lost due to the temperature difference in the water is sufficient, all the ice
will melt. If there is not enough energy to be lost, the final temperature will be 0 ℃, but the final mixture
will be water and ice since the water will reach 0 ℃, but no all the ice to melt.
Total energy that can be lost by the initial water:
(15℃ − 0℃) = 628500 𝐽
𝑄
=𝑚
𝑐
∆𝑇 = (10 𝑘𝑔) 4190
Total energy required to melt all the ice:
𝑄 = 𝑚𝐿 = (5 𝑘𝑔) 3.3𝑥10
= 1650000 𝐽
There is not enough energy available by the water changing temperature to melt all the ice, so the final
temperature of the ice/water mixture will be 0 ℃.
b) Set the total energy lost by the water equal to latent heat equation for ice to find the amount of mass
of ice that can be melted:
𝑄
𝑄 = 𝑚𝐿 → 𝑚 =
𝐿
628500 𝐽
→𝑚=
= 𝟏. 𝟗 𝒌𝒈
3.3𝑥10
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AP Physics 2
2.3 Gas Laws
Unit 2: Thermal Physics
Focus Question: What is kinetic theory?
Ideal Gas Model Assumptions

Composition - A large amount of particles travel in random directions at various speeds.

Distance of Particles – Particles are far apart compared to their size.

Attractive Force – Particles interact only in collisions (no attractive forces)

Collisions – Collisions of particles with container wall are perfectly elastic.
Gas Laws
The Gas Laws relate the temperature, pressure, and volume of a sample of a confined ideal gas. The
gas laws are explained by kinetic theory, which describes gas as many particles moving around
and colliding with each other and the container they’re in.




Pressure - Kinetic Theory assumes matter consists of small particles in motion. This can be used to show
how pressure exists in a container filled with gas. The particles collide with the walls of the container
(perfectly elastically) and bounce off, changing their direction and thus momentum. A force acting over
a time is required to change momentum, and this force causes pressure:
Thermodynamics definition of pressure:
When a particle collides with the wall of a container and rebounds, it undergoes a
change in momentum. The rate at which the momentum is change is the force:
𝐹=
.
The wall of the container the particle is hitting has a certain area. Pressure is force
over area.
In thermodynamics, pressure is the average rate of change of momentum of particles hitting the
container per unit area.
Higher pressure means more and faster collisions occur between particles and the walls of the container.
Example A: An ideal gas with molecules of mass m is contained in a cube with sides of area A. The pressure
exerted by the gas on the top of the cube is P, and N molecules hit the top of the cube in time t. What is the
average vertical speed of the gas molecules?
Kinetic theory assumes collisions are elastic, so the speed before and after the molecules hit
the top of the container is constant.
∆
𝑚 𝑣 − (−𝑣)
𝐹
2𝑁𝑚𝑣
𝑃= =
=
→𝑃=
𝐴
𝐴
𝐴𝑡
𝐴𝑡
→𝒗=
𝑷𝑨𝒕
𝟐𝑵𝒎
23

Boyle’s Law – The volume of a gas is inversely proportional to the pressure applied when temperature is
constant.
𝑷𝟏 𝑽𝟏 = 𝑷𝟐 𝑽𝟐
*Kinetic Theory - If the volume of a gas is decreases, molecules collide with the container more
frequently, causing more pressure.

Charles’ Law – The volume of a gas is directly proportional to the absolute temperature when the
pressure is constant.
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐

The 3rd Law – The pressure applied is directly proportional to the absolute temperature when volume is
constant.
𝑷𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐
*Kinetic Theory - If temperature increases, molecules move faster so they collide with the container
more often.

The mole – The mass of a mole of a substance is grams is equal to its molecular mass:
A mole of a substance is the mass of one molecule of the substance.
Number of moles: 𝒏 =
𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
The number of individual particles in one mole of any substance is Avogadro’s number:
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟: 𝑁 = 6.02𝑥10

Ideal Gas Equation – By combining Boyle’s and Charles’ Law:
𝑷𝑽 = 𝒏𝑹𝑻
R = universal gas constant = 8.31
∙
Example B: The pressure of a gas is P. If the volume of its container is halved and the absolute temperature is
tripled, then what is the final pressure in terms of P?
*According to the ideal gas law,
is constant for a closed system
𝑉𝑇
𝑃𝑉 𝑃 𝑉
1
=
→𝑃 =𝑃
→𝑉 = 𝑉
→ 𝑇 = 3𝑇
𝑇
𝑇
𝑉𝑇
2
(𝑉 )(3𝑇 )
𝑃 =𝑃
→ 𝑷𝒇 = 𝟔𝑷
( 𝑉 )(𝑇 )
24
Example C: A 2.1 m3 container holds SO2 gas (molecular mass = 64 g/mol) at 3.5x105 Pa and 320 K.
a) How many kilograms of SO2 are in the container?
b) If the top of the container has an area of .50 m 2, at what rate is the moment of gas molecules changing
momentum in the container?
c) How many moles of SO2 should be added to the container to double the pressure if the volume and
temperature are constant?
a) 𝑃𝑉 = 𝑛𝑅𝑇 → 𝑛 =
=
( .
)( .
)
( .
)(
)
= 276 𝑚𝑜𝑙𝑒𝑠
𝑚𝑎𝑠𝑠 = 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒 𝑥 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
m = 64
(276 𝑚𝑜𝑙) = 17664 𝑔 = 𝟏𝟕. 𝟔 𝒌𝒈
b) The rate of change of momentum is force.
𝑃=
c) 𝑛 =
→ 𝐹 = 𝑃(𝐴) = (3.5𝑥10 𝑃𝑎)(. 5 𝑚 ) = 1.75𝑥10 𝑁 = 1.75𝑥10 𝑘𝑔 𝑚/𝑠
If V and T are held constant, n and P are directly proportional.
Example D: The density of helium gas at 0 ℃ is .179 kg/m3. The temperature is raised to 100 ℃ while the
pressure is kept constant. Calculate the new density when the temperature is raised.
Pressure is constant so is constant.
𝑉
𝑉
𝑇
=
→𝑉 =𝑉
𝑇
𝑇
𝑇
→𝑉 =𝑉
(273 + 100 𝐾)
= 1.37
(273 + 0 𝐿)
𝜌 = 𝑀/𝑉, so density and volume are inversely proportion. If volume increases by a factor of 1.37, density
decreases by a factor of 1.37.
𝜌 =
. 179
𝜌
𝒌𝒈
=
=. 𝟏𝟑 𝟑
1.37
1.37
𝒎
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AP Physics 2
2.4 The First Law of Thermodynamics
Unit 2: Thermal Physics
Focus Question: How does conservation of energy apply to thermodynamics processes?

Closed System – A system is a set of objects being studies. Everything else is considered the
environment. A closes cylinder is system that does not exchange matter with the environment.

Total Internal Energy – Temperature is a measure of average kinetic energy of the molecules in a
system. The total internal energy is the combined energy of all the molecules in the system:
The average kinetic energy of a molecule in a system is given by 𝑛𝑅𝑇.
The total internal energy is the number of molecules (N) multiplied by the kinetic energy of each
molecule:
3
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑈 = 𝑁( 𝑘 𝑇)
2
Boltzmann’s constant is the universal gas constant over Avogadros number, 𝐾 =
, the number of
molecules of a gas is the number of moles times Avodagro’s number, so the equation for internal
equation becomes:
𝑈 = (𝑛𝑁 )
In an ideal gas, T =

, so U = nR
3𝑅
𝟑
𝑇 → 𝑼 = 𝒏𝑹𝑻
2𝑁
𝟐
→ U = PV
Work Done in Thermodynamics Processes:
If the initial pressure of the gas is P in the piston shown and the crosssectional area of the piston is A, then the force with which one must push is
PA. If the piston is compressed some distance ∆𝑥, the work done is:
𝑊 = 𝐹∆𝑥 → 𝑊 = 𝑃𝐴(−∆𝑥)
*𝐴∆𝑥 is volume, so
𝑾 = −𝑷∆𝑽
Work is the area under a Pressure vs. volume graph
*Work is only done when volume changes
Example A: A gas is compressed at a constant pressure of 2.00 × 105 Pa from a volume of 2.00 m3 to a volume
of 0.500 m3. The temperature is initially 40.0 °C.
a) Find the work done.
b) Calculate the final temperature of the gas.
a) 𝑊 = −𝑃∆𝑉 = −(2𝑥10 𝑃𝑎(. 5 𝑚 − 2.0 𝑚 ) = 30000 𝐽
b) At constant pressure:
=
->𝑇 = 𝑇
= (40 + 273 𝐾)
.
= 1252 𝐾 = 979 ℃
Recall from Physics 1 that doing positive work increases a system’s energy. Positive work is done on the
gas, increasing its internal energy.
26

The First Law of Thermodynamics – The 1st law of Thermodynamics is a conservation of energy stated
in terms of thermal physics:
The internal energy of a system can be changed by doing work on the system or adding heat to the
system:
∆𝑼 = 𝑸 + 𝑾
∆𝑈 - change in internal energy
Q - heat added to system
W – work done on the system
*When heat enters a system, some of the energy is used to increase the temperature and some is used to
do work on the environment.
o
o
o
o
o
o
∆𝑈 > 0: The temperature increases.
∆𝑈 < 0: The temperature decreases.
𝑄 > 0: Heat enters the system.
𝑄 < 0: Heat leaves the system.
𝑊 > 0: Work is done on the system; system is compressed.
𝑊 < 0: Work is done by the system, system expands.
Example B: If the gas in a container absorbs 275 joules of heat, has 125 joules of work done on it, and then
does 50 joules of work, what is the increase in the internal energy of the gas?
o 275 J of heat is added to the system so Q=+275 K
o 125 of work is on the system, which is +125 J of work. The system does 50 J of work, which is -50 J, so
the net work is +75 J.
∆𝑈 = 𝑄 + 𝑊 = 275 𝐽 + 75 𝐽 = 𝟑𝟓𝟎 𝑱
Gas Laws


Isobaric Process – Pressure is constant.
∆𝑷 = 𝟎
𝑾 = 𝑷∆𝑽
Isothermal Process – Temperature is constant.

If temperature doesn’t change, there is no change in internal energy
∆𝑼 = 𝟎
Isochoric Process – Volume is constant.
If there is no volume change, force isn’t applied over a distance, so there is
no work done. 𝑊 = 0

Adiabatic Process – No heat enters of leaves the system.
𝑸=𝟎
𝑾 = ∆𝑼
*On a PV graph, adiabatic processes are steeper than isothermal processes.
27
Example C: An ideal gas expands adiabatically.
a) Explain why the temperature decreases.
b) Use your answer to a to explain why the adiabatic curve of an ideal gas, expanding from a given state,
is steeper than the corresponding isothermal curve from the same state.
a) ∆𝑈 = 𝑄 + 𝑊, since Q=0, the change in energy is equal to work. Since the gas is expanding, work is
being done by the gas, so the work is negative, so ∆𝑈 must be negative.
b) Since an adiabatic process reduces temperature and the isothermal process does not, the final pressure
after the adiabatic expansion will be lower than the final pressure after the isothermal expansion.
Example D: For the PV diagram on the right for one mole of an
ideal gas:
a) Determine the work done in each part of the process.
b) Determine the change in internal energy for each part
of the process.
c) Find the heat added or removed from the system for the
entire process.
a) 𝑾 = −𝑷∆𝑽
o
o
o
o
A − B: ∆𝑉 = 0, so 𝑊 = 0
𝐵 − 𝐶: 𝑊 = −𝑃∆𝑉 = −(1𝑥10 𝑃𝑎)(. 03 𝑚 − .01 𝑚 ) = −2000 𝐽
C − D: ∆𝑉 = 0, so 𝑊 = 0
𝐷 − 𝐴: 𝑊 = −𝑃∆𝑉 = −(2𝑥10 𝑃𝑎)(. 01 𝑚 − .03 𝑚 ) = 4000 𝐽
b) ∆𝑈 = 𝑛𝑅∆𝑇. Pressure and volume are known at all points, so temperature can be found at all 4
points using the ideal gas law.
o A-B: ∆𝑈 = 𝑛𝑅(𝑇 − 𝑇 ) = (1)(8.314)(120 − 241) = −1509 𝐽
o B-C: ∆𝑈 = (8.314)(361 − 120) = 3006 J
o C-D: ∆𝑈 = (8.314)(722 − 361) = 4502 J
o D-A: ∆𝑈 = (8.314)(241 − 722) = −5999 𝐽
c) The change in internal energy is zero, but the net work is negative, so 2000 J of heat had to be
removed from the system.
∆𝑈 = 𝑄 + 𝑊 → 𝑄 = −𝑊 = −(−2000 + 4000) = −2000 𝐽
28
Example E: Find the work done, change in internal energy, and heat added or
removed during each cycle in the given PV diagram, which describes the changes
in pressure and volume of one mole of an ideal gas.
𝑇 =
=
𝑇 =
=
𝑇 =
=
(.
( )( .
)
)
(.
( )( .
)
=722 K
)
(.
( )( .
=241 K
)
)
=2165 K
o A to B
Work: Work is negative since the gas is expanding:
𝑊 = −𝑃∆𝑉 = −(2𝑥10 )(. 03 − .01 ) = −4000 𝐽
Change in internal energy: ∆𝑈 = 𝑛𝑅(𝑇 − 𝑇 ) = (1)(8.314)(722 − 241) = 6000 𝐽
Heat added: ∆𝑈 = 𝑊 + 𝑄 → 𝑄 = 10,000 𝐽
10,000 J of heat is added. 6000 J goes to increase the internal energy, and 4000 J is done by the gas to do
work.
o B to C
Work: Work is zero since volume is constant.
Change in internal energy: ∆𝑈 = 𝑛𝑅(𝑇 − 𝑇 ) = (1)(8.314)(2165 − 722) = 18000 𝐽
Heat added: ∆𝑈 = 𝑊 + 𝑄 → 𝑄 = 18,000 𝐽
18,000 J of heat was added, all to increase temperature of the system as the volume was held constant.
o C to A
Work: Work is the area under the graph: 𝐴𝑟𝑒𝑎 = (. 02)(2𝑥10 ) + (4𝑥10 )(. 02) = 12000 𝐽
Change in internal energy: ∆𝑈 = 𝑛𝑅(𝑇 − 𝑇 ) = (1)(8.314)(241 − 2165) = −24000 𝐽
Heat added: ∆𝑈 = 𝑊 + 𝑄 → 𝑄 = −36,000 𝐽
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AP Physics 2
2.5 The Second Law of Thermodynamics
Unit 2: Thermal Physics
Focus Question: What is entropy?

Entropy – Entropy is the measure of the disorder in a system. The change in entropy, ∆𝑆, when heat is
added to a system by a reversible process at constant temperature is given by:
𝑸
∆𝑺 =
𝑻

The 2nd Law of Thermodynamics – The entropy of the universe
increases in all natural processes.
*Work can be done to decrease entropy.
*A disorderly arrangement is more probable than an ordered one.

Consequences:
-It is impossible for thermal energy to flow from a cold to a hot object without performing work.
-It is impossible, in a cyclic process, to completely convert heat into mechanical work.
Example A: A 1000 kg boulder at rock is dropped off a 100 m cliff into a lake. Calculate the lakes change in
entropy if both the rock and lake at 10 ℃.
The energy lost when the boulder falls off the cliff is: ∆𝑈 = 𝑚𝑔ℎ
𝑚
(100 𝑚) = 1𝑥10 𝐽
→ ∆𝑈 = (1000 𝑘𝑔) 10
𝑠
The energy is lost when the boulder hits the lake, and is transferred to internal energy in the lake in the form of
entropy:
𝑄
1𝑥10 𝐽
∆𝑆 = =
= 3533 𝐽
𝑇 (10 + 273) 𝐾
Energy in the form of entropy cannot be used to perform work. The 2 nd Law of Thermodynamics predicts the
eventual heat death of the universe, where all energy in the universe is converted to entropy. This is so sad.
Alexa, play Descpacito.
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AP Physics 2
Unit 3: Electrostatics
Section 3.1 – Electric Charge………………………………………………………….………..32
Section 3.2 – Electric Force………………………………………………………….…………33
Section 3.3 – Electric Fields…………………………………………………………….…...….36
Section 3.4 – Electric Potential……….…………………………………………………..….…40
Section 3.5 – Capacitors……………………..…………………………….……………...…….45
31
AP Physics 2
3.1 Electric Charge
Unit 3: Electrostatics
Focus Question: What causes an object to have a net electric charge?

Fundamental Forces of Nature:
1) Gravitational force: Gravity is a force of attraction between masses. Gravity has infinite range and is
caused by masses bending spacetime.
2) Electromagnetic force – The force given by Coulobm’s law, has infinite range and is occurs through
the movement of photons.
3) Weak nuclear force – Brings about beta decay. The weak nuclear interaction has a small range and is
caused by W and Z bosons.
4) Strong nuclear force – Hold neutrons and protons together in the atomic nucleus; has a very short
range and is caused by the exchange of gluons.

Electric Charge is a property of matter. Charge can be positive or negative. Negative charge is a
property of electrons. Positive charge is a property of protons.
Charge is conserved. Charge can be transferred between objects, but cannot be created or destroyed.
Charge is quantized. Something that is quantized can only be an integral multiple of some number. For a
charged objected, the total charge is:
Total negative charge = -(number of extra electrons x fundamental charge)


Total positive charge = -(number of missing electrons x fundamental charge)
The fundamental charge is the charge of one electron: 𝑒 = 1.6𝑥10
𝐶
*Charge is only due to flow of electrons since electrons can leave an atom since they do not interact
with the strong nuclear force. A negative charge is due to extra electrons. A positive charge is due to an
electron deficit.

Charge is able to flow when electrons move, which readily occurs in a conductor.
o Conductor – Materials with free electrons that can move easily.
As a result, charge is easily transferred due to the flow of
electrons.
*objects with high conductivity has low resistivity.
*charge gathers on the surface of a conductor.
o Insulator - Material in which electrons are tightly bound to nucleus and thus charge is not
transferred.
*An insulator can have charge polarization due to individuals atoms being oriented one way. An
insulator can be attracted to a charge object due to this polarization.
Example A: Two separated, identical conducting spheres are charged with charges of 4.0 µC and −12 µC,
respectively. The spheres are allowed to touch. Describe the resulting movement of electrons.
The total charge is (4.0 µC+(−12 µC))= -8 µC. The spheres split the charge, so they will each have a charge of
- 4 µC. This occurs by electrons flowing from the – 12 µC sphere to the 4 µC sphere.
32
AP Physics 2
3.2 Electric Force
Unit 3: Electrostatics
Focus Question: What is Coulomb’s Law?




If charge is evenly distributed on an object, it can thought of as being concentrated at its center, much
like the force of gravity. *This is case in all of Physics 2, continuous charges are a Physics C topic
The net charge on an object is determined by the difference between the number of electrons and the
number of protons the object contains.
Properties of the electrostatic force:
o Electrostatic force is a vector quantity directed along a line joining the particles.
o Opposite charges attract and like charges repel.
o It is directly proportional to the product of the charges and follows the inverse square law for
distance.
Coulomb’s Law
𝐹=
𝒌 = 𝟖. 𝟗𝟗𝒙𝟏𝟎𝟗

𝑘 |𝑄 ||𝑄 |
𝑟
𝑵𝒎𝟐
𝒌=
𝑪𝟐
𝟏
𝟒𝝅𝝐𝟎
𝜖 − permittivity of free space, measure of resistance to formation of electric field in a vacuum
𝜖 = 8.85𝑥10

If more than one charge is present, the net electrostatic force on a charge is the vector sum of the forces
due to the other charges. This is known as Superposition.
Example A: Two conducting spheres are arranged as shown. Sphere A, with charge 𝑄 = 70 𝜇𝐶 is fixed to an
insulating stand. Sphere B of charge 𝑄 = −80 𝜇𝐶 is suspended form an insulating string that is deflected with
𝜃 = 30°. The distance r = 0.3 m. Calculate the mass of sphere B.
There are 3 forces on sphere B, the electrostatic force of attraction towards
sphere A, gravity downwards, and tension along the rope leading away from
B. The problem can be solved by applying F=ma in both the horizontal and
vertical directions (both directions are in equilibrium.
𝐹 = 0 = 𝑇 sin 𝜃 − 𝐹
𝑚𝑔
cos 𝜃
Substituting the expression for T from y into x:
𝑚𝑔
→(
) sin 𝜃 − 𝐹 = 0 → 𝑚𝑔 cot 𝜃 = 𝐹
cos 𝜃
𝐹
tan 𝜃 𝑘|𝑄 ||𝑄 |
→𝑚=
=
𝑔 cot 𝜃
𝑔
𝑟
𝐹 = 0 = 𝑇 cos 𝜃 − 𝑚𝑔 → 𝑇 =
→𝑚=
tan 30° (9𝑥10
10
)(70𝑥10
(. 3 𝑚)
𝐶)(80𝑥10
𝐶)
= 𝟑𝟐 𝒌𝒈
33
Example B: 3 charges are arranged as shown.
a) Calculate the net force on Q2.
b) Calculate the net force on Q1.
a) Both forces are attractive, so the there is a force due to sphere 1 up and
a force due to sphere 3 to the right. First, find the force due to each
charge, and then add the two force vectors.
𝐹 =
9𝑥10
𝑘|𝑄 ||𝑄 |
=
𝑟
(5𝑥10
𝐶)(4𝑥10
𝐶)
𝐶)(6𝑥10
𝐶)
(. 1 𝑚)
= 18 𝑁
As a vector, 𝐹 =< 0,18 > 𝑁
𝐹 =
𝑘|𝑄 ||𝑄 | (9𝑥10
=
𝑟
)(4𝑥10
(. 2 𝑚)
= 5.4 𝑁
As a vector, 𝐹 =< 5.4,0 > 𝑁
Since one force is horizontal and the other is vertical, the net force can be found by Pythagorean theorem:
𝐹=
(5.4 𝑁) + (18 𝑁) = 18.8 𝑁
The direction of the net force is up and to the right. The angle can be found using tan
= 73° above the positive x-axis.
𝜃 = tan
b) The force due to charge 2 is attractive and downwards, so it’s all in the y-direction. The
force due to F3 is repulsive. The electrostatic force acts along the line connecting the charges,
so the charge is up and the left with 𝜃 = tan
.
.
= 27°. To find the net force due to charge
3, the magnitude of the force needs to be calculated using Coulomb’s law. Trigonometry will
then be used to find the x and y components of this force.
𝐹 =
𝑘|𝑄 ||𝑄 | (9𝑥10
=
𝑟
)(5𝑥10
(. 1 𝑚)
𝐶)(4𝑥10
𝐶)
= 18 𝑁
Since the force is fully vertical and down, 𝐹 =< 0, −18 > 𝑁 as a vector.
34
𝐹 =
9𝑥10
𝑘|𝑄 ||𝑄 |
=
𝑟
(5𝑥10
𝐶)(6𝑥10
𝐶)
= 5.4 𝑁
(. 1 𝑚) + (. 2 𝑚)
As a vector, the force is 𝐹 =< −5.4 cos 𝜃 + 5.4 sin 𝜃 >=< −5.4
.
(.
)
(.
)
+ 5.4
.
(.
)
(.
)
𝐹 =< −4.8, 2.4 > 𝑁
The total force as a vector is 𝐹 = 𝐹 + 𝐹 =< 0, −18 > +< −4.8, 2.4 >=< −4.8, −15.6 > 𝑁
The magnitude of F is 𝐹 =
The angle is 𝜃 = tan
.
.
(−4.8 𝑁) + (−15.6 𝑁) = 16.3 𝑁
= 72° below the negative x-axis.
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AP Physics 2
3.3 Electric Field
Unit 3: Electrostatics
Focus Question: What is a field force?




A field force is a force that acts at a distance. Gravity, the electrostatic force, and the magnetic force are
all field forces.
Any electric charge causes an electric field in the space around it. By convention, electric field is based
on the force a positive charge (known as a test charge) would experience would if placed in the field.
Suppose a positive test charge is placed near a charge Q. The force it
would experience is:
𝐹=
The electric field strength it would experience is the electric force per unit
charge:
𝑬=
→𝐸=
*units of electric field: N/C
𝑞
𝑭
𝒒
→𝑬=
𝑲𝑸
𝒓𝟐
*Electric field is a vector quantity
*A positive charge will go in the direction of the electric field. A negative charge goes against it.
Example A: How are electric fields and gravitational field similar? How are they different?
o In most cases, electric fields are much, much stronger than gravitational fields.
o The electric field can point towards or away from a charge. Gravitational fields only point
towards masses.
o The electric field is due to photons. The gravitation field is due the geometry of spacetime.
Example B: Tiny droplets of oil acquire a small negative charge while falling. An electric field of magnitude
5800 N/C points straight down. One particular droplet of mass is observed to remain suspended against gravity.
If the mass of the droplet 3x10-15 kg, find the number of excess electrons in the droplet.
The sum of the forces in the vertical direction needs to be zero for the droplet to be
suspended. The force on the droplet is up since negative charges experience a force
opposite the electric field.
𝐹 = 𝐹 − 𝑚𝑔 → 𝐸𝑞 − 𝑚𝑔 = 0 → 𝑞 =
→𝑞=
(3𝑥10
𝑚𝑔
𝐸
𝑘𝑔)(10 )
= 5.17𝑥10
𝐶
5800 𝑁/𝐶
This charge is due to excess electrons. The number of electrons can be found by dividing
the charge by the charge of one electron:
# 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 =
5.17𝑥10
1.6𝑥19
= 32 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
36
Example C: Find the electric field at P.
Like force, electric field is a vector
quantity. The net field is found by
founding the vector sum of the field to A
and the field due to B. Both fields are
away from the charge. The field due to A
is straight up away from A and is only a positive y-component. The
field due to B is along the straight line connecting B to P (along the
hypotenuse of 3-4-5 triangle so the trig isn’t bad). The x and y
components of this field needs to be found and added to A’s field.
Magnitude of the field due to A: 𝐹 =
=
(
)
= 600 𝑁/𝐶
As a vector, this field is 𝐸 =< 0,600 > 𝑁/𝐶
Magnitude of the field due to A: 𝐹 =
=
√
= 234 𝑁/𝐶
The field is negative in x and positive in y: As a vector, this field is 𝐸 =< −234 cos 𝜃 + 234 sin 𝜃 >
→=< −234
40
30
+ 234
>=< −187,140 > 𝑁/𝐶
50
50
The vector sum is < 0,600 >
+> +< −187,140 >
=< −187,740 >
The total magnitude of the electric field is found with Pythagorean Theorem:
𝐸=
(−187) + (740) = 763 𝑁/𝐶
The direction is tan
= tan
= 76° above the negative x-axis.
Electric Field Lines
 Electric field lines area vector field drawn to indicate the direction a positive test charge would
experience if placed near the charge.
37
o Field lines go towards negative charges and away from positive charges.
o Stronger fields have more field lines.
Example D: Draw field lines for the charge arrangements below:
a)
b)
a)
b)
Electric Field in Conductors


In a conductor with a net charge, the charges all repel each other.
Properties of electric field in a conductor:
o All the charge will gather on the surface of the conductor.
o The electric field everywhere inside the conductor will be zero.
o The electric field just outside the conductor is perpendicular to the conductor’s surface.
Example E: Draw field lines for the irregular conductor shown.
On an irregular shaped conductor, more charge gathers on the sharp turns. On a conductor, the field lines are
perpendicular to the surface.
Example F: A point of -Q is placed at the center of a hollow conducting spherical
shell. The spherical has a total charge of +4Q.
a) Determine the charge on the inner and outer surfaces of the conducting
sphere.
b) Draw electric fields for this scenario.
a) The charge on the inner surface of the conductor is +Q as it’s attracted to
the -Q charge. The excess charge will gather on the outer surface, which is
+3Q in this case.
Any conductor with two surfaces with a charge nearby will follow this
behavior due to the electrostatic force.
38
The electric field due to hollow sphere is zero inside the hollow sphere, so the electric field inside is only due to
the -Q point charge. Outside the hollow sphere, there is field away since the total charge is positive.
Electric Flux

Electric Flux
*Surfaces with more flux will have more field lines going through them.
*Electric flux is used to calculate electric fields in calculus based physics.
*Electric field can also be though as electric flux density.
Example G: Consider a uniform electric field oriented in the x - direction. Find the electric flux through each
surface of a cube with edges L oriented as shown and the net flux.
Only surfaces with field through have flux, so only surfaces 1 and 2 needs to
be considered:
o Flux due to 1: 𝐸 ∙ 𝐴 = 𝐸𝐴 cos 180 = −𝐸𝐴
o Flux due to 2: 𝐸 ∙ 𝐴 = 𝐸𝐴 cos 0 = 𝐸𝐴
The net flux is 0.
Net flux will always be zero through a close surface if the field source is from
outside of the surface since the field will enter one face of the surface and exit
another.
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39
AP Physics 2
3.4 Electric Potential
Unit 3: Electrostatics
Focus Question: What is electric potential

Potential is related to potential energy. Potential energy is the potential to do work and is due to an
object’s position. In mechanics, an object has high potential when it’s higher up in a gravitational field.
In electric physics, a positive test has higher potential when it is near other positive charges. In general,
an object has higher potential when it is at a location where forces would cause it to move away.

Relationship between work and potential energy:
𝑾 = −∆𝑼 → ∆𝑼 = −𝑭𝒓
*The signs are opposite since doing work to bring an object to its position gives it potential energy in
the opposite direction of the force.
Electric Potential Energy
 Electric potential energy between two likes charges:
𝑘𝑄𝑞
𝑈=
𝑟
The potential energy is positive since a positive test charge has
potential to do work in a direction away from the positive charge.

Electric potential energy between two opposing charges:
𝑘𝑄𝑞
𝑈=−
𝑟
The potential energy is negative since a positive test charge near a
negative charge is electrostatically “stuck” to the negative charge, so it
would take work to separate it.

General Formula for Electrical Potential Energy:
𝑘𝑄𝑞
(𝑟)
𝑟
𝑘𝑄𝑞
→𝑈=
𝑟
Example A: An electron is released from rest in a uniform electric field. What happens to the electric potential
energy of the system?
𝑈 = 𝐹𝑟 =
The electric potential energy will decrease since the electron gains speed. The electron converts electric
potential energy of the field into kinetic energy for it to move.
The electron will move opposite the field.
40
Electric Potential

Electric potential is defined as the electrical potential energy per unit charge:
𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑉 =
∆𝑈 = 𝑈 − 𝑈 = 𝑞(𝑉 − 𝑉 )
*units of Electric Potential: – J/C, Volts (V)

Potential difference between two points:
Potential difference between A and B: ∆𝑉 = 𝑉 − 𝑉 =
The potential is in the opposite direction of the work done to give the charge that potential. It takes work
towards plate A to moves a positive charge towards plate A, giving it potential away from plate A. This
is analogous to an object gaining gravitational potential downward when you pick it up.
*Potential difference is scalar quantity, but can be positive or negative.

Work done by a field on a charge: 𝑊 = 𝑉𝑞
*the work energy theorem can be used to find the speed of a charge when a potential difference is
applied: 𝑊 = ∆𝑉𝑞 = 𝑚𝑣 − 𝑚𝑣

Relationship between electric potential and electric field:
𝑊 = 𝑞𝑣 = 𝐹𝑑
𝑞𝑉 = 𝐸𝑞𝑑
→ 𝑽 = 𝑬𝒅
Example B: A charge particle of mass .006 kg and a charge of 1 mC is stationary between two plates that are
20 cm apart. What is the potential difference of the plates?
Since the charge floats, the force of gravity equals the electric force. The bottom
plate has to be positive and the top plate negative since the charge has an upward
force on it and is positive.
𝐹 = 𝐹 − 𝑚𝑔 → 𝐹 = 𝑚𝑔 → 𝐸𝑞 = 𝑚𝑔
𝑉
𝑚𝑔𝑑
𝑞 = 𝑚𝑔 → 𝑉 =
𝑑
𝑞
(. 006 𝑘𝑔) 10
(.20 𝑚)
→𝑉=
= 12 𝑉
(1𝑥10 𝐶)
The answer to this question does not depend on the distance to the plate. The electric field due to a plate does
not depend on the distance to the plate.
→
41
Example C: A positive plate is 5.0 m from a negative plate. A proton has an initial
velocity of 1.0𝑥10 𝑚/𝑠 and travels from the positive plate. It hits the negative plate
traveling at 4.0𝑥10 𝑚/𝑠.
a) Calculate the potential difference between the plates.
b) Calculate the magnitude of the electric field.
*the mass of a proton is 1.67x10-27 kg and a charge of 1.6x10-19 C
a) Applying the work-energy theorem:
𝑊 = ∆𝑉𝑞 → ∆𝐾
→ ∆𝑉 =
∆
→ ∆𝑉 =
=
.
.
= 78000 𝑉
.
b) 𝑉 = 𝐸𝑑 → 𝐸 =
=
= 16000 𝑁/𝐶
.
Electric Potential of a Point Charge
 The potential due a single charge is the equal to the amount of work required to bring a small test charge
from infinity (very far away) to a distance R from the charge:
𝑉=
=
=
Example D: Find the amount of work required to bring a +10 𝜇𝐶 charge infinity to
a) Point A
b) Point B
For both cases, the work is 𝑊 = 𝑉𝑞. The potential at A and B is the sum
of the potentials due to charge 1 and 2. Since potential is a scalar, only the
sum of the potential needs to be taken. Both potentials will simply be
scalar numbers, so there is no need to break anything up into components.
a) The potential at A = potential due to Q1 and Q2.
𝑉 =𝑉 +𝑉
→
9𝑥10
𝑘𝑄
𝑘𝑄
+
=
𝑟
𝑟
(50𝑥10
5𝑚
𝑊 = 𝑉 𝑞 = (25601 𝑉)(10𝑥10
𝐶)
+
9𝑥10
(−80𝑥10
(5 𝑚) + (10 𝑚)
𝐶)
= 25601 𝑉
𝐶) = .26 𝐽
b)
𝑉 =𝑉 +𝑉
→
9𝑥10
(50𝑥10 𝐶) 9𝑥10
(−80𝑥10 𝐶)
𝑘𝑄
𝑘𝑄
+
=
+
= −24149 𝑉
𝑟
𝑟
(5 𝑚) + (10 𝑚)
(5 𝑚) + (10 𝑚)
𝑊 = 𝑉 𝑞 = (−24149 𝑉)(10𝑥10
𝐶) = −.24 𝐽
42
Example E: 3 identical charges of charge 𝑄 = 10 𝜇𝐶 and mass of 2 𝑘𝑔 are
arranged along a line as shown with d = 0.05 m. The two left charges are
held in place and the right charge is released from rest. Calculate its
maximum speed.
First, calculate the potential at the point of the right most charge. The potential at the very right is the sum of the
potential of the left two charges:
𝑉→
𝑘𝑄 𝑘𝑄 9𝑥10
+
=
𝑑
2𝑑
(10𝑥10
0.05 𝑚
𝐶)
+
9𝑥10
(10𝑥10
0.1 𝑚
𝐶)
= 2700000 𝑉
Apply the work energy theorem for the work done on the rightmost charge. The charge starts at rest so its initial
kinetic energy is zero:
𝑊 = 𝑉𝑄 = ∆𝐾 → 𝑉𝑄 =
𝑣 =
1
1
𝑚𝑣 − 𝑚𝑣 → 𝑣 =
2
2
2(2700000 𝑉)(10𝑥10
2 𝑘𝑔
𝐶)
2𝑉𝑄
𝑚
= 5.2 𝑚/𝑠
Equipotential Lines
 Equipotential lines – Lines that indicate equal potential (every point on a given equipotential line has the
same electric potential as every other point on the line.
*No work is required to move from point to point along an equipotential line.
*Equipotential lines are drawn perpendicular to the electric field at any point.
Example F: Draw field lines and equipotential lines for two negative charges.
43
Example G: The diagram shown represents equipotential lines.
a) Draw the direction of the electric field at each point.
b) Calculate the work required to move a +10 𝜇𝐶 from to A to B.
c) Calculate the work required to move a +10 𝜇𝐶 from to A to E.
d) Calculate the work required to move a −10 𝜇𝐶 from to A to E.
a) The electric potential lines point towards lower potential and are perpendicular to the equipotential
lines:
b) A and B are on the same equipotential line, so no work is required. It does not take any work to move
a charge to a place with the same potential.
c) 𝑊 = ∆𝑉𝑞 = (10𝑥10 𝐶)(15 𝑉) = 1.5𝑥10 𝐽.
W = +1.5𝑥10 𝐽.
The work is positive since a force is required to move the object to a higher potential since the force is
applied in the direction of motion, the work is positive.
d) 𝑊 = ∆𝑉𝑞 = (10𝑥10 𝐶)(15 𝑉) = 1.5𝑥10 𝐽.
W = -1.5𝑥10 𝐽.
The work is negative since a negative charge would move from A to E naturally. To move it there with a
force would require a force against the direction of motion, making the work negative. This is because
moving the charge there with a force prevents it from acquiring kinetic energy it would naturally gain.
This process is similar to lowering an object by a force in a gravitational field.
Rate your understanding: Electric Potential
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44
AP Physics 2
3.5 Capacitors
Unit 3: Electrostatics
Focus Question: How is electrical energy stored in a capacitor?

Electric Field due to a charged plate
A charge plate has an electrical field coming straight out of it, perpendicular to
both surfaces (away from a positive plate, towards a negative plate. The electric
field is based on the charge density between the plates and the permittivity of
space:
𝜎
𝑄
𝐸=
=
2𝜀
2𝐴𝜀
Q – charge on either plate
A – surface area of plate
𝜎 = = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
*The electric field does not depend on the distance to the plate, assuming the to plate is large compared
to the distance to the plate.
Parallel Plates
 Parallel Plates of the same charge
Both plates are conductors, so charge gathers on the surface, in this case the charge
gathers on the outside surface as they repel. Both plates have fields of
𝝈
𝟐𝜺𝟎
.
Outside the plates, in either regions, the fields point in the same direction:
𝐸 =𝐸 +𝐸 =
𝝈
𝟐𝜺𝟎
+
𝝈
𝟐𝜺𝟎
=
𝝈
𝜺𝟎
In the region inside the plates, E1 and E2 are in opposite directions:
𝐸 =𝐸 +𝐸 =

𝝈
𝟐𝜺𝟎
−
𝝈
𝟐𝜺𝟎
=𝟎
Parallel Plates of opposite charge
In the region between the plates, both fields point from the positive plate to
the negative plates, so the contributions from each plate adds:
𝐸 =𝐸 +𝐸 =
𝝈
𝝈
𝝈
𝑸
+
=
=
𝟐𝜺𝟎 𝟐𝜺𝟎 𝜺𝟎 𝑨𝜺𝟎
Capacitor – Formed by two conductors seperated by an insulator. When a capacitor is
connected to power, it the two plates become opposite charged. An electric field is stored
between the plates.
45
Capacitance
 Capacitance is the measure of a capacitor’s ability to store charge when a potential difference is applied
to its plates:
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 =
→𝑪=

𝑸
𝑽
The capacitance of a parallel-plate capacitor is based on its structure:
𝐴
𝐶 = 𝜺𝟎
𝑑
A − area of each plate
d – distance between the plates
Electric Field between Capacitor Plates: 𝐸 =
o A capacitor with more surface area on its plates: Has more capacitance since more charge can fit
on the plate.
o A capacitor with greater plate separation: Has less capacitance.
Example A: A capacitor consists of two parallel circular plates of radius .05 m that are seperated by a distance
of 0.5 cm. How much charge does each plate when the capacitor is connected to a 12 V battery?
𝐶=
𝑄 = 𝜺𝟎
𝑄
→ 𝑄 = 𝐶𝑉
𝑉
)(𝜋)(. 05 𝑚)
(8.85𝑥10
𝐴
𝑉=
(. 005 𝑚)
𝑑
(12 𝑉) = 1.7𝑥10
𝐶
Energy Stored in a Capacitor
 Capacitors store electric energy in the field between the plates. The energy stored in a capacitor is equal
done to the work to charge it up:
A capacitor is charged by moving charge to a plate, which requires work. As a plate is charge to
potential difference of V, the charge on the plate goes from 0 to some charge Q.
1𝑄
𝑈=
2 𝐶
Using Q=CV, the above equation can also be used to derive the more common form: 𝑼 =
𝟏
𝟐
𝑪𝑽𝟐
Example B: A parallel-plate capacitor has plate area of A and distance between the plates of D. It is connected
to a battery of voltage V.
a) Determine an expression for the charge, electric field, and energy stored in the capacitor.
b) With the battery still connected, the plate separation is doubled. Determine the new charge, electric
field, and energy stored in the capacitor.
c) The capacitor is fully charged, and disconnected from the battery. The plate seperation is again
doubled. Determine the new charge, electric field, and energy stored in the capacitor.
a) Charge: 𝑄 = 𝐶𝑉 = 𝜀
𝑉=
Electric Field: 𝐸 = 𝑉/𝐷
Energy Stored: 𝑈 = 𝐶𝑉 =
𝜀
𝑉 =
46
b) When the plate seperation is doubled, the capacitance will halve. With the battery still connected, the
voltage will remain the same since it will maintain the battery voltage.
Charge: 𝑄 = 𝐶𝑉 = 𝜀
𝑉=
The charge stored halves since the plates are further apart.
Electric Field: 𝐸 = 𝑉/2𝐷
The electric field halves. Since the charge on the plate decreases, the electric field decreases.
Energy: 𝜀
𝑉 =
The energy halves. Though the voltage stays the same, capacitance decreases. Less charge stored leads
to less stored energy.
c) With the battery no longer connected, the charge on the plates will remain constant. The voltage will
no longer remain constant as there is no more battery.
Charge: 𝑄 =
The charge on the plate will stay the same as there is no where for the charge on the plates to go.
Electric Field: 𝐸 =
=
= 𝑉/𝐷
The electric field remains the same. The electric field is only based on the charge on the plates and the
plate area.
Energy: 𝑈 =
=
=
The energy doubled. The charge remained the same, but the capacitance decreased. The work done to
separate the plates is transferred to the energy stored in the plates.
Dielectrics
 In a capacitor, it’s important for the space between the plates to be non-conductive. Capacitors can use
air in the gap between the plates, but air can sometimes allow the flow of electricity.
 Dielectric Breakdown – Charge jumps the gap between the plates of a capacitor.

Dielectric – A dielectric is an insulator placed between the plates of a
capacitor. A dielectric is a material that has a greater permittivity than
free space and can tolerate a greater electric field without dielectric
breakdown. A dielectric increases the capacitance.

Dielectric Constant – The dielectric constant, 𝜅, measures how much
the dielectric constant increases the permittivity of the space between
the plates:
𝜿=
C – Capacitance with dielectric
C0 – Capacitance without dielectric
*𝜅 ≥ 1
The field decreases: 𝜅 =
=
𝑪
𝑪𝟎
=
Permittivity increases: 𝜀 = 𝜅𝜀
47
Example C: A parallel-plate capacitor has plate area of 𝐴 = 0.2 𝑚 and distance between the plates of . 001𝑚.
It is connected to a battery of voltage 10 V.
a) Determine the capacitance, charge, and energy stored in the capacitor.
b) With the battery still connected, a dielectric of dielectric constant of 𝜅 = 2 is inserted. Determine the
capacitance, charge, and energy stored in the capacitor.
c) The capacitor is fully charged, and disconnected from the battery. The same dielectric of 𝜅 = 2 is
then inserted. Determine the new charge, electric field, and energy stored in the capacitor.
a) Capacitance = 𝜀
= 8.85𝑥10
Charge: 𝑄 = 𝐶𝑉 = 𝜀
.
= 1.8𝑥10
.
𝐹 = 1.8 𝑛𝐹
𝑉 = (1.8𝑥10
𝐹)(10 𝑉) = 1.8𝑥10
𝐶
Energy Stored: 𝑈 = 𝐶𝑉 = (1.8𝑥10
𝐹)(10 𝑉) = 8.9𝑥10
𝐽
b) A dielectric increases the capacitance. What happens to the rest of the circuit depends on if the
battery is still connected. With the battery still connected, the battery voltage is constant.
Capacitance = 𝜅𝜀
.
= (2) 8.85𝑥10
Charge: 𝑄 = 𝐶𝑉 = 𝜀
𝑉 = (3.6𝑥10
.
= 3.6𝑥10
𝐹)(10 𝑉) = 3.6𝑥10
𝐹 = 3.6 𝑛𝐹
𝐶
Energy Stored: 𝑈 = 𝐶𝑉 = (3.6𝑥10 𝐶)(10 𝑉) = 1.8𝑥10 𝐽
The charge and energy stored both increased due to the dielectric. When a dielectric is inserted while
power is connect, charge, energy, and capacitance all increase by factor of the dielectric constant. The
electric field decreases by the same factor 𝜅 =
c)
Capacitance = 𝜀
= (2) 8.85𝑥10
.
.
= 3.6𝑥10
𝐹 = 3.6 𝑛𝐹
Charge: From part a), the charge on the capacitor was 𝑄 = 1.8𝑥10 𝐶 when fully charged by the
battery. That charge will remain on the plates will the battery is disconnected. Inserting the
dielectric will not affect the charges on the plates, so the charge is still:
𝑄 = 1.8𝑥10 𝐶
The voltage will change since capacitance decreases with constant charge:
.
𝑉= =
=5𝑉
.
*When the battery is not connected, the voltage decreases by a factor of the dielectric constant:
𝜅=
Energy Stored: 𝑈 = 𝐶𝑉 = (3.6𝑥10
𝐹)(5 𝑉) = 4.4𝑥10
𝐽
The energy decreases since the charge remains constant and the potential difference decreases. Negative
work is done to increase the dielectric, decreasing the energy stored in the capacitor.
48
Example D: A capacitor of plate area 𝐴 = 0.1 𝑚 and plate separation of d=.1 cm
has two dielectrics inserted, with each dielectric taking up half the gap as shown.
The dielectrics has constant of 𝜅 = 2 and 𝜅 = 4. Determine the capacitance of the
capacitor.
The electric field in a region is reduced by a factor of 1/k when a dielectric is inserted. The field when a
between capacitor plates with a dielectric is: 𝐸 =
, so the fields in the regions are 𝐸 =
and 𝐸 =
.
The two dielectrics have different electric fields, so they will not have different potentials across them.
The potential total difference for the entire region between the plates is:
𝑉
=𝑉 +𝑉 =
𝑄
4𝐴𝜀
𝑑
𝑄
+
2
2𝐴𝜀
𝑑
3𝑄𝑑
=
2
8𝐴𝜀
*Q is unknown, but will cancel
The capacitance can be found using Q=CV:
𝐶=
𝑄
𝑄
8𝐴𝜀
=
=
𝑉
3𝑑
→𝐶=
8(.1 𝑚 )( 8.85𝑥10
3(.001 𝑚)
= 2.4𝑥10
Rate your understanding: Capacitors
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Die, electrics.
I understand some
aspects of capacitors.
𝐹 = 2.4 𝑛𝐹
2
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with few errors.
3
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capacitors with no
errors.
4
I can explain and teach
capacitance.
49
AP Physics 2
Unit 4: Circuits
Section 4.1 – Current………………………………….……………………………………...…51
Section 4.2 – DC Circuits………………………………………………………..…….……..…55
Section 4.3 – Energy Cells…………………………………………………….....……….….....62
Section 4.4 – Capacitors…..……….…………………………………………………...…….…65
50
AP Physics 2
4.1 Current
Unit 4: Circuits
Focus Question:


Current is charge moving through a conductor.
When there is potential difference in a conductor, positive charge (current) will flow from an area with
high potential to low potential. Current is created by apply a potential difference across a conducting
wire:
*By convention, current is defined at the direction positive charge flows. Electrons flow opposite
conventional current.
Current is equal to the rate at which charge moves through the conductor:
𝐼=
𝑐ℎ𝑎𝑟𝑔𝑒
∆𝑄
→𝐼=
𝑡𝑖𝑚𝑒
∆𝑡
*units: Amperes (A)
Example A: A hollow conducting sphere is initially uncharged, with an electric potential of zero. A wire is
connected to the sphere.
a) What must be true of the potential at the other end of the wire for the sphere to become charged?
b) A current of 5 mA flows to the sphere. How many electrons pass a point on the wire in one second?
Do the electrons move towards or away from the sphere?
c) How long does it take the sphere to obtain a charge of 2 C?
a) The other end of the wire must be at non-zero potential. Current will only flow through the wire if
there is a potential difference between the two ends.
b) One ampere is one coulomb per second. A single electron has charge equal to the elementary charge,
e= 1.6𝑥10
𝐶.
# 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 = 5𝑥10
𝑥1𝑠𝑥
= 3.1𝑥10 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
.
Current is flowing into the sphere. Conventional current is the motion of positive charge. Since positive
charge flows into the sphere, that means negatively charged electrons are flowing away from the sphere.
c) 1 𝐴 = 1
time =
𝑡=
=
/
= 400 𝑠
51
Resistance

The rate of charge flow through a wire is based on the magnitude of a potential difference, the length of
the conductor, the cross-sectional area and the material is it is passing through:
𝐼=
∆𝑉𝐴
𝜌𝐿
A – cross section area of wire (a greater cross section allows more current)
L – length of wire (a longer wire resists current flow).
𝜌 −resistivity, a property of the conductor and its temperature.
*A perfect conductor has zero resistivity. A perfect insulator has infinite resistivity.
*Resistivity is the reciprocal of conductivity

The entire quantity is known at resistance, which measures a wires ability to resist current flow.
Resistance is higher for longer resistances, and resistors with smaller cross sectional areas. Resistance is
also based on the resistivity of the material. Resistance also increases with temperature.

Resistors - Electrons collide with metal nuclei when
traveling through a conductor. These collisions to
transfer electrical energy to heat. This is analogous to
friction in mechanical motion. Resistors impede current
flow by converting electric energy (voltage) to heat.
*Resistance of a theoretically perfect conductor is zero. *Resistance of a perfect inductor would be
infinite. Resistors are required in circuits to slow down current. Electrons moving too fast will cause a
circuit to become very hot and can burn out circuit elements.

Resistivity and Resistance - Charge drifts in the direction of the force on it. The charge makes inelastic
collisions with stationary particles and transfer energy to them. Thus, conductors get warmer as current
flows.
𝝆𝑳
𝑹=
*𝑑𝑅 =
𝑨
R - resistance
L – length of conductor
A – cross sectional area of conductor

Substituting R into the relationship between potential difference and
current gives Ohm’s Law:
𝐼𝜌𝐿 𝜌𝐿
∆𝑉 =
→
=𝑅
𝐴
𝐴
→ ∆𝑽 = 𝑰𝑹
*Materials through which R is constant (a ∆𝑉 𝑣𝑠 𝐼 graph is linear) are known as ohmic. If R varies with
current (usually due to R increasing with temperature), the material is nonohmic.
52
Power
 Power is the rate at which a potential difference does work on a charge:
As charge moves through an element as shown. The electric field does work on the charge as it passes
through:
𝑾 = 𝑽𝒒
𝑷𝒐𝒘𝒆𝒓 =
𝑾𝒐𝒓𝒌 𝑽𝒒
𝒒
=
= 𝑽( )
𝒕𝒊𝒎𝒆
𝒕
𝒕
 𝑷 = 𝑽𝑰


Using Ohm’s Law with the power equation above, you can also derive the following two expressions for
power:
𝑽𝟐
 𝑷 = 𝑰𝟐 𝑹 =
𝑹
How Power is used in a circuit:
Example A: Why do lightbulbs fail often fail immediately after power is connected?
When a bulb is first turned on, it is at a low temperature, so it has its minimum resistance at this time. When
voltage is first supplied, there is a sudden spike in current and power. As a bulb heats up while it’s turned on, so
its resistance rises, so the power delivered to it decreases, making it less likely to burn out.
Example B: A circuit provides 20 A at 220 V. How many 60 W lightbulbs can be powered?
The total power supplied is: 𝑃 = 𝑉𝐼 = (220 𝑉)(20 𝐴) = 4400 𝑊
The number of lightbulbs this power supplied can power is:
= 73.3
73 light bulbs can be powered.
The lightbulbs can be wired in series or parallel to get the same power. In series, all the lightbulbs will get the
same current, but each will get less volts. If the bulbs are all in the parallel, they will all get the full voltage of
the battery, but the current will be slower so they will get those volts at a faster rate.
Bulbs and other devices are usually wired in parallel. If 73 lightbulbs are in series and one burns out, the entire
series will have a short-circuit and all light-bulbs will stop working. In parallel, if one light bulb burns out, the
branch with that bulb only will short-circuit, and the other bulbs will still work.
53
Example C: An electric heater is power by applying 50 V to a wire of 8 Ω.
a) Find the power supplied to the wire.
b) How much air could it heat (specific heat of air = 1
) from 10 ℃ to 25 ℃ in one hour?
℃
c) How much would it cost to run the heater for 4 hours if electricity cost $.10 per kilowatt hour?
a) 𝑃 =
=
(
)
( Ω)
= 312.5 𝑊
b) Power is the rate of energy transfer. The energy delivered to the air is the power multiplied by time:
𝐸 = 𝑃𝑡 = 312.5 (1 ℎ𝑟)
= 1.13𝑥10 𝐽
This energy is transferred to the air in the form of heat, Q, so the mass can be found form the specific
heat formula:
𝑄 = 𝑚𝑐 ∆𝑇 → 𝑚 =
∆
→𝑚=
.
(
℃
)(
℃
℃)
c) 𝐶𝑜𝑠𝑡 = (312.5 𝑊)(
= 75000 𝑘𝑔
)(4 ℎ𝑟)( $0.10
) = $0.13
Example D: The graph at right shows a plot of voltage versus current for a
filament lamp. The 12 V dashed line shows the voltage setting of the power
source.
a) Why does the graph not obey Ohm’s Law?
b) Why would the filament not obey Ohm’s Law?
c) How would you determine the resistance of the filament for point P on the
graph?
a) The voltage does not vary linearly with the current through the filament.
b) As the current increases, the resistance of the circuit changes (likely due to the increasing temperature).
c) Find the slope at point P. (V=IR is of the same form as y=mx)
Rate your understanding: Current
0
1
My physics knowledge
is not current.
I can understand
circuits with help.
2
3
4
I can analyze circuits
with only minor errors.
I can analyze circuits
with no errors.
I can explain and teach
circuits.
54
AP Physics 2
4.2 Circuits
Unit 4: Circuits
Focus Question: How does current move around a full circuit?



DC current – current that flows in only direction. Current for most electrical devices is AC current,
which constantly switches direction. This unit is concerned with DC current only.
Transient current – Temporary current that flows and then stops (such as when something is grounded).
Continous current – current that flows in a continous loop to a power supply that provides electromotive
force that supplies a constant electric potential difference to charge passing through it.
A circuit is a way for electric potential (also called voltage) to be transferred from a source of electric potential
energy to a load (something that uses electrical energy to function). The source and load are connected through
a conducting wire that carries electrons. The electrons carry electrical potential with them.



Current flows down a “voltage hill”.
Current from high electrical potential (positive charge) to low electrical potential (negative charge).
In a solid conductor, current is due to the flow of free electrons. As
electrons are negatively charged, they are attracted to flow towards
positive charge. However, in circuit analysis, we pretend that current is
due to the flow of positive charge in the opposite direction of electron
movement. This is called conventional current.
(Conventional) Current flows from positive charge to negative charge in a conductor.
+high potential
-low potential
*Potential (measured in volts) is electric potential energy per unit charge.

Charge flows from a source to a load in a circuit. A load is an object that uses electrical energy such as a
motor or light. Charge arrives at a load with high energy and leaves with lower energy.

After leaving a source, current circulates to a load. The load consumes electric potential but not
current.
Resistors in Series
 2 circuit elements are in series when current must flow through both. Since both
are in the same current flow, they have the same current. In series, potential is
dropped in both resistors.
𝑉 = 𝑉 + 𝑉 = 𝐼𝑅 + 𝐼𝑅
→ 𝐼(𝑅 + 𝑅 ) = 𝐼𝑅
→ 𝑹𝒆𝒒 = 𝑹𝟏 + 𝑹𝟐
*Req = Equivalent resistance.
55
Resistors in Parallel
 When a circuit splits into parallel, current splits up since the charges can only go down
one path. The current that splits carries the same potential.
𝐼 +𝐼 =𝐼 → 𝐼 +𝐼 =
→
→



+
𝟏
𝑹𝒆𝒒
=
=
𝟏
𝑹𝟏
+
𝟏
𝑹𝟐
When 2 resistors are in parallel, the over resistance goes down, so current increases. Adding a resistor in
paralllel adds another path for current, so more current is able to flow, even though the new path also
has resistance.
Current splits proportionally down branches, with more current takign the path with less resistance. By
V=IR, resistance and current are inversely with constant voltage. If one branch has no resistance
whatsover, all the current will go down that branch.
Power is shared in both and parallel. In parallel, the same potential is carried to each branch, but the rate
which potential is carried (current) is decreased.
Meters
Measuring Current - Ammeters have very low resistance and thus are connected in series with the element
whose current is to be measured.
Since the ammeter is in series with the resistor, it will measure the same current as
the resistor.
An ideal ammeter has a very small (negligible) resistance, so it will not drop any
volts or alter the current and will therefore not affect the circuit by it being there.
*A non-ideal ammeter will have resistor, and therefore decrease the current lower
than what it should be.
Measuring Voltage – Voltmeters have very high resistance and thus an accurate voltage reading would not be
recorded if it was connected in series with the element with which current is to be measured. Thus, voltmeters
are connected in parallel with the element that voltage is to be measured in.
Since the voltmeter is in parallel with the resistor, it will measure the same
voltage.
An ideal voltmeter has a very high (infinite) resistance, so only a negligible
amount of current will flow through it.
56
Circuit Networks
 Combination Series/Parallel Circuits - In the circuit shown, R 1 and
R2 are in series since current from the battery has to pass through
both resistors. However, R3 and R4 are not in series since the current
flowing through one is equal to the current in the other. After passes
through R2, current can either take the path through R3 or the path
through R4, making these two resistors in parallel. The majority of
the current will go through the resistor with least resistance.
Example A: Determine the current in the 2.0 Ω resistor and the
potential difference across the two marked points, A and B, when
the switch is a) open and b) closed.
a) With the switch open, all current through the 2.0 Ω resistor.
Current splits between the 4.0 Ω in parallel. The equivalent
resistance of the two 4.0 Ω resistors is 2 Ω ( + ).
When the switch is open, the total resistance is 4.0 Ω and the total current is 3.0 A (12/4). The potential
difference across the 2.0 Ω resistor is 2.0 × 3.0 = 6.0 V. The potential difference between A and B is therefore 6
Vas well.
b) When the switch is closed, no current flows through the 2.0 Ω resistor, since all the current takes the path
through the switch, which has no resistance. No voltage is dropped across the switch as there is no resistance,
making the potentional difference zero.
Example B: Solve the circuit shown.
To solve a circuit find the equivalent circuit and then work
backwards. The equivalent circuit is a circuit with one source
and one resistor (that has the equivalent resistance of the
circuit.
To find the equivalent, combine elements that are in series and parallel.
57
Step 1: Combine the 2 Ω and 10 Ω in series.
2 Ω + 10 Ω = 12
Step 2: Combine the two 12 Ω in parellel.
1
=6Ω
+
*When two identical resistors are in parellel, the equivalent resistance is half
the resistance.
Step 3 (Equivalent Circuit): Combine the two 8 Ω and 6 Ω in series.
8 Ω + 6 Ω = 14 Ω
The current leaving the battery (equivalent current can now be
found):
35
𝐼=
= 2.5 𝐴
14
Work backwards through the steps to the original circuit. If elements are in series when going backwards, they
have the same current. If elements are in parallel, they have the same voltage drop. Ohm’s Law can be used to
find the other quantity.
Back to Step 2: The 6 Ω and 8 Ω were combined in series, so keep
the current when splitting back up. The voltage in each can be
found using Ohm’s Law:
𝑉 = (2.5)(8) = 20 𝑉
𝑉 = (2.5)(6) = 15 𝑉
Back to Step 1: The two 12 Ω resistors were
combined in series, so keep the voltage of 15 V
across the 6 Ω they were combined into. The current
in each is found by Ohm’s Law:
15
𝐼 =
= 1.25 𝐴
12
58
Back to Equivalent Circuit: The 2 Ω and 10 Ω were combined in series, so keep the current when splitting back
up. The voltage in each can be found using Ohm’s Law:
𝑉 = (1.25)(2) = 2.5 𝑉
𝑉 = (1.25)(10) = 12.5 𝑉
The final solved circuit:
Example C: The circuit on the right is used to measue the current and voltage for a
resistor.
a) Suppose the ammeter is ideal and the voltmeter is non-ideal. Is the ammeter
reading an overestimate, underestimate or neither?
a) Suppose the voltmeter is ideal and the ammet is non-ideal. Is the
volmeter reading an overestimate, underestimate or neither?
a) If the voltmeter it is non-ideal, then it allows signficant current to run
through it. The ammeter will measure all the battery current, but not all of this current will run through
the resistor since much of it will go through the non-ideal voltmeter. This makes the ammeter reading an
overestimate.
*Negligible current runs through an ideal voltmeter.
b) If the ammeter is non-ideal, then it drops voltage. The voltage reading itself is the same voltage as the
resistor, it is neither an underestimater or overestimate. However, the actual voltage through the resistor
is less than it would be with an ideal ammeter.
*An ideal ammeter has no resistance, so it does not drop any voltage.
Example D: Three lightbulbs that are rated 60.0 W, 120 V lightbulbs are
connected across a 120 V power source, as shown.
a) Find the power delivered to each lightbulb.
b) Rank the brightness of the 3 bulbs.
c) The bulb labeled R3 goes out. What happens to the brightness of
the other two bulbs?
d) The bulb labeled R1 goes out. What happens to the brightness of the other two bulbs?
a) Since the power and voltage for each lightbulb are given, the resistance of each bulb can be solved for:
𝑃=
(120 𝑉)
𝑉
𝑉
→𝑅=
=
= 240 Ω
𝑅
𝑃
60 𝑊
→ 𝑅 = 𝑅 = 𝑅 = 240 Ω
59
To solve for the power, the circuit must be solved to get quantities for each bulb that can be used:
b) Lightbulbs with more power are brighter.
Power of R1 > (Power of R2 = Power of R3)
c) Bulb 2 gets brighter since it no longer has to split current with Bulb 3 and now gets all the battery
current. Bulb 1 gets dimmer. Getting rid of Bulb 3 gets rid of a parallel branch, making the equivalent
resistance go up. Thus the battery current goes down. Since Bulb 1 gets all the battery current regardless,
it gets less bright.
d) If R1 goes out, there is no longer a complete loop of current, and current ceases to flow. Thus, the
other two light bulbs go out.
Kirchhoff’s Loop Rule


The sum of the voltages around a loop in a circuit is zero.
Kirchoff’s Loop rule is due to conservaiton of energy. Potential difference
from a source is positive (it provides potential) and potential difference
from a load is negative (since it uses the potential).
60
Kirchoff’s Node Rule


The current going a node equals the current coming a node.
Kirchoff’s node rule is from conservation of charge.
Example E: Determinet the 3 unknown currents in the circuit shown using
Kirchoff’s Rules.
There are 3 unknowns, so 3 equations need to be set up using Kirchoff’s
laws using loops or nodes in the circuit.
 Node Rule node c:
𝐼 = 𝐼 + 𝐼 → 𝐼 − 𝐼 − 𝐼 = 0 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)
 Loop Rue for the loop shown through the 5 Ω resistor:
0 = +6 − 4𝐼 − 5𝐼
→ 4𝐼 + 5𝐼 = 6 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
 Loop Rue for the loop shown through the 4 Ω resistor:
0 = +6 − 4𝐼 − 9𝐼
→ 4𝐼 + 9𝐼 = 6 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛)
*A graphing calculator can solve this system*
Solving the 3 variable system using a solver: 𝑰𝟏 = −𝟎. 𝟖𝟒 𝑨, 𝑰𝟐 = 𝟎. 𝟓𝟑 𝑨, 𝑰𝟑 = 𝟎. 𝟑𝟎 𝑨
Rate your understanding: DC Circuits
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1
I have high resistance
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I can understand
circuits with help.
2
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with only minor errors.
I can analyze circuits
with no errors.
I can explain and teach
circuits.
61
AP Physics 2
4.3 Electric Cells
Unit 4: Circuits
Focus Question: What role do energy cells play in a circuit?

A battery is the most common source of positive potential
difference in a circuit. In a battery, current flows from low
potential to high potential due to an electromotive force.
The electromotive force of a circuit is the work done per unit charge.
𝜀=
𝑊
𝑞
*Power sources convert one form of energy to electrical energy. A battery uses chemical energy.
Internal Resistance

An ideal battery no resistance, and all the battery emf is delivered to the circuit. A real battery has an
internal resistance.
When a battery as an internal resistance, treat the internal resistance as
a small resistor inside the battery.
𝜀 − voltage provided by battery emf
r – internal resistance of battery
I – current through battery
The battery’s emf supplies 𝜀 volts. The internal resistance drops an
amount of voltage given to 𝑉 = 𝐼𝑟. The effective voltage leaving the battery is then: 𝑉 = 𝜀 − 𝐼𝑟.
 Effective Voltage of non-ideal (real) battey: 𝑉 = 𝜀 − 𝐼𝑟
 Effective Voltage of ideal battery: : 𝑉 = 𝜀
Example A: The graph shown gives the potential difference across the terminals of a battery as a function of
the battery current. Determine the internal resistance of the battery.
Comparing 𝑉 = 𝜀 − 𝐼𝑟 to 𝑦 = 𝑚𝑥 + 𝑏, a graph of V vs. I will
have a slope of -r and a y-intercept of 𝜀.
The y-intercept of the graph is 11 mV, so 𝜀 = 11 𝑚𝑉.
The slope of the line is -.25, so r = 0.25 Ω.
Example B: When two resistors, each of resistance 4.0 Ω, are connected in parallel with a battery, the current
leaving the battery is 3.0 A. When the same two resistors are connected in series with the battery, the total
current in the circuit is 1.4 A. Calculate the battery emf and internal resistance.
When there is a resistance in the circuit, the relationship between the circuit
resistance and internal resistance is:
0 = 𝜀 − 𝐼𝑟 − 𝐼𝑅
→ 𝜀 = 𝐼(𝑟 + 𝑅)
62
When the two 4 Ω resistors are in series, the circuit resistance is R=8 Ω. When they are in parallel, the circuit
resistance is R = 2 Ω. The following two equations can then be constructed using 𝜀 = 𝐼(𝑟 + 𝑅):
𝜀 = 3(𝑟 + 2)
𝜀 = 1.4(𝑟 + 8)
→ 3(𝑟 + 2) = 1.4(𝑟 + 1.8) → 𝒓 = 𝟑. 𝟐𝟓 Ω
The battery emf can be found using either equation: 𝜀 = 3(3.25 + 2) = 𝟏𝟓. 𝟕𝟓 𝑽
Circuits with Multiple Batteries
 Batteries in the same direction in series:
Two batteries with their terminal aligned in the same direction will
combined their electromotive force, leading to a greater total voltage
supplied
𝑉
=𝑉 +𝑉

Batteries in opposite direction in series:
If two batteries face in opposite directions, current will flow in the
direction of the battery with greater emf. The battery with lower emf will
take away from the total voltage:
𝑉
=𝑉 +𝑉
where 𝑉 > 𝑉
If the smaller battery facing the other direction is a rechargable cell, this will charge the battery.
Example C: Calculate the current in each circuit below.
Apply Kirchoff’s Voltage (Loop) Law for each circuit:
a) 0 = +12 − 10𝐼 + 6 − 8𝐼 = 0 → 𝐼 = 1 𝐴
b) 0 = +12 − 10𝐼 − 6 − 8𝐼 = 0 → 𝐼 = 0.33 𝐴
Example D: Calulate each current shown.
Since there are 3 unknown currents, 3 equations are needed.
Loop 1: The 12 volt battery supplies voltage, the loop flows
through the two resistors in the way current was defined so the
convention al signs are kept:
0 = +12 − 2𝐼 − 4𝐼
→ 4𝐼 + 2𝐼 = 12 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)
Loops 2: Current flows “backwards” through the 2 Ω resistor and
12 V battery, so the signs for these two are flipped.
0 = +6 − 1𝐼 + 2𝐼 − 12
→ 2𝐼 − 𝐼 = 6 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
63
Node: I2 and I3 both go into the node and I1 comes out of the node.
𝐼 +𝐼 =𝐼
→ −𝑰𝟏 + 𝑰𝟐 + 𝑰𝟑 = 𝟎 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)
Solving the equations using a solver: 𝑰𝟏 = 𝟏. 𝟕 𝑨, 𝑰𝟐 = 𝟐. 𝟔 𝑨, 𝑰𝟑 = −𝟎. 𝟗 𝑨
*I3 being negative means current flows opposite of the direction defined in the diagram.
*The direction of the current for problems with multiple batteries cannot be known until solving for the actual
currents, so the currents were given an arbitrary direction at the start of the problem. If the answer for a current
is negative, that means the actual direction of the current is opposite the defined direction.
Example E: Calulate each current shown.
Loop 1: Current flows “backwards” through the 3 V battery the
lower 20 Ω resistor, so the signs are flipped for these.
0 = +6 − 3 + 20𝐼 − 20𝐼
→ 20𝐼 − 20𝐼 = 3 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)
Loops 2: Just as in loop 1 current flows “backwards” through
branch that contains the 3 V battery the 20 Ω resistor.
0 = +8 − 3 + 20𝐼 − 15𝐼
→ −20𝐼 + 15𝐼 = 5 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
Node: By the directions defined in the diagram, all 3 currents flow
into the node.
𝐼 + 𝐼 + 𝐼 = 0 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)
Solving the equations using a solver: 𝑰𝟏 = 𝟎. 𝟎𝟏 𝑨, 𝑰𝟐 = −𝟎. 𝟏𝟒 𝑨, 𝑰𝟑 = 𝟎. 𝟏𝟒 𝑨
*I2 being negative means current flows opposite of the direction defined in the diagram.
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AP Physics 2
4.4 Capacitors in Circuits
Unit 4: Circuits
Focus Question: How do Capacitors act when placed in a circuit?


Capacitor Schematic Symbol –
When a capacitor is placed in a circuit, the gap between the plates acts a short circuit. Charge flows form
the battery until both plates are fully charged.

Electrons leaves the negative terminal of the battery and flow to the upper plate of the capacitor. The
electrons on the upper plate of the capacitor repel electrons from the lower plate, so those electrons flow
through to the battery.
Current flows through a circuit with a capacitor at a decreasing rate until the capacitor is fully charged.
A fully charge capacitor acts as a short circuit.

Capacitors in Parallel
 When capacitors are placed in parallel, they each reached the same potential difference. They share the
charge coming from the battery.
𝑉 = 𝑉 = 𝑉. The charges add, 𝑄 = 𝑄 + 𝑄 .
𝐶 𝑉 =𝑄 𝑉+𝑄 𝑉
Capacitors in parallel act like a single capacitor whose capacitance is the sum
of the capacitances.
𝐶
=𝐶 +𝐶
Example A: Two capacitors of capacitance 12 𝜇F and 4.0 𝜇F are connected in parallel to a source of potential
difference of 10 V. Calculate the charge on each capacitor.
Since the capacitors are in parallel they both have the full battery voltage across
their plates, so the charge can be calculated using Q=CV for each.
𝑄
= (12 𝑥10 𝐹)(10 𝑉) = 1.2𝑥10 𝐶
𝑄
= (4 𝑥10 𝐹)(10 𝑉) = 4.0𝑥10 𝐶
65
Capacitors in Series
 When capacitors are placed in series, the charge on each is the same since current in series is constant:.
The voltages add: 𝑉 = 𝑉 + 𝑉 →
= +
Finding the equivalent capacitance of capacitors in series works the same as
finding the equivalent resistance of resistors in parallel.
1
1
1
=
+
𝐶
𝐶
𝐶
Example B: Two capacitors are connected as shown. The capacitors have
capacitances of 𝐶 = 3 𝐹 and 𝐶 = 6𝐹. The first capacitor is fully charged by a 10
V battery and then disconnected from the battery. Afterwards, the switch shown is
closed and the first capacitor is allowed to discharge into the second capacitor.
Find the final charge on each capacitor.
When a capacitor discharges to another capacitor, it will give some of its charge of the other capacitor, with the
total amount of charge being conserved. Charge flows when there is a potential difference, so charge will flow
between the capacitors until both capacitors have the same potential difference across their plates.
The total charge once 𝐶 is charged is 𝑄 = 𝐶 𝑉 = (3 𝐹)(10 𝑉) = 𝟑𝟔 𝑪
At equilibrium the voltage on each is equal.
𝑄
𝑄
𝑄
𝑄
𝑉 =𝑉 →
=
→
=
→ 𝑄 = 2𝑄
𝐶
𝐶
3𝐹 6𝐹
The 36 C charge is split between the capacitors due to conservation of charge:
𝑄 + 𝑄 = 36 𝐶
Solving the system: 𝑸𝟏 = 𝟏𝟐 𝑪, 𝑸𝟐 = 𝟐𝟒 𝑪
Example C: The capacitors in the circuit all have a capacitance of 12 𝐹. Find the charge and potential
difference for each capacitor.
Problems with multiple capacitors can be solved by finding the equivalent
circuit by combing elements in series or parallel and then working backwards.
Step 1: Combine B and C in parallel
𝐶
= 12 𝐹 + 12 𝐹 = 24 𝐹
Step 2: Combine A with the equivalent 24 F capacitor in series. This is the
equivalent circuit.
1
1
1
=
+
→𝐶 =8𝐹
𝐶
12 24
The total charge is 𝑄 = 𝐶𝑉 = (8 𝐹)(12 𝑉) = 96 𝐶
66
Back to Step 1: The two capacitors that were split were in series, so they both
have the same 96 C since current is the same in series. The voltage on each is
found using Q=CV:
96 𝐶
𝑉 =
=8𝑉
12 𝐹
96 𝐶
𝑉 =
=4𝑉
24 𝐹
Back to original circuit: A is already solve. B and C split into parallel, so
the 4 V for each is kept since potential is constant in parallel, the charge on
each is then:
𝑄 = 𝐶𝑉 = (12 𝐹)(4 𝑉) = 48 𝐶
𝑸𝑨 = 𝟗𝟔 𝑪, 𝑸𝑩 = 𝟒𝟖 𝑪, 𝑸𝑪 = 𝟒𝟖 𝑪
Charging RC Circuit
The circuit shown has an ideal battery of emf V, a capiticator of capatiance C in series with a resistor of
resistance R, and a switch. At time t = 0, the switch is switch to position B:
 Charge on capacitor, Q(t): Initially, the capacitor is
uncharged. Over time, charge builds up on the capacitor.
The rate of the charge accumulating decreases over time as
the rate of current decreases. The capacitor has a certain
maximum charge (CV) it can hold, which it will approach
asympotically.

Current through resistor, I(t): As the capacitor gets charged
at a decreasing rate, the rate of charge flow, which is
current, the current will decrease. Once the capacitor is
fully charged, it acts as a short circuit, so current will stop.

Potential across capacitor, 𝑉 (𝑡): As more charge builds up on the capacitor, will have a potential
difference across its oppositely charged plates. Once the capacitor is fully charge, it will have a constant
potential equal to the battery voltage that was flowinn to it.

Potential across resistor, 𝑉 (𝑡): Initially, the capacitor acts a normal wire as charge flows normally, so
there will be potential (V=IR) across the resistor. As the current decreases to zero, so will the voltage
across the resistor.

Graphs for Charging RC Circuit
67
Discharging RC Circuit
The circuit is fully charged in position A. After a long the switch is
moved to position B.
When the switch is moved to B, the circuit is no longer connected to the
battery. The potential difference across the capacitor plates will no long
be countered by the battered, so the capacitor’s charge will flow away
from the capacitor. The charge will flow through in the opposite direction
it was charged and act as a temporary “battery” for that closed loop it is
in with the resistor. Over time, the rate of charge flow will decrease
asympotically.

Charge on capacitor, Q(t): As the capacitor discharges, charge leaves the plate. When the capacitor is
fully discharged, the circuit neutralizes so there is no net charge.

Current through resistor, I(t): As the rate of charge flowing from the plate decreases, the current
decreases.

Potential across capacitor, 𝑉 (𝑡): The capacito has a potential difference due to the opposing charges on
its plate. As the charges leave the plates, the potential difference decreases.

Potential across resistor, 𝑉 (𝑡): Since the capacitor acts as the power supply for the resistor when it
discharged, the voltage across the resistor will be equal in magnitude to the potential across the
capacitor.

Graphs for Discharging RC Circuit
Example D: In the circuit shown, V = 10 V, R = 10 Ω, and C =
10 𝜇𝐹.
a) The switch is initially open. Calculate the initial current
through both resistors if the switch is closed.
b) After a long time, the switch is opened. Calculate the current
through both resistors just after the switch is opened.
a) When the switch is initially opened, the capacitor will behave like a wire with no resistance, so no
current with flow through the resistor R. All current will flow through 2R to the capacitor:
Current in 2R: 𝐼 =
= 0.5 𝐴
Current in R = 0
Ω
68
b) When the capacitor is fully charged and the switch is opened, the
capacitor will discharge. Current only flows through a full circuit, so
current will only flow through the complete loop contain resistor R and
both plates of the capacitor. The capacitor will have the full battery
voltage of 10 V across its plates.
Current in 2R=0
Current in R: 𝐼 =
= 1.0 𝐴
Ω
Example E: Consider the circuit shown from the moment current starts
flowing.
a) Calculate the initial current in the 20 Ω resistor.
b) Calculate the current in the 20 Ω resistor after a long time.
c) If 𝐶 = 2.0 𝑚𝐹, calculate the charge stored in the capacitor.
a) Initially, charge goes to the capacitor until its plates are charged.
While the capacitor is charging, current flows to the capacitor and “skips” the 30 Ω resistor since the
capacitor has no resistance. The circuit at time t = 0 is from the battery, through the 20 Ω resistor,
through the capacitor, and then through the 10 Ω resistor.
𝐼=
=4𝐴
Ω
Ω
b) After a long time, no current will flow through the capacitor, so complete circuit will not be from the
battery and then through all 3 resistors:
𝐼=
=2𝐴
Ω
Ω
Ω
c) The capacitor is in parallel with the 30 Ω resistor, so it only gets the voltage that is across the 30 Ω
resistor at steady state. The voltage across the 30 Ω resistor at steady state is:
Ω
(12 𝑉) = 6 𝑉
𝑉 =
Ω
Ω
Ω
The charge on the capacitor using Q=CV is:
𝑄 = (6 𝑉)(2𝑥10 𝐹) = .012 𝐶
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AP Physics 2
Unit 5: Magnetism
Section 10.1 – Magnetic Field………………………………….………………….……….…..71
Section 10.2 – Magnetic Force on a Charge………………………………………….…….…..75
Section 10.3 – Magnetic Forces on Current-Carrying Wires…………………….………….….78
Section 10.4 – Faraday’s Law………………..……….………………………………..…….…81
70
AP Physics 2
5.1 Magnetic Fields
Unit 5: Magnetism
Focus Question: How does pressure change with depth?
Consider an electron traveling at velocity v next to a wire with current in it.

Rest Observer Reference Frame – The negative and
positive charges in the wire have the same density.

Frame fixed to the moving electron – Due to length
contraction, the positive charges are more dense than the
negative charge, so there is net negative charge. As a result,
the electron is attracted to the wire.
*The Magnetism Force is simply the electrostatic force applied to moving charges.
Magnetic Fields of Permanent Magnets

In permanent magnets, magnetism is due to the alignment of magnetic moments inside the
elementsmaking up the material. Only a few materials, most commonly iron, act as permanent magnets.

Non-magntized material
Magnetized material
Magnetic momentums unaligned
Aligned magnetic moments
Magnets have two poles, known as north and south. Magnetic field lines point from north to south:
By convention, magnetic field lines outwards from north and go into south.

Earth possesses a magnetic field due the amount of iron in the core, with the geographic north pole
acting as the magnetic south pole.
71
Magnetic Fields Around A Wire


A current-carrying wire causes a magnetic field around it. The field moves in a circle
around the wire. The direction of the current is given by the right hand rule. If you
point your right thumb in the direction of current and wrap your fingers around the
wire, the magnetic field is in the direction of your fingers.
Magnetic field due to a wire:
The magnitude of a magnitude field measures how strong
the field is.
Magnetic field at a distance r from a conducting wire with
current I running through it:
𝑩=
B – Magnetic field
𝝁𝟎 𝑰
𝟐𝝅𝒓
𝜇 – magnetic vacuum permittivity = 4𝜋𝑥10
*Magnetic Field units – Teslas (T)
*A circle represents a vector pointing out of the page. An x represents a vector into the page.
Example A: Find the direction at point P due to wires A and B for the arrangements shown.
a)
b)
a) By the right-hand rule, the field is clockwise around A and counter-clockwise
around B. At point P, both fields are down, so the net field at P is down.
b) By the right-hand rule, the field is counter-clockwise around A and clockwise around B. At
point P, both fields to the right at the point.
72
Example B: Find the magnitude and direction of the magnetic field due to point A,
which is half way between the two wires shown, which carry currents in opposite
directions.
First, the right hand rule is used to find the directions of the fields in the region of
point P:
At point P, both field are into the page, so the net field is into the page and the magnitude of the total field can
be found by adding the field due to the top and bottom wires:
𝐵
=𝐵
→𝐵
=
+𝐵
=
𝜇 𝐼
2𝜋𝑟
+
𝜇 𝐼
2𝜋𝑟
(4𝜋𝑥10 )(1 𝐴) (4𝜋𝑥10 )(2 𝐴)
+
= 6𝑥10
2𝜋(.1 𝑚)
2𝜋(.1 𝑚)
𝑇
The net magnetic field at P is 6 𝜇𝑇 into the page.
Magnetic Field due to a Loop of Wire

Magnetic Field due to a Single Loop of Wire
Current within the loop all points in the same direction by the right-hand rule, leading to a stronger
current.
73

A solenoid is wire wrapped in multiple loops in a cylindrical coil. Solenoids with current running them
create strong magnetic fields.
Magnetic Field due a solenoid of length L with N loops of wire with current I running through it:
𝐵=

𝜇 𝑁𝐼
𝐿
Right hand rule for solenoids – Curl fingers in direction of current, thumb points towards the magnetic
north pole.
Example C: You are given a length of conducting wire and want to create a solenoid that will produce a strong
magnetic field. Should you wind the wire in many small loops or few larger loops?
The magnetic field for a solenoid is stronger with loops of wire and for shorter lengths. The loop density per
length, N/L, is directly proportional to the strength of the field. The actual size of the loops does not affect the
field. It is best to use many small loops.
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AP Physics 2
5.2 Magnetic Force on Moving Charge
Unit 5: Magnetism
Focus Question: How does a magnetic field affect a moving charge?

A charge particle traveling in a magnetic field will experience a magnetic force perpendicular to the
field:
𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝐹𝑜𝑟𝑐𝑒 = 𝐹⃗
𝐹 ⃗ = 𝑞 𝑣⃗𝑥𝐵⃗
*charge times the cross product of speed and magnetic field
𝐹 ⃗ = 𝑞𝑣⃗𝐵⃗ sin 𝜃
𝜃: angle between velocity and magnetic field

The velocity of the charged particle, the magnetic field, and magnetic force are all perpendicular vectors.
The relation between these vectors is given by the right hand rule for a moving charge (also called the
fight palm rule:




Thumb points in direction of velocity.
Fingers point in direction of magnitude field.
Force comes out of the palm.
If the moving charge is negative, use the left-hand rule, which works the same way as the right-hand rule
(just obviously using the other hand).
Example A: Specify the direction of the magnetic force in each of the following situations:
a) up
b) right
c) up
d) out of page
e) out of page
f) no force (v and B are not perpendicular)
75
Example B: Can the magnetic force change a charged particle’s speed?
No. In order for a particle to change speed, there must in an acceleration acting in the same direction as
velocity. Since the magnetic force is always perpendicular to the particle’s velocity, there is no change in speed.
The magnetic force changes the particle’s direction only.
Example C: A proton moves at 5x106 m/s along the positive-axis. It enters a
region in which there is a magnetic field of magnitude 2.50 T, directed at an angle
of 30.0° with the x – axis in the xy plane. Find the initial magnitude and direction
of the magnetic force on the proton.
By the right hand rule, the force is in the +z direction.
𝐹 ⃗ = 𝑞𝑣⃗𝐵⃗ sin 𝜃 = (1.6𝑥10
𝐹 = 1𝑥10
𝐶)(5𝑥10
𝑚
)(2.5 𝑇) sin 30°
𝑠
𝑁
Motion in a Circle


A charge traveling in a uniform magnetic field will experience a constant force perpendicular to its
motion. When a force is always perpendicular to velocity, the direction of the velocity constantly
changes, causing the object to travel in circular motion.
The magnetic force provides the centripetal force .
𝐹
=𝐹
𝑚𝑣
𝐵𝑞𝑣 =
𝑟
𝑚𝑣
→ 𝐵𝑞 =
𝑟
If the initial velocity is not perpendicular to the field than only the component of velocity perpendicular
to the field affects the radius of the orbit. In this case, the charge travels in a helical path.
Example D: A charged particle of mass m and charge q moves in a magnetic field with a flux density of B in
circular motion. Derive an expression for the frequency of its circular motion.
𝐹 = 𝐹 → 𝐵𝑞𝑣 =
𝑇=
𝑚𝑣
𝐵𝑞𝑟
→𝑣=
𝑟
𝑚
2𝜋𝑟
𝑣
𝐵𝑞
→𝑓=
→𝑓=
→𝑓=
𝑣
2𝜋𝑟
2𝜋𝑟
2𝜋𝑚
Lorentz Force

A Lorentz force is the total force on a charge particle moving both a
magnetic and electric field simultaneously.
𝑭⃑ = 𝑭⃑𝑬 + 𝑭⃑𝑩
𝑭⃑ = 𝒒𝑬⃑ + 𝒒𝒗⃑𝒙𝑩⃑
76
Example E: The potential difference between two parallel plates is 850 as
shown. There is a magnetic field of .76 T into the page between the plates. The
plates are 0.8 m and the particle passes undeflected between the plates.
a) Can the sign of the charge be determine?
b) What is the speed of the particle?
a) The charge could be positive or negative. The magnetic force on a positive charge would be up and
the electric force would be down. The magnetic force on a negative charge would be down and the
electric force would be up. Either charge would pass undeflected between the plates.
b) For the charge to travel undeflected, 𝐹⃑ = 𝐹⃑
𝑞𝐸 = 𝑞𝑣𝐵 → 𝐸 = 𝑣𝐵 →
𝑣=
= (.
)(.
)
= 𝑣𝐵
= 1400 𝑚/𝑠
Example F: A positively charged particle of mass m = 2 𝑛𝑔 and charge of 4 𝜇𝐶 is accelerated from rest and
reaches and speed of 2𝑥10 𝑚/𝑠 when it enters region 1, after which is has constant speed. It travels
undeflected in region 1 and then enters region 2, where it travels in a semicircle. The magnetic of the magnetic
field in both region is 0.5 T.
a) Calculate the magnetic of the potential difference that
must have been applied before it entered region 1.
b) Calculate the magnitude of the electric field in region 1.
c) Calculate the radius of the circular path the object
travels in region and specific which way the particle
curves.
a) 𝑊 = ∆𝐾
𝑚𝑣
(2𝑥10
1
1
𝑘𝑔)(2𝑥10 )
→ 𝑉𝑞 = 𝑚𝑣 − 𝑚𝑣 = 𝑉 =
=
= 1𝑥10 𝑉
2
2
2𝑞
2(4𝑥10 𝐶)
b) 𝐹⃑ = 𝐹⃑
𝑚
𝑁
(0.5) = 1𝑥10
→ 𝑞𝐸 = 𝑞𝑣𝐵 → 𝐸 = 𝑣𝐵 → 𝐸 = 2𝑥10
𝑠
𝐶
c) 𝐹⃑ = 𝐹⃑
𝑚𝑣
𝑚𝑣 (2𝑥10 )(2𝑥10 )
→
= 𝑞𝑣𝐵 → 𝑟 =
=
=2𝑚
(4𝑥10 𝐶)(.5 𝑇)
𝑟
𝑞𝐵
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AP Physics 2
5.3 Magnetic Forces on Conductors
Unit 5: Magnetism
Focus Question: What is the force on current-carrying conductor in a magnetic field?

A conducting wire carries a larger number of individual charges, each with a magnetic force acting:
A current carrying wire has many charged particles, each with a
magnetic force on them.
𝐹 = 𝐵𝑞𝑣
The charge is the total charge of all the electrons:
𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑞 = 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑐ℎ𝑎𝑟𝑔𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
→ 𝑞 = 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑐ℎ𝑎𝑟𝑔𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑖𝑟𝑒 = 𝑒(𝑛𝐴𝐿)
where A = cross section of the wire and L = length of the wire
𝐹 = 𝐵(𝑒𝑛𝐴𝐿)𝑣 → 𝐹 = 𝐵(𝑛𝑎𝑉𝑒)𝐿
𝑭 = 𝑩𝑰𝑳
𝑭 = 𝑩𝑰𝑳 𝐬𝐢𝐧 𝜽
Example A: A copper wire has a mass per unit length of 100 g/cm and has a current of 3 A to the right in the
plane of this page. What is the direction and magnitude of the magnetic field requierd to counteract the wire’s
weight?
The force needs to be up and the direction of positive charge velocity is the right, so the magnetic
field would need to be into the page by the right-hand rule.
𝐹 − 𝑚𝑔 =→ 𝐹 = 𝑚𝑔
𝑚𝑔
𝑚 𝑔
→ 𝐵𝐼𝐿 = 𝑚𝑔 → 𝐵 =
→ 𝐵 = ( )( )
𝐼𝐿
𝐿 𝐼
10
𝑘𝑔
→ 𝐵 = .1
= .33 𝑇
𝑐𝑚
3𝐴
Example B: A mass 𝑚 is suspended by an insulating string connected to a circuit
that is partically in a magnetic field of magnitude 𝐵 as shown. The wire has a length
𝐿 horizontally in the field as shown. The circuit has a resistor of resistance 𝑅.
Determine an expression for the battery voltage required for the mass to remain
suspended.
The section of wire in the field will experience an upward force by the right-hand
rule that needs to balance out the force from the weight of the mass. There are also
forces on the two vertical sections of the wire, but these sections cancel out.
𝐹 − 𝑚𝑔 =→ 𝐹 = 𝑚𝑔
𝑉
→ 𝐵𝐼𝐿 = 𝑚𝑔 → 𝐵
𝐿 = 𝑚𝑔
𝑅
𝑚𝑔𝑅
𝑉=
𝐵𝐿
78
Force between Conductors

Wires flowing in same direction
By the right hand rule, both wires have a field that is out of the page above them and into the page below them.
Consider the magnetic force on wire 1 due to wire 2. Wire 1 is above
wire 2 so the field due to wire is out of the page. Positive charge in
wire 1 moves to the right, so by the right hand rule, the force is down.
For wire 2, it experiences field into the page due to wire 1 and have
positive charge moving to the right, so the force on wire 2 is up by the
right hand-rule, meaning the wires attract each other.
The magnitude of the force on wire 1 is: 𝐹 = 𝐵 𝐼 𝐿 → 𝐹 =

𝐼 𝐹→𝐹=
Wires flowing in opposite direction
By the right hand rule, wire 1 has a field out of the page above it
and into the page below it. Wire 2 has a field into the page above it
and out of the page below it.
Consider the magnetic force on wire 1 due to wire 2. The field due
to wire 2 at the wire 1 is into the page. Positive charge in wire 1
moves to the right, so the force on wire 1 is up due to the right-hand
rule. Likewise, the force on wire 2 is down, so the wires repel each other.
𝜇𝐼
𝜇𝐼 𝐼 𝐿
𝐼 𝐿→𝐹=
2𝜋𝑟
2𝜋𝑟
𝝁𝑰 𝑰 𝑳
In either case, the force between two wires is 𝑭 = 𝟏 𝟐 . If two wires flow in the same direction, the force is
𝟐𝝅𝒓
attracive, if they flow in oppose directions, they repel each other. “If they flow together, they go together”.
𝐹=𝐵 𝐼 𝐿→
Example C: Wire A floats 1 cm above wire B. The wires carry currents in opposite directions as shown. Both
wires have a length of 10 m and a mass of 300 g. What current must wire A carry for it to remain suspended in
the air directly over wire B?
In the region between the wires, the field due to wire B is into the field. The field
due to B in this region is:
𝐵 =
where d is the distance between the wires.
By the hand-rule rule, this causes a force up on wire A. This force needs to cancel
out A’s weight.
𝐹 − 𝑚 𝑔 =→ 𝐹 = 𝑚 𝑔
𝜇 𝐼
𝑚 𝑔2𝜋𝑑
(𝐼 )(𝐿) = 𝑚 𝑔 → 𝐼 =
→ 𝐵𝐼𝐿 = 𝑚 𝑔 →
2𝜋𝑑
𝜇 𝐼 𝐿
→𝐼 =
(. 3 𝑘𝑔) 10
(4𝜋𝑥10
(2𝜋)(.01 𝑚)
)(100 𝐴)(10 𝑚)
= 150 𝐴
79
Torque on a Loop of Wire

A magnetic field applied to a loop of wire causes a torque on the wire.
Each side of the square loop of wire above has current in different directions. The two horizontal portions have
current parallel to the magnetic field, so there is no magnetic force. The two sections have forces into and out of
the page on opposing sides by the right-hand rule. This causes the loop of wire to spin around an axis in its
center.
The total torque is 𝜏 = 𝜏

+𝜏
= 𝐵𝐼𝑎
+ 𝐵𝐼𝑎
→ 𝜏 = 𝐵𝐼𝑎𝑏
Torque on a loop in a magnetic field
𝝉 = 𝑩𝑰𝑨 𝐬𝐢𝐧 𝜶
A – area of loop
𝜶 – angle between field and plane of loop
*For multiple loops: 𝜏 = 𝑛𝐵𝐼𝐴 sin 𝛼
Example D: A 200-turncoil of wire has a radius coil of wire has a radius of 20 cm and the normal to the area
makes an and the normal to the area makes an angle of angle of 30 o with a 3 mT magnetic field. What is the
torque on the loop if the current is torque on the loop if the current is 3 A?
𝜏 = 𝑛𝐵𝐼𝐴 sin 𝛼
→ 𝜏 = (200)(3𝑥10
𝑇)(3 𝐴) sin 30° = 0.9 𝑁𝑚
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AP Physics 2
5.4 Faraday’s Law
Unit 5: Magnetism
Focus Question: How does a changing magnetic field induce a current?
Magnetic Flux


Magnetic Flux – Magnetic flux is a measure of the amount of magnetic
field passing through a surface. It’s directly related to the number of field
lines that pass through a surface.
Formula for magnetic flux:
𝜙 = 𝐵𝐴 cos 𝜃
𝜙 : 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑓𝑙𝑢𝑥
𝐴 − 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑢𝑟𝑓𝑎𝑐𝑒; there is more flux when the field passes through a
larger surface
𝜃 − 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑓𝑖𝑒𝑙𝑑 𝑙𝑖𝑛𝑒 𝑎𝑛𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒; there is more flux when the field lines pass straight
through the surface rather than passing through at an angle.
∗units of magnetic flux – Webers, Wb
*Magnetic field is also called magnetic flux density
Example A: A loop of area 0.8 m2 is in a constant magnetic field of B = 0.15 T. What is the magnetic flux
through the loop when:
a) the loop is perpendicular to the field
b) the loop is parallel to the field
a) When the fields are perpendicular, 𝜃 = 90°
b) When the fields are perpendicular, 𝜃 = 0°
𝜙 = 𝐵𝐴 cos 𝜃 = (. 15 𝑇)(. 8 𝑚 ) cos 90 = 0
𝜙 = 𝐵𝐴 cos 𝜃 = (. 15 𝑇)(. 8 𝑚 ) cos 0 = .12 𝑊𝑏
Induced emf

When a rod is given a velocity perpendicular to a magnetic field, a current flows through the wire:
When the bar is moved to the right in a field that outs into the page as shown,
positive charge is pushed up by the right-hand rule. Since there is current in the
wire, there must be a potential difference. Charge accumulates until the electric
force caused by the potential difference cancels out the magnetic force:
𝐹 =𝐹
𝜀
𝑞𝐵𝑣 = 𝐸𝑞 → 𝐵𝑣 =
𝑙
→ 𝜺 = 𝑩𝒗𝒍
*induced emf in a bar of length moving at speed in a magnetic field of B
Example B: A Boeing 787, having wingspan of 60 m, is flying through Earth’s magnetic field near Tokyo,
Japan (B = 56 T) at 265 m/s. Treating the wing as a straight wire, find the induced emf from wingtip to
wingtip.
𝜀 = 𝐵𝑣𝑙 = (56𝑥10
𝑇) 265
(60 𝑚) = 0.90 V
81
Faraday’s Law

If a bar moving in an electric field is connected to a closed circuit, the inducted emf causes current to
flow:
As the bar moves to the right, it covers an area of A as shown in a time ∆𝑡.
𝜙 = 𝐵𝐴 = 𝐵𝑣∆𝑡𝑙
The time rate of change of the flux is

= 𝐵𝑣𝑙 since velocity is constant. This is equal to the emf.
Faraday’s Law – The induced emf due to a changing magnetic field is equal to the time rate of change of
the magnetic flux through the surface bounded by the circuit
Induced emf =
𝜺=−
∆𝝓𝑩
∆(𝑩𝑨)
=−
∆𝒕
∆𝒕
Magnetic flux can be changed by changing the magnitude of the field, the area the field goes through, or
the angle between the flux and the surface it passes through.

Lenz’s Law – “An induced current will have a direction such that it will oppose the change in flux
that produced it.” (the minus sign in Faraday’s law)
Example C: Determine the direction of the current in the wire in each of the cases below:
a) B is increasing
b) B is increasing
c) B is decreasing
a) The field is out of the page and increasing, so the change in flux is out of the page. The induced current
resists the change in flux by inducing a current that creates a field into the page to oppose the change in flux, so
the current is clockwise.
b) The field is into the page and increase, so the change in flux is into the page. The induced current creates a
field out of the page to oppose the direction of the change in flux, so the current is counter-clockwise.
c) The field is into the page and decreasing, meaning the change in flux is out of the page. he induced current
resists the change in flux by inducing a current that creates a field into the page to oppose the change in flux, so
the current is clockwise.
82
Example D: A magnetic is dropped through a conducting loop as shown. Find the
direction of the current in the loop as the magnet falls through it.
a) The flux in the loop is down (since magnetic field comes out of
north) and increasing because the magnetic field at the loop is
getting larger as the magnet approaches. The induced current must
then oppose the increase in the flux by inducing a field in the
opposite direction. This can be done with counter-clockwise current
b) As the magnet leaves the loop from the other side, the flux is decreasing. This makes the
change in the flux up, so the induced current needs to induce a field down, making the
induced current clockwise.
Example E: A conducting loop is placed next a current-carrying wire as
shown. Determine the direction in the loop if:
a) The current in the wire increases
b) The current in the wire decreases
By the right-hand rule, the field is into the page on the right side of the wire.
a) The field is into the page and increasing, so the change in flux is into the page. The induced current is to
produce a field of the page, which corresponds to counter-clockwise current.
b) The field is into the page and decreasing, so the change in flux is out of the page. The induced current is to
produce a field into the page, which corresponds to clockwise current.
83
Example F: A square loop side length s = .70 m is perpendicular to a magnetic field of 1.5
T. The field suddenly decreases to zero is 0.3 s. What current flows through the circuit as a
result?
A change in flux causes an induced current. The flux is into the page and decreasing, the
current to the right through the resistor. Gauss’s law is used to find the induced current,
which is the induced emf divided by the resistance.
∆(𝐵𝐴)
𝐵∆(𝐴)
𝜀=−
→𝜀=−
∆𝑡
∆𝑡
(1.5 𝑇) 0 − (. 7 𝑚)(. 7 𝑚)
→𝜀=−
= 2.45 𝑉
.3 𝑠
𝜀 2.45 𝑉
𝐼= =
= 0.31 𝐴
𝑅
8Ω

If there are multiple loops with flux through them, they each have the induced
emf:
An emf is induced in each loop of wire, so multiply the induced emf for one loop
by the number of loops-
𝜺 = −𝑵
∆𝝓𝑩
∆𝒕
Example G: A coil of loop with 30 turns an radius of .25 m is perpendicular to a magnetic field of flux density
2.2 Wb/m2. The coil is given a quarter turn in .2 s. The total resistance of the coil is 20 Ω.
a) What current flows through the coil?
b) What charge flows through the coil during this time?
c) How much energy is dissipated in the process?
a) The flux changes due to the change in angle. A quarter-turn is 90 o
𝜀=−
𝜀=−
𝐼=
∆
=
b) 𝐼 =
=−
∆
( .
) (.
∆
)
° ( .
) (.
)
°
.
.
Ω
= 2.15 𝑉
= 0.11 𝐴
→ 𝑞 = 𝐼𝑡 = (0.11 𝐴)(. 2 𝑠) = 0.22 𝐶
c) Energy dissipated is equal to 𝑃 = 𝐼 𝑅
→ 𝑃 = (. 11 𝐴) )(20 Ω) = .24 𝐽
84
Example H: The cart shown in the figure above has mass 2.0 kg and moves towards a uniform magnitude field
of magnitude B=2.0 T into the page. Attached to the cart is a rectangular loop of wire that is 0.10 m by 0.20 m
and have a resistance of 2.0 Ω.
a) Indicate the direction of current in the loop on the cart when its front edge enters the field.
b) When the front of the cart reaches the field, it has a speed of 5 m/s. Calculate the force on the cart.
c) Calculate the force on the cart when it is entirely inside the field.
a) The direction of positive charges in the wire is to the right and the field is
into the page, which pushes current up in the front section of wire, this
causes a counter-clockwise current in the wire.
b) 𝐹 = 𝐵𝐼𝐿
The induced current is the induced emf in the front wire, which is BLv, divided by the resistance in the
wire.
𝐹 = 𝐵𝐼𝐿 → 𝐹 = 𝐵
→𝐹=
(
) (.
) (
Ω
𝐿=𝐵
)
𝐿=
= 0.1 𝑁
c) The magnetic force on the left and right vertical sections of wire will induce current in opposite
directions, making the net force zero.
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AP Physics 2
Unit 6: Light Waves
Section 6.1 – Properties of Light Waves…………………..……………….……….………..…87
Section 6.2 – Refraction………………………………………………….……………….…….89
Section 6.3 – Mirrors……………………………………………………….…….......................93
Section 6.4 – Lenses……….……………………………………………………………..…...100
Section 6.5 – Diffraction ……….………………………………………………...…...………106
Section 6.6 – Thin-Film Interference ……….……………………………………………...…110
86
AP Physics 2
6.1 Properties of Light Waves
Unit 6: Light Waves
Focus Question: What is an electromagnetic wave?


Mechanical Wave – A wave that requires a medium to travel through. In a mechanical wave, a
disturbance in a medium travels due to the particles of the medium colliding with each other.
Electromagnetic Waves – A wave that does not require a medium to travel through.
All electromagnetic waves are transverse waves. An electric field
propagates though space, with a magnetic field perpendicular to it.
Since moving charges cause magnetic fields, and changing magnetic
fields cause electric fields, the fields are self-sustaining. In a vacuum,
electromagnetic wave travel at the speed of light, so the relationship
between frequency and wavelength of an electromagnetic wave can
be expressed using the wave speed formula from Physics 1:
𝑤𝑎𝑣𝑒 𝑠𝑝𝑒𝑒𝑑 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑥 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ
→ 𝑐 = 𝑓𝜆
c = speed of light in a vacuum = 3𝑥10 𝑚/𝑠
As the wavelength of an electro-magnetic increases, its frequency decreases. The various types of
electro-magnetic waves make-up the electromagnetic spectrum:

The Doppler Effect for Electro-magnetic Waves
o Red Shift - When light is moving away from an observer, the apparent
frequency decreases, increasing the wavelength, which shifts the
light towards higher-wavelength light, which is red on the visible
light part of the spectrum.
o Blue Shift - When light is moving towards an observer, the apparent
frequency increases, decreasing the wavelength, which shifts the
light towards blue.
Example A: A blue light of frequency 6.8𝑥10
light will the observer see?
is moving away from an observer at 8𝑥10 𝑚/𝑠. What color
Doppler effect formula from Physics 1: 𝑓 = 𝑓 (
Apparent frequency of light: 𝑓 = 𝑓
±
∓
±
∓
)
= (6.8𝑥10 𝐻𝑧)(
/
87

Properties for all types of waves.
o Rectilinear Propagation – Waves in a constant medium travel in straight lines.
o Reflection – When a wave reaches a barrier, light can reflect off.
o Refraction – When a wave enters a new medium, its speed and direction change.
o Diffraction – Waves spread out around an obstacle.
o Superposition – When waves interfere, their amplitudes at the point of interference add.

Polarization - In most electromagnetic waves, fields vibrate
in random directions. This can be corrected through
polarization.
A polarizer produces plane-polarized light. This is done
through filters that use reflection or absorption.
Malus’s Law - When two polarizers are used one after the
other, the first called the “polarizer” and the second is the
“analyzer”. The angle between their polarizing axes can be used to find the intensity of the polarized
light:
Transmitted Intensity:
𝑰 = 𝑰𝟎 𝒄𝒐𝒔𝟐 𝜽
I – transmitted insensity in Wm2
Io – Incident intensity (before passing filters)
𝜃 − 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑙𝑡𝑒𝑟𝑠
Example B: Unpolarized light of intensity I is incident on a polarized filter. The light leaves the filter with
intensity I/2. A second filter is to be added so the light intensity is reduced to I/4. What angle should the filter be
oriented at with the first?
For the second polarizer, to reduce from I/2 to I/4
𝐼 = 𝐼 𝑐𝑜𝑠 𝜃 →
𝜃 = 45°
1
√2
= 𝑐𝑜𝑠 𝜃 → cos 𝜃 =
2
2
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AP Physics 2
6.2 Refraction
Unit 6: Light Waves
Focus Question: What happens to light waves when they enter a new medium?
Index of Refraction
 Light travels at the speed of light in a vacuum since there is no interaction with matter.
 Optical density of material determines the speed of light in a material. Light slows down in transparent
materials since it interacts with the matter in the material.
 The index of fraction determines how much the light slows down:
Index of refraction for material =
→ 𝒏 = 𝒗𝒄
Law of Refraction
 When light enters a new medium, it changes direction.
 Refraction – A wave changes direction due to its change in speed when transferring to a new medium.
*frequency stays constant when a wave switches mediums
o Going from fast medium to slower medium:
Light waves turn towards the normal when traveling from a fast medium to a slower medium.
o Going from slow medium to faster medium:
Light waves turn away from the normal when traveling from a fast medium to a slower medium.

Snell’s Law for Refraction – Snell’s Law relates the index of
refraction between two mediums light is traveling between to angle
the light wave make with the normal:
𝒏𝟏 𝐬𝐢𝐧 𝜽𝟏 = 𝒏𝟐 𝐬𝐢𝐧 𝜽𝟐
*the angles in snell’s law reference the angle with the normal
𝜃 : angle of incidence
𝜃 : angle of refraction
n1: index of refraction in first medium
*n2: index of refraction in new medium
89
Example A: A light wave traveling in air strikes a piece of glass as shown. The frequency of the incident wave
is 5.11014 Hz. The angle of incidence is 30° and the angle of refraction is 20°.
a) Find the index of refraction of the glass.
b) Find the speed of light in the glass.
c) Find the frequency and wavelength of the refracted light wave.
a) The index of refraction in air is 1.0, which is where the wave starts.
𝑛 sin 𝜃 = 𝑛 sin 𝜃 → 𝑛 =
b) 𝑛 =
→𝑣 =
=
=
/
.
°
°
= 1.46
= 2.05𝑥10 𝑚/𝑠
c) The frequency of the light is the same in the glass as it was in air, 𝑓 = 5.1𝑥10
𝐻𝑧
𝑣 = 𝑓 𝜆 (wave equation from Physics 1)
𝜆 =
=
.
/
.
= 4.9𝑥10
𝑚
Example B: A 2.0 person stands at the edge of a pool 10 m wide that is 3 m deep. The person looks at the
center of the water’s surface and sees a coin on the bottom of the pool. What is the actual location of the coin?
The index of refraction of light in water is 1.33.
Find the angle the light wave makes with the water. This is equal to the angle below the man’s eyes as the
angles are alternate interior angles.
5
𝜃 = tan ( ) = 68°
2
Find the angle of incidence using Snell’s Law.
𝑛 sin 𝜃 = 𝑛 sin 𝜃 → 𝑠𝑖𝑛𝜃 =
𝑛 sin 𝜃
→ 𝜃 = sin
𝑛
(1) sin 68
= 44°
1.33
Find d using the tangent function.
𝑑 = 3 tan 44° = 2.93
The coin is 7.93 m from the edge of the pool.
90
Example C: Light travels through a vacuum when it encounters a .1 m
thick transparent material with n =1.5. Calculate how far the light is
displaced.
When the light travels through the material, it will refract. When the light
goes back into the air, it will make the same angle with the normal when its
exiting as it did when it enters. It will still be displaced a distance d due to
the different angle inside the material
𝑛 sin 𝜃 = 𝑛 sin 𝜃 → 𝜃 = sin
d=
(. )
1 sin 30
= 19°
1.5
cos(90 − 30 + 19) =.02 m
Total Internal Reflection
 The incident angle that results in an angle of refraction of 90 o. If the angle of incidence is larger than the
critical angle, no light escapes and all light is reflected. This is called total internal reflection.
Critical angle - The incident angle that results in an angle of
refraction of 90o. If the𝜃 > 𝜃 , no light escapes.
𝑛 sin 𝜃 = 𝑛 sin 90°
→ sin 𝜃 = 𝑛 /𝑛
Example D: Aquaman is swimming underwater at a depth of 2.0 m and looks upward.
a) Explain why Aquaman can only see the world outside the
water through a circle on the surface?
b) Calculate the radius of this circle given that n=1.33 for water.
c) How will the answer to b) change if Aquaman looks up from a
greater depth?
91
a) When looking the surface, the radius drawn from
the surface to Aquaman’s eye makes an angle with
the normal. When that angle exceeds the critical
angle, he will no longer be able to see light above the
surface.
b) 𝑛 sin 𝜃 = 𝑛 sin 90°
→ 𝜃 = sin
𝑛
= sin
𝑛
1
= 49°
1.33
The radius is the hypotenus of a right triangle with a
height of 2.0 (the distance below the surface). Using
trigonometry, the radius is:
𝑟=
2𝑚
=3𝑚
cos 49°
c) At a greater depth, the circle will have a greater radius, leader to a greater visible circle. The actual angle
value will not be affected.
Dispersion



The index of refraction for light varies slightly based on the
frequency of light.
When white light is incident on a surface, light of all frequencies
enters. When the light is refracted, different frequencies of light are
refracted at different angles.
When white light is refracted through a prism, the resulting light is
dispersed, resembling a rainbow pattern.
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AP Physics 2
6.3 Mirrors
Unit 6: Light Waves
Focus Question: How does the shape of a mirror affect the reflected image?
When light passes through a medium to another, light can be reflected, transmitted, or absorbed. More reflected
occurs when there is a greater difference between two mediums. The reflected with have an amplitude less than
the incident wave. The amount percentage of light reflected is known as albedo. Mirrors have high albedo.
Law of Reflection
Angle of incidence = angle of reflection
*only applicable for a flat mirror
Flat Mirror
On a flat mirror, every ray for the object will bounce off the mirror at the angle of incidence. Extending the
reflected eye from the observer through the mirror shows and image that is a distance from the mirror on the
opposite side as the object.
𝑑 :distance of object to mirror
ℎ :size of object
𝑑 :distance of image to mirror
ℎ : size of image
Properties of Plane Mirrors:



Image distance - The image is the same distance from the mirror as the object.
Image orientation – The image is upright and equal in size to the object.
The image in this case is a virtual image since the light ray does not actually pass through the image.
Obviously, the part of the diagram behind the mirror is perceived space and isn’t actually there.
o Virtual Image – Image formed when light only appears to pass through image. Virtual images
are also upright
o Real Image – Light actually goes through image. Real images are always inverted.
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Example A: A man is 2.0 tall and his eyes are a distance of .15 m from the top of his head. How long must a
mirror be for him to see his full image in the mirror?
Since the reflected and incident rays are symmetrical,
the mirror only needs to be half the person height (1
m).
Concave Mirrors
A concave is made from the inner surface from part of a sphere.
o R is the radius of the curvature.
o C is called the center of curvature and is a distance of R from the mirror
along the central axis of the mirror.
o f is the focal length, which is half the radius of curvature (R/2).
*Many mirrors are concave, but only curved slightly compared to their
size..
Types of Reflected Rays on a Concave Mirror

Ray that comes from the object and travels parallel to the axis of the
mirror:
The ray reflects back so that it passes through the focal point.

Ray that goes through the focus:
The ray is reflects back parallel to the central axis of the mirror.
94

Ray that goes through center of curvature:
The ray is reflects back along its original path.
The image is formed at the point where two reflected rays intersect.
Image Construction for Concave Mirrors
The type of image on a concave mirror depends on the location of the image.
 Object located beyond C
The image is located between C and f. The image is real, inverted, and diminished in size compared to the
object.
95

Object located at C
The image is located at C. The image is real, inverted, and the same size as the object.

Object located between C and f.
The image is located beyond C. The image is real, inverted, and magnified.
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
Object located at f.
There is no image formed.

Object located between f and mirror.
The image is virtual, upright, and magnified.
97
Mirror Equations

The mirror equation relates the distance of the object to the mirror (𝑠 ) to the image of the image to the
mirror (𝑠 ):
𝟏
𝟏
𝟏
=
+
𝒇 𝒔𝑶 𝒔𝒊
o s is positive if: the image/object is located in front of the mirror (image is real).
o s is negative if: the image is located behind the mirror (image is virtual)

Magnification, M:
𝑴=
o M is positive if: the image is upright.
𝒉𝒊
𝒔𝒊
=−
𝒉𝟎
𝒔𝒐
o M is negative if: the image is inverted.
Example B: A concave mirror has a focal length of 10 cm. An object is a distance of 25 cm away.
a) Is the image real or virtual?
b) Is the image upright or inverted?
c) Is the image larger or smaller than the object?
d) Repeat a – c for an object distance of 5 cm.
a)
=
+
→
=
+
→ 𝑠 = 16.7 𝑐𝑚
b) The si is positive, the image is real, which then makes the image inverted.
c) The image is smaller than the object because numbers. The magnification is 𝑀 = −
d)
=
+
→
=
+
.
= −.67
→ 𝑠 = −10 𝑐𝑚
The image is virtual since the si is negative, which makes the image upright. The image is smaller.
The magnification is 𝑀 = −
= −.4
Example C: An object is place 7.0 cm in front of a mirror, producing a smaller virtual image 9.5 cm away. Find
the focal length of the mirror
Since the image is virtual, 𝑠 = −9.5 𝑐𝑚
1
1
1
1
1
1
= + → =
+
𝑓 𝑠
𝑠
𝑓 7.0 𝑐𝑚 −9.5 𝑐𝑚
→ 𝑓 = 27 𝑐𝑚
98
Concave Mirror
A concave mirror is when the object is on the other side of curvature of the mirror.
*focal length is negative for convex mirrors.
Properties of Convex Mirrors

Image orientation - The image is upright.

Image size – The image is diminished in size compared to the object.

The image is virtual as it’s located behind the mirror.
Example D: A object 2.0 cm high is placed 10 cm from a convex mirror with a radius of curvature of 8 cm.
a) What is the position of the image?
b) What is the magnification of the image?
The same mirror applies for concave and convex mirrors. The focal length is 4 cm.
a)
=
+
→
=
+
→ 𝑠 = −4 𝑐𝑚, The image is 4 cm on the opposite side of the mirror,
making it a virtual image.
b) 𝑀 = −
=−
=2
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AP Physics 2
6.4 Lenses
Unit 6: Light Waves
Focus Question: How does the path of light change when it passes through a lens?

A lens is a transparent object with two non-parallel curved surfaces.
Converging Lens
 A Converging lens, always called a biconvex lens, has two convex surfaces.
 Ray Tracing for Converging Lens
o Ray 1 – Ray 1 is drawn horizontally. The resulting ray passes through the focal length on the
opposite side.
o Ray 2 – Ray 2 is dawn through the center of the lines: The resulting ray is parallel to the central
axis on the opposite side.
o Ray 3 – Ray 3 is drawn through the focal point on the side of the object. The resulting ray
continues in the same direction.
The location of the image produced from a convex lens depends on the location of the object.

An object located further than 2f1 from the lens:
Properties of image: located between f2 and 2f2, inverted, smaller, real.
100

An object located at 2f1:
Properties of image: located at f2, real, inverted, same size as object.

An object located further than between 2f1 and f1:
Properties of image: The image is beyond 2f2. The image is real, larger, and inverted.
101

An object located at f1:
Properties of image: There is no image as the refracted rays never intersect.

An object closer to the lens than f1:
Properties of image: The image is located between f1 and 2f1. The image is upright, magnitude, and in
front of the lens, making it virtual.
102
Diverging Lens

A diverging lens, always called a biconcave lens, has two concave surface. The properties formed does
not depend on the distance of the object to the lens.
Properties of image: The image is upright, virtual, smaller, and in front of the lens. This is the same for any
biconcave lens regardless of image location.
Lens Equations

The equation for magnification for a lens is the same as the equation for the magnification of a mirror:
𝑴=

𝒉𝒊
𝒔𝒊
=−
𝒉𝟎
𝒔𝒐
The thin-lens equation works for both converging and diverging lens:
𝟏
𝟏
𝟏
=
+
𝒇 𝒔𝑶 𝒔𝒊
*Converging mirrors have positive focal lengths.
*Diverging lenses have negative focal lengths
*Real images have positive distances and negative heights
*Virtual images have negative distances and positive heights
103
Example A: A 4 cm object is placed 15 cm in front a biconvex lens whose focal length is 6.0 cm.
a) Where is the image formed?
b) How large is the image?
a)
=
+
b) 𝑀 = −
→
=
+
→ 𝑠 = 10 𝑐𝑚
= −.625 → −.625 =
→ ℎ = −2.5 𝑐𝑚
Example B: A 10 cm object placed 12 cm in front of a concave lens produces an image 4.0 cm tall.
a) Where is the image produced?
b) What is the focal length of the lens?
a) Since the image is virtual, it’s size is positive. It’s position should be negative.
ℎ
𝑠
10 𝑐𝑚
12 𝑐𝑚
𝑀=
=− →
=−
→ 𝑑 = −4.8 𝑐𝑚
ℎ
𝑠
4 𝑐𝑚
𝑑
b)
=
+
→
=
+
.
→ 𝑓 = 8 𝑐𝑚
Example C: A convex lens with focal length of 20 cm produces an image 2 times the size of the object.
a) Where is the object located if the image is real?
b) Where is the object located if the image is virtual?
a) A real image will have a positive distance.
= +
→ =
→ 𝑠 = 30 𝑐𝑚
b) A virtual image will have a negative distance.
= +
→ =
→ 𝑠 = 10 𝑐𝑚
104
Lens Combinations

When light passes through 2 lens, the image of the first lens acts as the objects for the second lens.
Example D: Two converging lenses are placed 20 cm apart, with an object 30 cm from one of the lens. The
lens closest to the object has a focal length of 10 cm and the other has a focal length of 20 cm. Locate the final
image.


Find image after passing through first lens:
1
1
1
1
1
1
= + →
=
+ → 𝑠 = 15 𝑐𝑚
𝑓 𝑠
𝑠
10 𝑐𝑚 30 𝑐𝑚 𝑠
The image is 15 cm past the focal lens and is inverted with respect to the original image. This is now
treated as the object for the second lens. Since the image is 15 cm from the first lens, it is 5 cm from the
second lens, so so is 5 cm for this case.
1
1
1
1
1
1
= + →
=
+ → 𝑠 = −6.67 𝑐𝑚
𝑓 𝑠
𝑠
20 𝑐𝑚 5 𝑐𝑚 𝑠
Since the result is negative, the image of the first image is virtual and is 6.67 to the left of the second
mirror.
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AP Physics 2
6.5 Diffraction
Unit 6: Light Waves
Focus Question: How does the pattern of light on a screen change as a result of diffraction?
A wave spreads out, or diffracts when it spreads through a small opening, or aperture. When
doing so, the plane wave passes through the slit acts like a point source after passing through
the slit.
*Smaller openings lead to more pronounced diffraction.
Double-Slit Diffraction

Young’s Double-Slit Experiment – Proved that light is
a wave.
Light was shone through two slits that were a distance
𝑑 apart. The resulting light hit a screen a distance of 𝐿
away. The light on the screen has bright and dark
fringes. Where the two light waves from the two slits
interfere constructively, the light on the screen is bright.
When the two waves interfere destructively, there is a
dark fringe.

Double-Slit Experiment – Relates 𝑑: distance between slits to 𝑦: the distance from the center of the
screen to a dark fringe.
- Assume screen is very far from the slit so that rays are parallel when they interfere destructively at the
first minimum.
d – distance between slits
L = distance to screen
Y – distance from center of screen to 1st bright spot
𝜃 −angle from horizontal to point on screen
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The path difference is the phase difference between the two waves coming out of the two slits. Bright spots will
occur on the screen for waves coming from the slits that interfere constructively on the screen.

Constructive interference corresponds to a path difference of 𝜆.
The path difference forms the leg of a right triangle with the distance between the slits as a hypotenuse.
𝑑 sin 𝜃 = 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 → 𝑑 sin 𝜃 = 𝜆
The formula for double slit diffraction assumes that sin 𝜃 ~𝜃 since the angle is assumed to be very small
since the distance to the screen is very large compared to the distance between the slits.
𝝀
𝑑𝜃 = 𝜆
→ 𝜽 = 𝒏 where n is an integer
𝒅
n =1 is the first bright spot, n = 2 is the second bright spot, etc. The formula gives the angle to bright
spots on the screen.

Destructive interference corresponds to a path difference of .
𝑑 sin 𝜃 = 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 → 𝑑 sin 𝜃 =
→𝜽=𝒏

𝝀
𝟐𝒅
Intensity of Diffraction Pattern
Example A: Light with 𝜆 = 590 𝑛𝑚 is shone through two slits 1.5 mm apart. A screen is placed 6.0 m away.
a) How far from the central maximum does the first bright fringe appear?
b) How far from the central maximum is the second dark fringe?
c) How would moving the slits closer together change the diffraction pattern?
a) 𝜃 =
→𝜃=
.
= .0004
107
Right-triangle trigonometry can be used to find the distance to the first fringe:
𝐿 tan 𝜃 = 𝑦
Since theta is small, it can be assumed that tan 𝜃 ~𝜃
→ 𝑦 = 𝐿𝜃 = (6 𝑚)(. 0004) = .0024 𝑚
b) 𝜃 =
→𝜃=
( .
)
= .0002
→ 𝑦 = 𝐿𝜃 = (6 𝑚)(. 0002) = .0012 𝑚
c) The slits moving closer together would increase the angles,
making the fringes further apart.
Example B: Monochromatic light passes through 2 slits that are 1.9𝑥10 m apart. A screen is placed .600 m
away. The first maximum on the screen is found to be 21.1 mm from the central maximum. What color is the
light?
The angle needs to be found using right-triangle trig before finding using the diffraction formula.
𝑦 21.1𝑥10 𝑚
𝑦 = 𝐿𝜃 → 𝜃 = =
= .035
𝐿
. 60 𝑚
𝜆
𝜃 = → 𝜆 = 𝜃𝑑 = (. 035)(1.9𝑥10 ) = 6.65𝑥10 𝑚 = 665 𝑛𝑚
𝑑
This corresponds to red light.
Single-Slit Diffraction
 Huygen’s Principle – A wave can be thought of as consisting of a collection
many point wavelets. A diffraction pattern occurs when light is shone through a
single slit due to individual wavelets interfering with each other.
*Individual wavelets interfering with each causes a diffraction pattern when
light is shone through a single slit.

Single-Slit Diffraction Formula
Wavelets pairs that are have a slit width apart and have a path
difference of 𝜆/2 interfere destructively. The formula for singleslit diffraction formula gives the angle to the first dark fringe:
sin 𝜃 = 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 → sin 𝜃 =
𝑑𝜃 = 𝜆
→𝜽=𝒏
𝝀
𝒅
108
*A single-slit diffraction pattern has wider and less fringes compared to the double-slit pattern:
The blue graph shows the single-slit diffraction pattern intensity. Shown in red is the graph for the equivalent
double-slit pattern. The single slit has a much large bright central fringe and further apart fringes.
Example C: Light passes through a single slit and shines on a flat screen that is located 0.40 m away. The
width of the slit is 4.0×10-6 m. Determine the width of the central bright fringe when the wavelength of the light
in a vacuum is λ = 690 nm (red).
The width of the central bright fringe is twice the distance to the first dark
fringe:
𝜃=
𝜆 690𝑥10 𝑚
=
= 0.18
𝑑
4𝑥10 𝑚
The distance to the first dark fringe: 𝐿𝜃 = 𝑦 → 𝑦 = (0.4 𝑚)(0.18 𝑚) = .072 𝑚
The width of the central bright spot is 2(0.72 m)=0.14 m
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AP Physics 2
6.6 Thin Film Interference
Unit 6: Light Waves
Focus Question: How does light reflect off of thin films?

Phase Shift When Reflecting
o If a wave is reflected in going from a medium with lower refractive index to a higher
reflective index: The wave is reflected and experiences a half-wavelength phase shift.
o If a wave is reflected in going from a medium with higher refractive index to a lower
refractive index: The wave is reflected in phase.

Thin Film Interference – When travels through a film of a transparent medium (such as oil), some light
is reflected and some light is reflected.
o A - Light is incident on the film. Some is reflected out of phase
(phase change of ) and some is refracted and slows down,
bending towards the normal.
o B - The refracted ray reflects in phase at B.
o C - At C, the reflected wave enters the air again, with the same
angle that it entered. The rays at A and C interfere.
The path difference between the waves at A and C is 2d. They
interfere destructively at A and C:
𝑡𝑖𝑚𝑒 𝑡𝑜 𝑔𝑜 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑓𝑖𝑙𝑚 =
= 2𝑑𝑛/𝑐(𝑖𝑛𝑠𝑖𝑑𝑒) = 𝑚𝜆/𝑐(𝑜𝑢𝑡𝑠𝑖𝑑𝑒)
2𝑑𝑛 = 𝜆
Formula for destructive interference: → 𝟐𝒅𝒏 = 𝒎𝝀 where m is an integer.
Since the waves are out of phase, they interfere constructively for a path difference of :
𝑡𝑖𝑚𝑒 =
=
→ 2𝑑𝑛 = 𝜆
𝟏
Formula for constructive interference: → 𝟐𝒅𝒏 = (𝒎 + )𝝀
𝟐
Example A: A film of oil having a refractive index of 1.40 floats on a puddle of rain water having a refractive
index of 1.33. The puddle is illuminated by sunlight. When viewed at near-normal incidence a particular region
of the oil film has an orange color, corresponding to a wavelength of 575 nm. Calculate the minimum possible
thickness of the film in the orange region.
Since the oil appears orange, the interference has to be constructive.
2𝑑𝑛 = 𝑚 + 𝜆, for the minimum thickness, m = 0:
2𝑑𝑛
1
𝜆
= 𝜆→𝑑=
2
2𝑛
=
575𝑥10 𝑚
= 1.03𝑥10
4(1.40)
𝑚
110
Example B: A solar cell must be coated to ensure as little as possible of the light falling on it is reflected. A
solar cell has a very high index of refraction (about 3.50). A coating of index of refraction 1.50 is placed on the
cell. Estimate the minimum thickness needed in order to minimize reflection of light of wavelength 524 nm.
The formulas on the first page are only for one phase change.
Since the light wave goes to a higher refractive index, and
then back to a higher refractive index again, the conditions for
constructive and destructive interference flip. The formulas for
two phases become:
Constructive Interference
2𝑑𝑛 = 𝑚𝜆
Destructive Interference
2𝑑𝑛 = (𝑚 + )𝜆
Since reflection is to minimized, destructive interference is desired. Since the minimum thickness is need, m =
0.
1
𝜆
524𝑥10 𝑚
2𝑑𝑛
= 𝜆→𝑑=
=
= 8.7𝑥10 𝑚
2
2𝑛
4(1.40)
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AP Physics 2
Unit 7: Particle Physics
Section 7.1 – Discrete Energy…………………………………..…………………….…….…113
Section 7.2 – The Photoelectric Effect………………………………………….………….….115
Section 7.3 – Wave-Particle Duality……………………………………………………......…117
Section 7.4 – Atomic Energy Levels……….……………………………………………….…119
Section 7.5 – Nuclear Physics……….………………………………………………..…….…123
Section 7.6 – Nuclear Reactions……………………………………………………..………..128
112
AP Physics 2
7.1 Discrete Energy
Unit 7: Particle Physics
Focus Question: What does it mean for energy to be discrete?
 Continuous Variable -
Continuous
Discrete
A variable that can take on any value in a range.
ex: height, temperature
 Discrete Variable A variable that can only take on certain values that are
an integer multiple of some number. Discrete variables
can not on a value between these multiples.
ex: shoe size, grade level
Ultraviolet Catastrophe



When an object is heated, it emits radiation which consists of electromagnetic waves.
A blackbody absorbs all electromagnetic radiation on it and re-emits the energy.
Radiation emitted from a black body only matches the understanding of light as a wave as high
wavelengths. At low wavelengths, energy radiated should have a high intensity compared to
observation, a problem known as the ultraviolet catastrophe.

The problem of blackbody radiation was solved by Max Planck, who assume energy can be emitted or
absorbed in discrete bundles, known as quanta.
The Planck model works since at all frequencies, all energy is used to cause vibration. At higher
frequencies, it cost a lot more energy to make a “bundle” or quanta of energy.

Quantization
Because energy can only have certainly values, we say that the energy is quantized.

Quantum number - Light is a collection of quanta, or bundles of energy each of energy E = hf.
Any amount of energy in a collection of light must be a multiple of hf.

Oscillators emit or absorb energy when making a transition from one quantum state to another –
𝑬 = 𝒏𝒉𝒇
*n must be an integer
h: Planck’s constant, relates a photon’s frequency to its energy.
h = 6.63x10-34 m2kg/s
A single photon has an energy of hf (E=hf). Hf is a quanta, or bundle of energy. Any energy
associated with that photon has to have a value that is an integer multiple of hf.
113
Example A: A 2.00-kg block is attached to a massless spring that has a force constant of k = 25.0 N/m. The
spring is stretched 0.400 m from its equilibrium position and released from rest.
a) Assuming the energy of the oscillator is quantized, find the quantum number n for the system
oscillating.
b) Suppose the oscillator makes a transition from the state above to the state below it. By how much does
the energy of the oscillator change?
a) The total energy of the system is 𝑈 = 𝑘𝑥 =
(0.4 𝑚) = 2 𝐽
25
*the frequency of a mass spring system is 1/period, where the period is 𝑇 = 2𝜋
(
𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑛ℎ𝑓 → 𝑛 =
=
=
)(
= 5.36𝑥10
.
In situations on a larger scale like this, there are so many quanta, and the spacing between them is so
small, that the amount of energy appears to be a continuous variable. Quantum effects are only
significantly for small particles.
b) Each quantum state if hf apart.
ℎ𝑓 = ℎ
= 6.63𝑥10
= 3𝑥10
𝐽
Compared to the energy of the system, this amount of energy is negligible, which is why the amount of
energy can be treated as continuous even though the possible energies for the spring system is discrete.
Example B: Estimate how many photons of wavelength 500 nm are emitted per second by a 60 W lamp,
assuming that 1% of the energy of the lamp goes into photons of this wavelength.
The energy of a single photon of this wavelength is 𝐸 = ℎ𝑓 =
=
( .
)(
)
= 4𝑥10
𝐽
In one second, 60 J is emitted (60 W=60 J/s). 1% goes into photons of the specified wavelength, which is
. 01𝑥60 = .06 𝐽.
The number of photons is
.
= 1.5𝑥10
𝑝ℎ𝑜𝑡𝑜𝑛𝑠
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114
AP Physics 2
7.2 The Photoelectric Effect
Unit 7: Particle Physics
Focus Question: What is the photoelectric effect?

The photoelectric effect – Light incident on a metallic surface causes electrons to be emitted from the
surface. *electrons are “knocked out” of the metal due to the light colliding.
*only certain metals are photo-electric
*The photoelectric effect can be explained by light behaving as a particle

Kinetic Energy of photo-electron
Work function – minimum amount of energy needed to free an electron
from a metal surface: ∅ = ℎ𝑓
f0 = a photon having at least this frequency is needed to free an electron.
If an electron was freed by an incoming photo having energy of E = hf,
and it had more energy than the work function, then the electric would
have maximum kinetic energy given by:
 Stopping Potential - Voltage required to stop emitted electrons.
𝑽𝒔 =
𝑾
𝒒
Example A: A photosensitive metal has a work function of 3.5 eV. The metal is illuminated with light of
wavelength 0.20 𝜇𝑚.
a) Calculate the energy of the incident photons.
b) Calculate the maximum speed of the ejected electrons.
c) Calculate the maximum wavelength of the incident light to eject photo-electrons.
a) 𝐸 = ℎ𝑓 =
=
b) 1 eV = 1.6𝑥10
( .
)(
)
.
= 9.9𝑥10
𝐽
𝐽
𝐾
= ℎ𝑓 − 𝜙 = 9.9𝑥10
𝐾
= 𝑚𝑣
→𝑣
− (3.5 𝑒𝑉) 1.6𝑥10
=
( .
=
= 4.3𝑥10
)
.
𝐽
= 9.7𝑥10 𝑚/𝑠
c) At minimum, the ejected electrons will have zero kinetic energy: ℎ𝑓 − 𝜙 = 0 → 𝑓 =
threshold frequency: 𝑓 =
=
( .
)
.
= 8.5𝑥10
.
wavelength at threshold frequency: 𝜆 =
=
/
.
𝐻𝑧
= 3.5𝑥10
𝑚, 350 nm
115

Photoelectric Effect Graph
Example B: The graph below shows the stopping
potential of emitted photoelectrons vs. the
frequency of light incident on a metallic surface.
a) Calculate the work function of the metal.
b) If the stopping potential is 1.5 V,
determine the maximum kinetic energy of
the emitted photoelectrons.
c) Calculate the wavelength of light that
will eject photoelectrons with this
maximum kinetic energy.
a) The work function is equal to the
threshold frequency times plank constant: 𝐾 = 0 = 𝑓 ℎ − 𝜙 → 𝜙 = ℎ𝑓
(4.5𝑥10
𝜙 = 6.63𝑥10
𝐻𝑧) = 2.9𝑥10
In electron volts, this is equal to (2.9𝑥10
𝐽)𝑥
𝐽
.
= 1.8 𝑒𝑉
b) When the stopping potential is 1.5 V, the frequency is 8x10 14 Hz.
𝐾 = ℎ𝑓 − 𝜙 = 6.63𝑥10
c) 𝜆 =
=
/
= 3.75𝑥10
(8𝑥10
𝑚
𝐻𝑧) − 2.9𝑥10
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𝐽
375 nm
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AP Physics 2
7.3 Wave-Particle Duality
Unit 7: Particle Physics
Focus Question: How does pressure change with depth?
Light has characteristics of both a wave and a particle.
 Characteristics of light that behave like a wave:
Light creates an interference pattern when shone on a double slit.

Characteristics of light that behave like a particle:
Electrons are emitted from metal when light shines on it (photoelectric effect).
Photons can participate in collisions (Compton effect)
Matter Waves


Compton shot an X-Ray photon at a graphite target to observe the
collision between the photon and a graphite atom’s electron.
Compton Effect - Photons follow the laws of conservation of
momentum and conservation of energy despite being massless.
When a photon was shot at an atom by Compton, the electron was
scatter and the photon after the collision had a lower frequency.

de Broglie Hypothesis – If waves exert properties of particles, than particles should exhibit
wave-like properties. “Matter waves” have wave functions that determine their position. The
wavelength of a matter wave is called the de Broglie wavelength and is given by:
𝝀=
h – Planck’s constant
𝒉
𝒑
p – momentum of the particle
Example A: An electron is accelerated from rest through a potential difference of 100 V. What is its expected
de Broglie wavelength?
Find speed with work-energy theorem.
𝑉𝑞 =
𝜆=
1
1
𝑚𝑣 − 𝑚𝑣 → 𝑣 =
2
2
2𝑉𝑞
=
𝑚
2(100 𝑉)(1.6𝑥10
𝐶)
= 5.9𝑥10 𝑚/𝑠
9.11𝑥10
𝑘𝑔
6.63𝑥10
ℎ
ℎ
→𝜆=
=
𝑝
𝑚𝑣 (9.11𝑥10
𝑘𝑔)(5.9𝑥10
)
= 1.2𝑥10
𝑚
Example B: The kinetic energy of an electron is halved. What happens to its de Broglie wavelength?
Kinetic energy halving means that the velocity was reduced by a factor of √2, which will cause momentum to
decrease by a factor of √2. As a result, the deBroglie wavelength will increase by a factor of √2.
117

Wave Function – Any particle has a wave function that represents its
position. The particle is most likely to be found at the peaks of the
wave function.
Heisenberg Uncertainty Principle

Heisenberg Uncertainty Principle – It is not possible to precisely the position and momentum of a
particle:
ℎ
∆𝑥∆𝑝 ≥
4𝜋
∆𝑥 − uncertainty in position
∆𝑝 − uncertainty in momentum
The more accurately the position is known, the less accurately the momentum is known and vice versa.
* The Heisenberg uncertainty principle does not come from experimental error, it is a fundamental property of
nature:
Example C: Suppose an electron is stationary and we want to observe its position. In order to observe it, the
electron is lit up with a single photon. What happens to the electron?
In order to measure the position of a particle, light must be incident on the particle, which means that photons
will collide with the particle. The collisions of photons with the particle will impart momentum to it, causing the
electron to move as result of it being observed.
Example D: An electron and a small aircraft are observed to have equal speeds of 500 m/s, accurate to within
0.0200%. What is the minimum uncertainty in the position of each if the mass of the jet is eight metric tons?
Electron: ∆𝑥∆𝑝 ≥
∆𝑥 =
→ minimum ∆𝑥 =
∆
6.63𝑥10
4𝜋(9.11𝑥10
Jet: ∆𝑥∆𝑝 ≥
𝑘𝑔)(.0002𝑥500 )
→ minimum ∆𝑥 =
=
∆
= 5.8𝑥10
𝑚
.
(
)(.
)
= 6.6𝑥10
𝑚
The uncertainty for the electron’s position is very significant compared to the size an atom. The jet’s
uncertainty in position is negligible compared to its size.
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AP Physics 2
7.4 Atomic Energy Levels
Unit 7: Particle Physics
Focus Question: What happens when an atom absorbs or emits a photon?


Electrons exists in an atom at different energy-levels. Lower energy electrons are closer to the nucleus
(less potential energy).
Light emitted from a black body emits light of all wavelengths. Passing the light through a prism to
disperse it into different wavelength would yield a continuous spectrum.
Emission Spectrum


When a container of gas is exposed to an electric field, light is emitted in the form of a photon. The light
can be analyzed by passing it through a diffraction grating.
Emission Spectrum – Bright bands seen when looking at a light source through diffraction grating.
Created when electrons jump to lower energy levels, show light emitted form element.
*no two elements have the same spectra. Emission spectra can be used to identify elements.
Absortion Spectrum


By sending a beam of white light through a gas, photons may transfer to the gas, causing the atoms to
move to a more excited state and the transmitted light will be missing photons.
Absorption Spectrum - Dark bands in continuous spectrum. Created when electrons jump to higher
energy levels. Shows light absorbed by element; inversion of emission spectrum.
Bohr’s Hypothesis

For a given element, electrons can only exist at certain energy levels. These energy levels vary element
to element.
119

Within an atom, an electron can jump to a higher orbit
(transitions to a higher energy level) if: the atom absorbs
a photon (the energy of the photon increases the potential
energy of a photon).

An atom falls to a lower orbit (transitions to a lower
energy level) if: the atom emits a photon (the potential
energy lost by the photon is released as a photon)

The photon frequency associated with a transition form
an initial state to a final energy state is given by:
∆𝑬 = 𝑬𝒇 − 𝑬𝒊 = 𝒉𝒇
Energy Level Diagrams - Permissible energy levels for an atom are
represented using horizontal lines.
- n : principal quantum number
Individual elements have defined permissible levels.
- ground state:
Lowest level for an electron, n = 1. The ground state is where the
electron has the lowest energy (since it is most electrostatically
bound to the nucleus).
- ionized atom:
Electron is so far from the nucleus that U=0
*since other states of the atom are less energetic than an ionized atom (U=0), energy levels for an
unionized atom have negative values.
Example A: An electron jumps from energy level n = 3 to energy n = 2 in the hydrogen atom.
a) Find the atom’s change in energy.
b) Find the energy of the emitted photon.
c) Find the frequency and wavelength of the emitted photon.
a) From the given diagram, n=3 corresponds to -1.51 eV in hydrogen. n=2 is at -3.40 eV. When the
electron goes from 3 to 2, it loses 1.89 eV.
b) Since the electron lost 1.89 eV, the emitted photon has 1.89 eV of energy.
120
c) 1.89 eV ~(1.89 𝑒𝑉) 1.6𝑥10
𝐸 = ℎ𝑓 → 𝑓 =
𝜆=
=
=
/
.
.
= 3.0𝑥10
= 4.5𝑥10
.
= 6.67𝑥10
𝐽
𝐻𝑧
667 nm
𝑚
Example B: In the fictional land known as Canada, the aurora borealis provides a colorful display in the night
sky. What is the origin of the various colors seen in the auroras?
High speed particles hit molecules in the atmosphere and excite the moles, causing them to emit photons
characteristic of their emission spectra.
Example C: An atom’s ground state is at -12.0 eV. The absorption spectrum for the atom has two dark lines,
one at 207 nm and another at 146 nm. Each dark line corresponds to a transition from ground state to excited
state.
a) Construct an energy level diagram showing the excited states.
b) What wavelength photon would be emitted in a transition from the higher excited state to the lower
excited state?
c) Suppose a photon of wavelength of 70 nm interacts with the hypothetical atom, ionizing it. If the atom is
initially in the ground state, what is the speed of this electron?
a) In order to construct the diagram, the energies of the two exited states need to be found.
𝐸 = ℎ𝑓 = ℎ( )
𝐸 = (6.63𝑥10
)(
)(
)(
)(
.
)
𝐸 = −8.5 𝑒𝑉
𝐸 = (6.63𝑥10
.
)
𝐸 = −6.0 𝑒𝑉
The energies are negative since the electrons are electrostatically bound to the nucleus.
121
b) When an electron goes from n=3 to n=2, a photon with 1.5 eV is emitted.
𝐸 = ℎ𝑓 → 𝑓 =
→𝜆= =
=
(
)( .
( .
)( .
)
)
= 8.28𝑥10
𝑚
828 nm
c) It takes 12 eV volts of energy to ionize the atom. This is equivalent to:
= (12 𝑒𝑉) 1.6𝑥19
𝐸
= 1.92𝑥10
𝐽
The energy of the photon is:
𝐸
= ℎ𝑓 = ℎ
= 6.63𝑥10
= 2.84𝑥10
𝐽
Some of the energy of the photon goes to ionizing the electron. The rest of the energy goes to kinetic
energy of the election.
𝐸
=𝐸
−𝐸
= 2.84𝑥10
𝐽 − 1.92𝑥10
𝐽 = 9.2𝑥10
𝐽
𝐾 = 𝑚𝑣 → 𝑣 =
→𝑣=
( .
.
)
= 1.4𝑥10 𝑚/𝑠
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AP Physics 2
7.5 Nuclear Physics
Unit 7: Particle Physics
Focus Question: How are mass defect and binding energy related?

General Symbol of the nuclide

The atomic mass unit – Defined as 1/12 the mass of an atom of carbon (which as 12 nucleons).
1 𝑢 = 1.66𝑥10
𝑘𝑔
The mass of a proton or neutron are equal to be about one atomic mass unit.
 Mass of proton = 1.007276u
 Mass of neutron = 1.008665u
 Mass of electron = 0.0005486u
Solving problems involving mass defect and binding energy.


Energy is released when nucleons (protons and neutrons) are brought together to form a nucleus.
Due to general relativity, mass is lost in order to supply this energy.
Mass-Energy Equivalence: Energy is equivalent to mass according to:
𝑬 = 𝒎𝒄𝟐
Energy is released when nucleons are brought together in a nucleus, causing a loss of mass. The mass of
an atomic nucleus is less than the mass of its constituent nucleons.
Example A: The mass of an atom of Helium is 4.0026u. Find the mass of the Helium nucleus and compare it to
what you would expect it to be based on the number of protons and neutrons.
The mass of the nucleus is the atomic mass minus the mass of the two electrons:
𝑀
= 4.0026𝑢 − (2𝑥. 0005486𝑢) = 4.0015𝑢
The nucleus of the helium atom has 2 protons and 2 neutrons. The mass of these constitute nucleons is:
2𝑚
+ 2𝑚
= 2(1.007267𝑢) + 2(1.008665𝑢) = 4.0319𝑢
The mass of the actual nucleus is 0.0304u lower than what would be expected based on the constitute nucleons.
This is because energy was lost to energy to bring the nucleons together. The .0304u difference is known as the
mass defect of the nucleus.
123

mass defect – The mass of a nucleus is generally less than the sum of the masses of the nucleons.
∆𝑚 = (mass of protons + neutrons) – (mass of nucleus)
∆𝑚 = 𝑚

−𝑚
binding energy - Work required to completely separate a nucleus.
𝐸 = ∆𝑚𝑐
*The energy equivalent of 1 atomic mass unit is:
𝐸 = 𝑚𝑐 = (1.66𝑥10

𝑘𝑔) 3𝑥10
= 1.5𝑥10
𝐽. This is equal to about 932 MeV.
binding energy per nucleon – Work required to separate a nucleus per nucleon. Elements with higher
binding energy per nucleon are much more stable. Elements with less binding energy per nucleon are
more unstable and more likely to decay.
Example B: Find the binding energy per nucleon of carbon-12.
The mass of carbon 12 is exactly 12u (the atomic unit is defined based on carbon 12). The nuclear mass is this
atomic minus the mass of the electrons. Carbon has 6 electrons:
Nuclear mass= 12𝑢 − (6𝑥. 0005486𝑢) = 11.99671𝑢
Carbon 12 has 6 protons and 6 neutrons. The mass of 6 protons and 6 neutrons is:
6𝑚
+ 6𝑚
= 6(1.007267𝑢) + 6(1.008665𝑢) = 12.095559𝑢
The mass deficit is: 12.095559𝑢 − 11.99671𝑢 = .09894𝑢
The total binding energy is: E=. 09894𝑢(932 𝑀𝑒𝑉) = 92 𝑀𝑒𝑉
There are 12 nucleons, so to find the binding energy per nucleon, divide this number by 12:
Binding energy per nucleon:
= 7.7 𝑀𝑒𝑉
Example C: The binding energy per nucleon for
separate the nucleons of
11
𝐵 is 7MeV. What is the minimum energy needed to
5
11
𝐵?
5
There are 11 nucleons, so to find the total energy, multiple the binding energy per nucleon by 12:
𝐸= 7
𝑀𝑒𝑉
(12 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠) = 77 𝑀𝑒𝑉
𝑛𝑢𝑐𝑙𝑒𝑜𝑛
124

Binding energy per nucleon curve
o Most stable element – Iron
o Nuclear fusion –Nuclear fusion is the combining of two small nuclei into one larger nucleus.
Nuclear fusion will occur with the elements below Fe; joining smaller nuclei forms bigger ones
(more stable).
o Nuclear fission – Nuclear Fission is the splitting of a large nucleus into two smaller nuclei.
Nuclear fission will occur with the elements above Fe; splitting bigger nuclei forms smaller
ones(and thus more stable).

To determine if energy is lost in a reaction, calculate the mass difference.
∆𝑚 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Example C: Radium-226 (226Ra) undergoes natural radioactive decay to disintegrate spontaneously with the
emission of an alpha particle to form radon (Rn). Calculate the energy in Joules that is released.
Mass of radium = 226.0254u
Mass of radon = 222.0176u
Mass of 𝛼 particle = 4.0026u
Total mass of products: 222.0176𝑢 + 4.0026𝑢 = 226.0202𝑢
Mass of Reactant – Mass of Products = 226.0254𝑢 − 226.0202𝑢 = .0052𝑢
The energy released in Joules is:
𝐸 = (. 0052𝑢) 1.66𝑥10
𝑘𝑔
(3𝑥10 ) = 7.8𝑥10
𝑢
𝐽
125
Example D: Consider the following reaction:
𝑁 + 𝐻𝑒 →
𝑂+ 𝑝
Measurement of rest masses shows that:
rest of mass of ( 𝑂 + 𝑝) > rest mass of (
In what condition can this reaction take place?
𝑁 + 𝐻𝑒)
Mass is gained in the reaction. This can be done if the reactants has sufficient kinetic energy that were
converted to mass in the reaction.
Estimating nuclear radius

Suppose an alpha particle is fired at a nucleus. The alpha particle begins far away that there is electric
potential energy between them. As the alpha particle approaches the nucleus, the nucleus will repel the
alpha particle. At closes approach, the alpha particle will momentarily stop before switching direction:
The particle starts far away so it has negligible potential
energy. When it gets close enough the atom, it’s kinetic
energy goes to zero and it has maximum potential energy.
The distance at this point is assumed to be the nuclear
diameter (giving a very rough estimate).
𝑈 +𝐾 = 𝑈 +𝐾
𝑚𝑣 =
→ 𝑚𝑣 =
2q is the charge of the alpha particle. Zq is the charge of
the atom, where Z is the number of protons.

Nuclear diameter can also be determined by measuring the diffraction of a beam of high-energy
electrons or neutrons having a de Broglie wavelength of . The nuclear barrier acts like a single-slit
having a width D.
𝐬𝐢𝐧 𝜽 =
𝝀
𝑫
Θ – angle to first minimum in
diffraction pattern
λ- de Broglie wavelength of
electron
D – nuclear diameter
126
Example D: A beam of 80.0 MeV neutrons are diffracted upon passing through a thin lead foil. The first
minimum in the diffraction pattern is measured at 12.6. Estimate the radius of the lead nucleus.
First, the momentum of the neutron needs to be found to find the de Broglie wavelength:
1
𝐾 = 𝑚𝑣 → 𝑝 = 𝑚𝑣 → 𝑝 = √2𝑚𝐾
2
𝜆=
ℎ
ℎ
=
=
𝑝 √2𝑚𝐾
6.63𝑥10
2(1.67𝑥10
sin 𝜃 =
𝜆
𝜆
→𝐷=
𝐷
sin 𝜃
→𝐷=
2.3𝑥10
𝑚
= 1.1𝑥10
sin 12.6°
= 2.3𝑥10
𝑚
𝑘𝑔)2(80𝑥10 𝑒𝑉) 1.6𝑥10
𝑚
Example E: Why can electrons be used to measure atomic diameter using diffraction?
Electrons can pass through the nucleus since they do not interact with the strong nuclear force that bounds
protons and neutrons together.
Neutrons are also used since they do not interact with the electromagnetic force.
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AP Physics 2
7.6 Radioactive Decay
Unit 7: Particle Physics
Focus Question: How do unstable nuclei decay?
Most nuclides are unstable, and thus decay in various ways by emitting particles and energy in a process known
as radioactivity.

Alpha Decay – In alpha decay, an atomic nucleus emits an alpha particle. An alpha particle consists of 2
protons and 2 neutrons (the nucleus of a helium atom).
Example:
241
237
4
𝐴𝑚 →
𝑁𝑝 + 𝐻𝑒
95
93
2
*An alpha particle can be blocked easily. A piece of paper can stop an alpha particle.

Neutrinos - In contrast to the alpha particle, it was discovered that beta particles could have a large
variety of kinetic energies. In order to conserve energy it was postulated that another particle called a
neutrino  was created to carry the additional EK needed to balance the energy.

Beta Minus Decay – In - decay, a neutron becomes a proton and an electron is emitted from the
nucleus. To conserve energy, an anti-neutrino (𝑣̅ ) is also emitted
Example: 14C  14N + 𝑣̅ + e-

Beta Plus Decay – In + decay, a proton becomes a neutron and a positron is emitted from the nucleus.
To conserve energy, a neutrino (v) is also emitted.
Example: 10C  10B +  + e+
128
*Both positrons and anti-neutrinos are examples of anti-matter. Anti-matter has the same properties of
matter, but each corresponding particle has the opposite sign.
*A beta particle is an electron or positron. A thin sheet of lead can stop a beta particle.

Gamma Decay – Gamma decay occurs when an atom emits high-energy photons due to electrons
dropping to lower energy levels.
*A gamma particle is a photon. It takes a thick layer of lead to block a gamma particle. Gamma
particles are very high energy compared to beta and alpha particles.
Example A: Consider the following nuclear decays.
14
14
𝐶 to 𝑁?
6
7
14
b) What nucleus decays by beta minus emission to 𝐶?
6
4
14
1
c) Solve for the nucleus X in the following equation: 𝐻𝑒 + 𝑁 → 𝑋 + 𝐻.
2
7
1
a) What particle is emitted in the decay of
a) A proton became a neutron, making this beta-plus decay. A positron and neutrino are emitted.
b) In beta-minus decay, a neutron becomes a proton, so the element would be boron, which has 5
14
protons pre-decay ( 𝐵).
5
c) The number of protons and neutrons need to balance, so the missing product must have 17 total
17
nucleons and 8 protons.
𝑂
8
210
𝑃𝑜) decays into a nucleus of lead by emitting an alpha particle. The
84
rest mass of the alpha particle 4.0026u and the kinetic energy of the particle is 5.5 MeV.
Example B: A nucleus of polonium (
a) Find the atomic mass and atomic number for the resulting lead nucleus.
b) Determine the mass difference between the polonium and the lead nuclei. Assume each travel with
low speed.
c) What is the deBroglie wavelength of the alpha particle?
a)
210
4
𝐴
𝑃𝑜 → 𝑃𝑏 + 𝐻𝑒
84
𝑍
2
Solve the reaction yields A=206 and Z=82.
129
b) 𝑚
−𝑚
=𝑚
+ 𝑚𝑎𝑠𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑎𝑙𝑝ℎ𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑠 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦
The mass equivalence of 5.5 MeV is
𝑚
−𝑚
c) 𝜆 =
.
.
= (4.0026𝑢) 1.66𝑥10
→ 𝑝 = √2𝐾𝑚 → 𝜆 =
+ 9.8𝑥10
( .
𝑘𝑔 = 6.6𝑥10
= 6.1𝑥10
)( .
𝑘𝑔
𝑘𝑔
√
.
𝜆=
= 9.8𝑥10
)(( .
)
𝑚
.
Half-Life


Law of Radioactive Decay – Radioactive decay is random and spontaneous.
-Random: It cannot be predicted which unstable nuclear in a sample will decay.
-Spontaneous: Decay cannot be effected in any way.
The law of radioactive decay states that the rate of decay is proportional to the number of nuclear that
have not yet decay. The higher the initial population of a radioactive material, the more decays there will
be in a time interval.
Half-life: time for half of a sample to decay.
Example C: Radioactive tritium has a half-life of 12.3 years. A sample of tritium contains 200 g.
a) What mass remains after 36.9 years?
b) How long will it be until only 12.5 g of tritium remain?
a) This is equal to 3 half-lives.
Amount remaining = (200 𝑔)𝑥
= 25 𝑔
b) Calculate the number of half-lives: 12.5 𝑔 = (200 𝑔)𝑥
This is equal to 4 half-lives, which is 49.2 years
→𝑛=4
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Half-Life 3 confirmed.
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