AP Physics C
Electricity & Magnetism
Prepared by:
Jeff Laufer
AP Physics C
Electricity & Magnetism
Table of Contents
Unit 7: Electrostatics…………….……..………..3
Unit 8: Electric Potential……………….……….21
Unit 9: Circuits………………………...………..35
Unit 10: Magnetism………………………......…48
Unit 11: Electromagnetism……………….……..67
2
AP Physics C
Electricity & Magnetism
Unit 7: Electrostatics
Section 7.1 – Electric Charge……………………………………………………….……………4
Section 7.2 – Charge & Electric Force…………………………………………….……..………7
Section 7.3 – Electric Field……………………………………………………………….…….10
Section 7.4 – Introduction to Gauss’s Law……….………………………………………….…14
Section 7.5 – Applications of Gauss’s Law…………………………………………………….17
3
AP Physics C: Electricity & Magnetism
7.1 Electric Charge (Chapter 23)
Unit 7 - Electrostatics
Focus Question: What is charge?
Electric Charge
▪
▪
▪
▪
▪
Charge is a fundamental property of objects, like mass.
Units of charge: Coulombs (C).
There are two types of charge, positive charge (due to protons) and electrons (caused by electrons).
Protons and electrons have equal but opposite charges. The magnitude of the charge of a proton or
electron is the fundamental charge is equal to 1.60x10−19 bC. Any net charge on an object is due to an
unequal amount of protons/electrons and the net charge is a multiple of the fundamental charge.
Classically demonstrating that there are two types of charge:
➢ When rubbing a glass rod with silk, electrons are transferred from the glass rod to the silk. The glass
rod loses electrons and becomes positively charged.
➢ When rubbing a rubber rod with fur, the rod gains electrons and becomes negatively charge.
*Charge is always transferred by the flow of electrons. A Positive charge means an electron deficit.
Atomic Structure
▪
▪
▪
The atomic nucleus contains protons and neutrons. Protons have positive charge
while neutrons have no charge. Almost all the mass of an atom is in the nucleus.
Protons and neutrons are held together by the strong force.
Electrons exist far outside the nucleus of an item. They have little mass, but have a
negative charge, equal to the positive charge of protons. Electrons are scattered out over region of space
outside the nucleus (an orbital). You can’t precisely state where an electron is at a given moment, but
can state where it is likely to be found (electron locations are given by probability).
Conductors vs. Insulators
▪
▪
Conductors – Materials with free electrons that can move easily. As a result,
charge is easily transferred due to the flow of electrons.
o Most metals are good conductors
o The atomic structure of metals causes electrons to move as far away
from each other as possible. Free electrons in metal are what causes conductivity from metal.
Insulators - Material in which electrons are tightly bound to nucleus and thus charge
is not transferred.
o Glass, rubber, and plastic are examples of non-conductors (insulators)
o Electrons in insulators are tightly bound to the nucleus, so there are no free
electrons to transfer.
4
The Electroscope
▪
The electroscope – Electroscopes are used to detect charge. An electroscope consists
of two “leaves” made of foil. When the electroscope is charged, the leaves repel each
other and diverge. When the charge has more magnitude, the leaves divege more and
more.
Charging by Conduction
▪
▪
Both insulators and conductors can be charged by direct contact in a process known as conduction.
When a charged rod touches an uncharged electroscope charge from the rod is transferred to
electroscope.
A negatively charged rod will transfer electrons to the electroscope. A positively charged rod will cause
electrons form the electroscope to transfer to the rod.
▪
*when the rod is remvoed, the electroscope retains a charge that is the same sign as the rod.
Charging by Induction
▪
▪
In induction, conductors are charged without coming into direct contact with another charged object
To charge an electroscope by induction:
a) Bring a charged object close to the knob of the electroscope, but do not touch it.
*When a conductor is close to a charged object, free electrons on the conductor will be attracted to the
charged object.
b) Ground the electroscope by touching it.
*In electricity, “ground” means to make an electrical connection between the earth and the object being
“grounded”. When an object is grounded, free electrons between the object and earth. Because earth is
large, it can make an object charge neutral by either absorbing or supplying an essentially limitless
amount of electrons.
c) Break the connection with ground.
*with the grond connection broken, there will be no way for electrons to be transferred.
d) Remove the charged object.
*the charge on the conductor will redistribute itself evenly on the object as soon as the charged
rod is removed leaving it with a charge opposite the rod.
5
Example A: Three metal spheres rest on insulating stands as shown in the diagram at
right. Sphere Y is smaller than Spheres X and Z, which are the same size. Spheres Y
and Z are initially neutral, and Sphere X is charged positively. Spheres Y and Z are
placed in contact near Sphere X, though Sphere X does not touch the other spheres.
a) Does sphere Y have a positive, negative, or neutral charge?
b) Does sphere Z have a positive, negative, or neutral charge?
c) Sphere X is moved away Spheres Y and Z are separated. Compare the magnitude of the charge on Spheres Y
and Z.
a) Y has a negative charge. X has a positive charge, so negative charge is attracted
towards it. Therefore, electrons will flow from Z to Y.
b) Z has a positive charge since it lost electrons to Y.
c) Equal and opposite charge.
Example B: A student investigates how a negatively charged ebonite rod influences an
uncharged electroscope as shown in the diagram at right.
a) Briefly describe how the ebonite rod can be given a negative charge.
b) Write a procedure of how the negatively charged ebonite rod could produce a positive
charge on the electroscope. Other materials can be used, but none can be charged, and none
can alter the charge of the ebonite rod.
c) Assume the student charges the electroscope. Describe how the negatively charged ebonite rod could be used
to test whether the electroscope is positively charged. Also explain what the student should expect to observe
and why
a) The ebonite rod can be rubbed with a fur cloth. Compared to fur, ebonite is a strong insulator, so electrons
from the fur will transfer to the ebonite, giving it excess electrons and a subsequent negative charge.
b) Bring the ebonite rod close to, but not touching, the electroscope knob. Ground the electroscope. Break the
ground connection. Move the rod away.
c) Touch the rod to the scope. If the leaves collapse, they were positively charge. If they further diverge, they
were negatively charge.
Rate your understanding: Electrostatics
0
1
Physics and I must
have the same charge.
I can apply Coulomb’s
Law with help.
2
3
I can apply Coulomb’s
Law with minor errors.
I can apply Coulomb’s
Law with no errors.
4
I am charged up about
physics.
6
AP Physics C: Electricity & Magnetism
7.2 Charge & Electric Force (Chapter 23)
Unit 7 - Electrostatics
Focus Question: How is the electrostatic force similar to gravity?
▪
▪
If charge is evenly distributed on an object, it can thought of as being concentrated at its center, much
like the force of gravity.
The net charge on an object is determined by the difference between the number of electrons and the
number of protons the object contains.
Calculate electrostatic force.
Coulomb’s Law
▪
▪
Charges attract/repel each other by the electrostatic force.
Electrostatic force is a vector quantity; the direction of the force depends on the sign (+/-) of the charges
exerting force on each other. The direction of the force is along the line connecting the charges.
𝒌|𝑸𝟏 𝑸𝟐 |
⃑|=
|𝑭
𝒓𝟐
𝒌 = 𝟖. 𝟗𝟗𝒙𝟏𝟎𝟗
▪
𝑵𝒎𝟐
𝑪𝟐
𝟎
𝜖0 − permittivity of free space, measure of resistance to formation of electric field in a vacuum
𝜖0 = 8.85𝑥10−12
▪
𝟏
𝒌 = 𝟒𝝅𝝐
𝐶2
𝑁𝑚2
Superposition Principle: The net electrostatic force on a charge is the sum of all the forces from all the
individual charges acting on it.
Example A: Example A: Positive point charges of equal magnitude Q are located at (0,𝑎) and (0,−𝑎).
a) Find the total force the two charges exert on a charge -Q at (a,0).
b) Find the total force the two charges exert on a charge -Q at (0,2a).
c) Find the total force the two charges exert on a charge -Q at (0, a/2).
a) The y-component of the force cancels due to symmetry so only the xcomponent needs to be calculated. Both forces point in the negative xdirection and are equal in magnitude:
𝑘|𝑄||−𝑄|
𝑘𝑄 2
𝑎
𝐹 = −2
cos
𝜃
=
−2
2
2
2
𝑟2
(√𝑎2 + 𝑎2 ) √𝑎 + 𝑎
→𝐹=−
2𝑘𝑄 2 √2
𝒌𝑸𝟐 √𝟐
→
𝑭
=
−
2𝑎2 2
𝟐𝒂𝟐
b) Both forces act downwards in the y-direction. The negative charge is
attracted to the positive charges, both of which are below it.
7
𝑘|𝑄||−𝑄|
𝑘|𝑄||−𝑄|
𝑘𝑄 2
𝑘𝑄 2
𝑘𝑄 2 𝐾𝑄 2
𝐹=−
−−
2 = − 𝑎 2 − (3𝑎)2 = − 𝑎 2 − 9𝑎 2
(2𝑎 − 𝑎)2
(2𝑎 − (−𝑎))
𝟏𝟎𝒌𝑸𝟐
→𝑭=−
𝟗𝒂𝟐
c) Both forces are in the y-direction. The force due to the charge at (0,a) is positive and the charge due to (0,-a)
is negative.
𝐹=
𝑘|𝑄|| − 𝑄|
𝑎 2
(2𝑎 − 2)
→𝑭=
−
𝑘|𝑄||−𝑄|
𝑎 2
(−2𝑎 − 2)
=
𝑘𝑄 2
3
(2 𝑎)
2−
𝑘𝑄 2
5
(− 2 𝑎)
2
=
𝑘𝑄 2
𝑘𝑄 2
−
9 2
25 2
𝑎
𝑎
4
4
𝟔𝟒𝒌𝑸𝟐
𝟐𝟐𝟓𝒂𝟐
Example B: Two small pith balls of mass .5 kg each are suspended by strings of length L = .3 m from a
common point. They are both negatively charged and repel each other, remaining d = 0.1 m apart at
equilibrium. Find the excess number of electrons on each pith ball.
Apply F=ma for equilibrium of one pith ball:
∑ 𝐹𝑦 = 0 = 𝑇 cos 𝜃 − 𝑚𝑔 → 𝑇 =
𝑚𝑔
cos 𝜃
𝑚
→𝑇=
(0.5 𝑘𝑔)(10 𝑠2 )
√(.3 𝑚)2 −(.1𝑚)
2
= 5.08 𝑁
2
.3 𝑚
.1
∑ 𝐹𝑥 = 0 = 𝑇 sin 𝜃 − 𝐹𝐸 → 𝐹𝐸 = (5.08 𝑁)
𝐹𝐸 =
2
𝑚
.3 𝑚
= 0.85 𝑁
𝑘𝑄 2
𝐹 𝑟2
√ 𝐸 = 9.71𝑥10−7 𝐶
→
𝑄
=
𝑟2
𝑘
Charge is due to an integer number of excess electrons. To find this amount of electrons, divide the total charge
by the fundamental charge, which the charge of a single electron (1.6𝑥10−19 𝐶).
𝑛=
9.71𝑥10−7 𝐶
= 𝟔. 𝟏𝒙𝟏𝟎𝟏𝟐 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔
1.6𝑥10−19 𝐶
8
Example C: A charge Q is uniformly distributed from 𝑥 = −𝑎 to 𝑥 = 0. Find the force that charge distribution
exerts on a charge Q located at 𝑥 = 𝑎.
*dQ is an infinitesimally small charge, the charge is equal to 𝜆𝑑𝑥, where 𝜆 is the linear charge density of the
𝑄
line of charge. The charge density is the total charge over the length so 𝜆 = 𝑎 .
*r, the distance between the point charge and dQ, is a variable that changes as a function of x, in this case, 𝑟 =
𝑎 − 𝑥, as x is always negative.
𝑄
𝑘𝑄 𝑎 𝑑𝑥
𝑘𝑄𝜆𝑑𝑥
𝑑𝐹 =
→
𝑑𝐹
=
(𝑎 − 𝑥)2
𝑟2
0
𝐾𝑄 2
𝑑𝑥
𝐾𝑄 2
1
𝑘𝑄 2 1 1
0
→𝐹=
∫
→𝐹=
(
)
→𝐹=
( − )
(𝑎 − 𝑥)2
𝑎
𝑎 𝑎 − 𝑥 −𝑎
𝑎 𝑎 2𝑎
−𝑎
→𝑭=
𝒌𝑸𝟐
𝟐𝒂𝟐
Rate your understanding: Electrostatics
0
1
Physics and I must
have the same charge.
I can apply Coulomb’s
Law with help.
2
3
I can apply Coulomb’s
Law with minor errors.
I can apply Coulomb’s
Law with no errors.
4
I am cooler than
Coulomb.
9
AP Physics C: Electricity & Magnetism
7.3 Electric Fields (Chapter 23)
Unit 7 - Electrostatics
Focus Question: How is a field force different from a contact force?
We think of an electric charge as creating an electric field in the space around it. Any other charge in this space
will have an electric force exerted on it.
Physics 2 Review – Electric Field Basics
▪
To test electric fields, we model the behavior of how a positive test charge (of magnitude +q) would
respond to being placed in the field:
𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝑡𝑒𝑠𝑡 𝑐ℎ𝑎𝑟𝑔𝑒
𝐸=
𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡𝑒𝑠𝑡 𝑐ℎ𝑎𝑟𝑔𝑒
𝑬=
𝐸=
𝑘𝑄𝑞
𝑟2
/q
𝑭
𝒒
→
𝑬=
𝒌𝑸
𝒓
*units of field – N/C
▪
Electric field is vector quantity that measures the force per unit charge a charge would experience in the
space around a charge:
𝑬=
▪
𝑭
𝟏 𝑸
=
𝒒 𝟒𝝅𝝐𝟎 𝒓𝟐
If 𝑟⃗ is a unit vector drawn from Q to q:
𝑭
𝟏 𝑸
=
𝒓̂
𝒒 𝟒𝝅𝝐𝟎 𝒓𝟐
By convention, the direction of electric field is in the direction a positive test charge would experience if
placed in the field.
When Q is positive, then field at q points away from Q.
When Q is negative, the field at q points towards Q.
⃑𝑬
⃑ =
▪
▪
By the principle of superposition, the total electric field is the vector sum of the fields due to all charges
present.
Example A: The distance from q1 to q2 is 3.9 m.
a) If point X is 1.3 m from q1, then what is the magnitude
of the electric field at point X?
b) At what distance from q1 is the electric field intensity zero?
10
▪
Electric Field lines – drawn to indicate the direction of the force a test charge would experience.
-Drawn outward from a positive charge and inwards towards a negative charge.
-Stronger fields are drawn with more lines.
Example B: For each case of point charges below, draw the resulting electric field lines.
a)
b)
Example C: Two charged, parallel plates are separated by a distance, d.
The electric field intensity is 200 N/C. An electron with a speed of 4.0 ×
106 m/s comes in from the left just above the negatively charged plate.
If the plates are 10 cm in length, how far has the electron moved in the ydirection during the time it takes to travel across the plates?
The field goes from the top plate to the bottom plate. However, an electron goes against the field so it goes up in
a parabolic path due its acceleration upwards. In the x-direction, it has a constant velocity.
11
Electric Field of a Continuous Charge Distribution (Calculus)
▪
For a continuous charge distribution, the total electric field is the vector sum of
all infinitely small elements, ∆𝑞, in the distribution (this is another way of
stating the superposition principle):
𝒌
𝒅𝑬 = 𝟐 𝒅𝒒
𝒓
𝒅𝒒
⃑ = 𝒌∫
𝑬
𝒓𝟐
Where 𝑑𝑞 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 (𝑎𝑟𝑒𝑎, 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑟 𝑙𝑒𝑛𝑔𝑡ℎ)
𝜌 − 𝑣𝑜𝑙𝑢𝑚𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑑𝑞 = 𝜌dV
𝜎 − 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, dq= 𝜎dA
λ-𝑙𝑖𝑛𝑒𝑎𝑟 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, dq=λdl
Example D: A Wire – Find the electric field at some distance, s, from a long straight
insulating rod of charge +Q and length L at a point P.
The electric field in y cancels due to symmetry. The rod
is broken into a bunch of dQ’s of thickness dy where
𝑄
𝑑𝑄 = 𝜆𝑑𝑦 = 𝐿 𝑑𝑦.
𝑄
𝑘 𝑑𝑦
𝑘𝑑𝑄
𝑠
𝐾𝑄𝑠
𝑑𝑦
𝑑𝐸𝑥 = 2 cos 𝜃 = 2 𝐿 2
=
𝑟
𝑠 + 𝑦 √𝑠 2 + 𝑦 2
𝐿 (𝑠 2 + 𝑦 2 )32
𝐿
2
𝐿
𝐿
𝐿
−
𝑘𝑄𝑠
𝑑𝑦
𝑘𝑄𝑠
𝑦
𝑘𝑄
2
2
→𝐸=
∫
(
) 2 =
(
−
)
3 → 𝐸 =
𝐿
2
2
2
2
2
𝐿
𝐿
𝑠𝐿
2 + 𝑦 2 )2
𝑠
√𝑠
+
𝑦
𝐿
𝐿
(𝑠
𝐿
−
√( ) + 𝑑 2 √(− ) + 𝑑 2
−
2
2
2
2
→𝑬=
𝒌𝑸
𝟐
𝑳
𝒔√ 𝟒 + 𝒔𝟐
12
Example E: A Ring of Charge – A ring shaped conductor of radius R a carries
total charge Q. Find the electric field at point P, a distance.
The ring is broken up into a bunch of dQ’s,
which are small arc’s of length 𝑅𝑑𝜙. The
variable 𝜙 is the angle a length from the
center of the ring to the arc makes with the
vertical. 𝜙 varies from 0 to 2𝜋.
𝑑𝑄 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ → 𝑑𝑄 =
𝑄
𝑅𝑑𝜙
2𝜋𝑅
Again, only the x-component is important since y-cancels to symmetry.
𝑄
𝑘𝑥 2𝜋𝑅 𝑅𝑑𝜙
𝑘𝑑𝑄
𝑘𝜆𝑅𝑑𝜙
𝑥
𝑑𝐸𝑥 = 2 cos 𝜃 = 2
=
𝑟
𝑅 + 𝑥 2 √𝑅 2 + 𝑥 2 (𝑅 2 + 𝑥 2 )32
2𝜋
𝐸=∫
𝑘𝑥𝑄𝑑𝜙
2𝜋(𝑅 2 + 𝑥 2 )
0
3
2
𝑘𝑥𝑄
=
2𝜋
𝑑𝜙 → 𝑬 =
3∫
2𝜋(𝑅 2 + 𝑥 2 )2
0
𝒌𝒙𝑸
𝟑
(𝑹𝟐 + 𝒙𝟐 )𝟐
Example F: A Uniformly Charged Disk – Find the electric field due to a uniformly charge insulating disk of
radius R at a point P perpendicular to the disc as shown.
Example E determined the field due to a
thin ring of charge. The solid disc is
broken into a bunch of infinitely small
rings of radius r.
𝑑𝑄 = 𝜎𝑑𝐴 = 𝜎𝑑(𝜋𝑟 2 ) = 𝜎2𝜋𝑟𝑑𝑟
𝜎=
𝑅
𝐸=∫
0
𝑘(2𝜋𝑟𝜎𝑑𝑟)𝑥
3
(𝑟 2 + 𝑥 2 )2
=
𝑄
𝜋𝑅2
𝜎𝑥 𝑅 2𝑟𝑑𝑟
𝜎
𝜎𝑥
𝑸
𝑸𝒙
∫
−
=
−
3 =
𝟐
4𝜀0 0 (𝑟 2 + 𝑥 2 )2 2𝜀0 2𝜀0 √𝑅 2 + 𝑥 2 𝟐𝜺𝟎 𝝅𝑹
𝟐𝜺𝟎 𝝅𝑹𝟐 √𝑹𝟐 + 𝒙𝟐
Rate your understanding: Electric Fields
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I understand some
aspects of electric
fields.
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electric fields problems
with few errors.
3
4
I can set and solve
electric fields problems
with no errors.
I can explain and teach
electric fields.
13
AP Physics C: Electricity & Magnetism
7.4 Intro to Gauss’s Law (Chapter 24)
Unit 7 - Electrostatics
Focus Question: What is the flux?
▪
Electric Flux – Electric flux (Φ) is a measure of the amount of electric field that penetrates a
surface.
▪
Flux is a scalar quantity that is the dot product of electric field
and the surface area the field passes through.
*only the perpendicular component of the surface area will have flux going through it.
Calculate flux for a non-uniform electric field.
▪
For a non-uniform field, break the surface up into tiny elements, each of area dA:
▪
For a closed, surface, the direction of dA is perpendicular outward.
Example A: E is uniform (constant) from left to right, and our surface is a cube
with face area A. What's the total flux through the (closed) cube shown?
o Flux due to A1: 𝐸 ∙ 𝐴1 = 𝐸𝐴 cos 180 = −𝐸𝐴
o Flux due to A2: 𝐸 ∙ 𝐴2 = 𝐸𝐴 cos 180 = 𝐸𝐴
o Flux due to A3: 𝐸 ∙ 𝐴3 = 𝐸𝐴 cos 90 = 0 *E is perpendicular to dA (the
field does not pass through the surface.
o Flux due to A4: 𝐸 ∙ 𝐴4 = 𝐸𝐴 cos 270 = 0
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.
*If a closed surface has no net charge, then the net flux through it is zero.
14
Gauss’s Law
Example B: Derivation of Gauss’s Law – Consider a point charge +Q inside a spherical shell of radius R.
Determine the flux through the sphere.
*Every point on the sphere has E radially outward. dA is radially outward by
definition, so 𝜃 = 0 everywhere.
𝑑𝜙 = 𝐸 ∙ 𝑑𝐴 cos 𝜃 → 𝑑𝜙 = 𝐸 ∙ 𝑑𝐴 cos 0 = 𝐸 ∙ 𝑑𝐴
𝜙 = 𝐸 ∫ 𝑑𝐴
*Every point on the sphere is the same distance from the charge, so the field on the
sphere is equal at all points to 𝐸 =
𝜙=
𝑅2
. ∫ 𝑑𝐴 is simply the surface area of the shell.
𝑘𝑄
1
1
𝑄
(4𝜋𝑟 2 ) → (𝑘 =
)→𝜙=(
) 2 (4𝜋𝑟 2 )
2
𝑟
4𝜋𝜀0
4𝜋𝜀0 𝑟
→𝝓=
▪
𝑘𝑄
𝑸
𝜺𝟎
Gauss’s Law - The total of the electric flux out of a closed surface is equal to the charge enclosed
divided by the permittivity:
∅ = ∮ ⃑𝑬 ∙ ⃑⃑⃑⃑⃑
𝒅𝑨 𝐜𝐨𝐬 𝜽 =
𝑸𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅
𝝐𝟎
*The 𝑐𝑜𝑠𝜃 can be removed if the angle between the field vector and dA is constant.
Gauss’s Law is used to find the electric field due to various types charge distributions. When using Gauss’s
Law, an imaginary surface is drawn in the region you want to find the electric field. As long as the Gaussian
surface has appropriate symmetry in regard to the field, properties of the flux through that surface can be used
to calculate the field.
Example C: Thin Hollow Shell – a) Find the electric field inside and outside a hollow shell of uniform charge
+Q and b) sketch a graph of electric field from distance of the center of the shell.
Inside the Shell: The electric field is to be found inside the shell, so a Gaussian surface is drawn as shown inside
the sphere. The Gaussian surface needs to be symmetric with respect to the sphere, so a Gaussian sphere of
variable radius r, where r<R is used.
⃑⃑⃑⃑⃑ 𝐜𝐨𝐬 𝜽 =
⃑ ∙ 𝒅𝑨
∅ = ∮𝑬
𝑸𝒆𝒏𝒄
𝝐𝟎
*𝑸𝒆𝒏𝒄 is the charge enclosed by the Gaussian surface. All the charge is outside of
the surface (>r), so the charge enclosed is zero, making the field zero as well.
→𝐸=0
15
Outside the Shell: The electric field is to be found outside the shell, so a Gaussian surface is drawn as shown
outside the sphere. The Gaussian surface needs radially outward (spherical) symmetry, so a Gaussian sphere of
variable radius R, where r>R is used. *the Gaussian surface will always be a sphere for a sphere of charge.
∅ = ∮ ⃑𝑬 ∙ ⃑⃑⃑⃑⃑
𝒅𝑨 =
𝑸𝒆𝒏𝒄
𝝐𝟎
∗The angle between E and dA is always zero, and E is constant along the
Gaussian surface, so ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 cos 𝜃 is simply EA here, where A is the surface are
of the Gaussian surface. (∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 cos 𝜃 will usually be EA or zero when using
Gauss’s law in Physics C).
For this problem, the Gaussian surface encloses all the charge, so 𝑄𝑒𝑛𝑐 = 𝑄.
𝜙 = 𝐸 ∫ 𝑑𝐴 =
𝑄
𝑄
𝑄
→ 𝐸(4𝜋𝑟 2 ) = → 𝐸 =
𝜀0
𝜀0
4𝜋𝜀0 𝑟 2
→𝑬=
Rate your understanding: Gauss’s Law
0
1
I don’t know what
the flux is going on
anymore.
I understand parts of
Gauss’s Law and flux.
𝑲𝑸
𝒓𝟐
2
I understand most of
Gauss’s Law and flux.
3
I understand all of
Gauss’s Law and flux.
4
I can explain and teach
Gauss’s Law.
16
AP Physics C: Electricity & Magnetism
7.5 Applications of Gauss’s Law (Chapter 24)
Unit 7 - Electrostatics
Focus Question: What is Gauss’s Law?
▪ To use Gauss’s Law, a surface called a Gaussian surface must be selected (pretty much always a
▪
▪
▪
▪
cylinder or sphere). It is not necessarily a surface of the object being analyzed.
The symmetry of the closed surface makes it possible to evaluate ∮ 𝐸 ∙ 𝑑𝐴, often without scary calculus.
If E is perpendicular to the Gaussian surface, and of constant magnitude, then ∮ 𝐸 ∙ 𝑑𝐴 = 𝐸𝐴.
If E is parallel to the Gaussian surface, then ∮ 𝐸 ∙ 𝑑𝐴 = 0.
The following problems look at various charge distributions. Charge can be distributed along any
conducting surface (this is not a Gaussian surface, which is something we make up to solve problems).
Example A: Infinite Sheet – An infinite plane has uniform charge density of 𝜎. Find the electric field due to
the plane.
A large plane of charge has field going outwards on both sides, so the
chosen Gaussian surface needs to have symmetry going out from two
ends. A cylinder will be used here since there the dA symmetry of the caps
matches the field. The flux is then:
𝜙 = 𝜙𝐿𝐸𝐹𝑇 𝐶𝐴𝑃 + 𝜙𝑅𝐼𝐺𝐻𝑇 𝐶𝐴𝑃 + 𝜙𝑆𝐼𝐷𝐸
For the side, dA is perpendicular to E, so there is no flux. For the two
caps, E and dA are in the same direction so ∫ 𝐸 ∙ 𝑑𝐴 cos 𝜃 = 𝐸𝐴 for both.
𝑄𝑒𝑛𝑐
𝜙 = ∫ 𝐸 ∙ 𝑑𝐴𝐿𝐸𝐹𝑇 + ∫ 𝐸 ∙ 𝑑𝐴𝑅𝐼𝐺𝐻𝑇 =
𝜀0
The charge enclosed for both is the charge density, 𝜎, multiple by the cap area, A.
𝜙 = 𝐸𝐴 + 𝐸𝐴 =
𝜎𝐴
𝝈
→𝑬=
𝜀0
𝟐𝜺𝟎
*The field lines of the sheet are straight, parallel, uniformly spaced and perpendicular to the sheet. The
calculation assumes an infinitely large sheet, but the results are a reasonable approximation for a large sheet.
When a conducting plate is given a charge, the charge distributes itself uniformly over the entire surface of the
plate. The electric field is directed away from the plate.
Example B: Parallel plates of equal charge – Find the electric field a) outside the plates
and b) inside the plates.
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
as found in Example A.
a) Outside the plates, in either regions A or C, the fields point
𝝈
𝝈
𝝈
in the same direction: 𝐸 = 𝐸1 + 𝐸2 = 𝟐𝜺 + 𝟐𝜺 = 𝜺
𝟎
𝟎
𝟎
b) In the region inside the plates, E1 and E2 are in opposite directions:
𝝈
𝝈
𝐸 = 𝐸1 + 𝐸2 = 𝟐𝜺 − 𝟐𝜺 = 𝟎
𝟎
𝟎
17
▪
▪
When parallel plates are given equal and opposite charges, the field is nearly uniform in the space
between the plates. Most of the charge is in the inner surface of the plates, which very little charge on
the outside of the plates.
The electric field is the resultant of the fields due to the two sheets of 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 = 𝐸1 + 𝐸2 =
▪
𝝈
𝝈
𝝈
+
=
𝟐𝜺𝟎 𝟐𝜺𝟎 𝜺𝟎
Using Gauss’s Law
A Gaussian cylinder is drawn that goes through the surface of one of the
plates: 𝜙 = 𝜙𝐿𝐸𝐹𝑇 𝐶𝐴𝑃 + 𝜙𝑅𝐼𝐺𝐻𝑇 𝐶𝐴𝑃 + 𝜙𝑆𝐼𝐷𝐸
The side has no flux since E is perpendicular to dA of the lateral side of the
cylinder. The left cap has no field through it. For the right cap, E and dA
are in the same direction so ∫ 𝐸 ∙ 𝑑𝐴 cos 𝜃 = 𝐸𝐴.
The charge enclosed by the Gaussian cylinder is the charge density of the
plate times the area of the cap of the cylinder passing through the plate:
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝑄𝑒𝑛𝑐
𝜖0
→ 𝐸𝐴 =
𝜎𝐴
𝜀0
𝝈
→𝑬=𝜺
𝟎
Example C: Infinite Line Charge – Determine the electric field a distance R from an infinitely long
conducting wire with a uniform charge density of 𝜆.
The line of charge has field going radially outward
along the line. The lateral ends of a cylinder have
symmetry to match this, so a Gaussian cylinder will
be used. Field does not go through the caps of the
cylinder, so only the lateral edge will be used:
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
𝑄𝑒𝑛𝑐
𝜺𝟎
∫ 𝑑𝐴 is the lateral surface area since that is the surface with flux through it. The charge of the line inside the
cylinder is the linear charge density, 𝜆, multiplied by the length of the cylinder, L.
𝜙 = 𝐸 ∫ 𝑑𝐴 =
𝜆𝐿
𝜆𝐿
𝝀
→ 𝐸(2𝜋𝑟𝐿) =
→𝑬=
𝜺𝟎
𝜺𝟎
𝟐𝝅𝒓𝜺𝟎
18
Example D: A sphere of radius R is surrounded by a concentric spherical shell of
inner radius 2R and outer radius 3R, as shown The inner sphere is an insulator
containing a net charge +Q distributed uniformly throughout its volume. The
spherical shell is a conductor containing a net charge +2Q. Use Gauss's law to
determine the electric field for the following values of r, the distance from the center
of the insulator.
a) 0 < r < R
b) R < r < 2R
c) 2R < r < 3R
d) r > 3R
Spheres obviously have spherically symmetry, so Gaussian spheres will be used for all regions.
a) A charge insulator has its charge uniformly distributed throughout. The charge
enclosed by the Gaussian surface is the charge density of the entire region multiplied
by the volume enclosed by the Gaussian surface.
𝑄𝑒𝑛𝑐
𝜌𝑉𝐺𝐴𝑈𝑆𝑆𝐼𝐴𝑁
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
→ 𝐸 ∫ 𝑑𝐴 =
𝜺𝟎
𝜺𝟎
∫ 𝑑𝐴 is the surface area of the Gaussian surface.
𝐸(4𝜋𝑟 2 ) =
𝑄
4
4
𝜋𝑅 3
3
3
𝜋𝑟 3
→𝑬=
𝜀0
𝑸𝒓
𝟒𝝅𝜺𝟎 𝑹𝟑
b) The charge enclosed is total charge of the insular, +Q.
𝑄𝑒𝑛𝑐
𝑄
→ 𝐸(4𝜋𝑟 2 ) =
𝜺𝟎
𝜺𝟎
𝑸
→𝑬=
𝟒𝝅𝒓𝟐 𝜺𝟎
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
c) The outer shell is a conductor, so all charge gathers on the outer and inner surfaces.
A charge of -Q gathers on the inner surface as it’s attracted to the insular. The excess
charge, +3Q, gathers on the outer shell. The charge enclosed by the Gaussian surface is
zero.
𝑄𝑒𝑛𝑐
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
→ 𝑄𝑒𝑛𝑐 = 0
→𝑬=𝟎
𝜺𝟎
d) The total charge enclosed in this case if +3Q.
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
𝑄𝑒𝑛𝑐
3𝑄
→ 𝐸(4𝜋𝑟 2 ) =
𝜺𝟎
𝜺𝟎
→𝑬=
𝟑𝑸
𝟒𝝅𝒓𝟐 𝜺𝟎
19
Example F: A very long, solid, non-conducting cylinder of radius R has a positive charge of uniform volume
density 𝜌. A section of the cylinder far from its ends is shown below. Let 𝑟 represent the radial distance from
the axis of the cylinder.
a) Derive an expression for the magnitude of the electric field at a radius 𝑟 < 𝑅.
b) Derive an expression for the magnitude of the electric field at a radius 𝑟 > 𝑅.
A Gaussian cylinder will be used for both cases as the radially linear symmetry of a cylinder obviously matches
the field produced by a cylinder.
a) The cylinder is a conductor, so all charge is on
the outer surface, so there is no charge enclosed
by the Gaussian cylinder:
𝑄𝑒𝑛𝑐
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
→ 𝑄𝑒𝑛𝑐 = 0
→𝑬=𝟎
𝜺𝟎
𝑄
⃑⃑⃑⃑⃑ = 𝑒𝑛𝑐
b) ∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝜺𝟎
𝜌𝑉𝑒𝑛𝑐
𝜙 = 𝐸 ∫ 𝑑𝐴 =
𝜺𝟎
Since the charge is only contained in the actual cylinder of
radius R, the volume of the cylinder of radius R will be used
when finding the charge enclosed:
𝜌𝜋𝑅 2 𝐿
→ 𝐸(2𝜋𝑟𝐿) =
𝜺𝟎
𝟐
𝝆𝑹
→𝑬=
𝟐𝒓𝜺𝟎
Rate your understanding: Gauss’s Law
0
1
Weird flux but okay.
I understand parts of
Gauss’s Law and flux.
2
I understand most of
Gauss’s Law and flux.
3
I understand all of
Gauss’s Law and flux.
4
I can explain and teach
Gauss’s Law.
20
AP Physics C
Electricity & Magnetism
Unit 8: Potential & Capacitance
Section 8.1 – Electric Potential………………………………….………………….………..…22
Section 8.2 – Capacitance…………………………………….………………….………….….27
Section 8.1 – Dielectrics……………………………………….…………………….…………32
21
AP Physics C: Electricity & Magnetism
8.1 Electric Potential (Chapter 25)
Unit 8 – Electric Potential & Capacitance
Focus Question: How is electric potential different from electric field strength?
▪
▪
The change in electric potential is the change in electric potential energy when a charge, q, moves from
one point to another. The change in electric potential is also equal to the amount of work the electric
field would have to do to move the charge.
Charges tend to move away from like charges and towards opposite charges. When charges move in a
field, there is a change in electrical potential energy.
*The change in electric potential energy is also equal to the amount of
work the electric field would have to do to move the charge.
𝑈𝐴 − 𝑈𝐵 =Work=Fr=
𝐾𝑄𝑞
𝑟2
𝑟=
𝐾𝑄𝑞
𝑟
𝑈𝐴 − 𝑈𝐵 =Eqr
*Units of electric potential energy: Joules (J)
Electric potential from point charges.
Example A: A particle of mass 6.3𝑥10−3 𝑘𝑔 and charge 9.0𝑥10−4 C floats between two horizontal plates that
are 25 cm apart. What is the potential difference between the plates?
Since the charge floats, the electrostatic force cancels out the force of gravity.
∑ 𝐹𝑦 = 0 → 𝐹𝐸 − 𝑚𝑔 = 0 → 𝐹𝐸 = 𝑚𝑔
→ 𝐹𝐸 = 𝐸𝑞 =
𝑉
𝑉𝑞
𝑚𝑔𝑑
𝑞→
= 𝑚𝑔 → 𝑉 =
𝑑
𝑑
𝑞
𝑚
→𝑉=
(6.3𝑥10−3 𝑘𝑔) (10 2 ) . 25 𝑚
𝑠
9.4𝑥10−4 𝐶
= 𝟏𝟔. 𝟖 𝑽
22
Example B: Two charges are located 6 m from each other as shown. Point P is located 6 m from each of the
two charges. 𝑞1 = −4 𝜇𝐶, 𝑞2 = 1 𝜇𝐶
*the superposition principle still applies for potential*
a) Calculate the electric potential at point P.
b) Calculate the work required to bring a charge of magnitude of
20 𝑛𝐶 from infinity to point P.
a) 𝑉 = 𝑉1 + 𝑉2 =
𝑘𝑞1
𝑟
+
𝑘𝑞2
𝑟
=
(9𝑥109
𝑵𝒎𝟐
)(−𝟒𝒙𝟏𝟎−𝟔 𝑪)
𝑪𝟐
6𝑚
+
(9𝑥109
𝑵𝒎𝟐
)(𝟏𝒙𝟏𝟎−𝟔 𝑪
𝑪𝟐
6𝑚
=
−4500 𝑉
b) 𝑊 = 𝑉𝑞 = (−4500 𝑉)(20𝑥10−9 𝐶) = −9𝑥10−5 𝐽
*the work in this case is negative. The potential at the point is negative so a positive charge would be attracted,
but it takes negative work to bring it here, similar to the way negative work is done when an object is lowered in
a gravitational field.
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 C: Draw electric field lines and equipotential lines for the following two oppositely charge point
charges:
The Electron Volt,
▪
▪
▪
An electron volt is a common unit of energy.
The joule is a large unit for measuring energies of many charge
particles. Thus, the electron volt is used.
An electron volt is the energy acquired by an electron as a result of
moving through a potential difference of 1 V.
1 𝑒𝑉 = 𝑊 = 𝑉𝑞 = (1 𝑉)(1.6𝑥10−19 𝐶) = 1.6𝑥10−19 𝐽
1 𝑒𝑉 = 1.6𝑥10−19 𝐽
23
Example D: What is the speed of an electron that has been accelerated through a potential difference of 4 V?
Calculate its final kinetic energy in eV. (an electron has a mass of 9.11x10-31 kg and a charge of 1.6x10-19 C).
𝑊 = ∆𝐾 → 𝑉𝑞 =
1
1
2𝑉𝑞
𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 → 𝑣𝑓 = √
2
2
𝑚
2(4 𝑉)(1.6𝑥10−19 𝐶)
→ 𝑣𝑓 = √
9.11𝑥10−31 𝑘𝑔
= 1.2𝑥106 𝑚/𝑠
1
1
𝑚 2
1 𝑒𝑉
𝐾 = 𝑚𝑣 2 = (9.11𝑥10−31 𝑘𝑔) (1.2𝑥106 ) = 6.4𝑥10−19 𝐽 𝑥
= 4 𝑒𝑉, 𝑎𝑠 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑
2
2
𝑠
1.6𝑥10−19 𝐽
Electric potential due to a continuous charge distribution.
▪
Consider a continuous charge distribution to be made of up an infinite number of
point charges with charge 𝑑𝑞. Superposition can be used to add up all the 𝑑𝑉
from each contributor, 𝑑𝑞.
Example E: Find an integral for the potential at point A due to the uniform line of
charge as shown The charge density is 𝜆 C/m.
*x is coordinate of A and is constant. y is a variable that gives the location of a typical
element of charge 𝑑𝑞.
𝑑𝑉 =
𝑑𝑄 = 𝜆𝑑𝑦 =
𝑎
𝑄
𝑘𝑑𝑄
𝑟
𝑄
𝑑𝑦
2𝑎
𝑘 2𝑎 𝑑𝑦
𝑎
𝑘𝑄
𝑑𝑦
𝑉= ∫
=
∫
√𝑥 2 + 𝑦 2 2𝑎 −𝑎 √𝑥 2 + 𝑦 2
−𝑎
𝑉=
→
𝑘𝑄
𝑘𝑄
𝑎
ln |𝑦 + √𝑦 2 + 𝑥 2 |
→𝑉=
(ln |𝑎 + √𝑎2 + 𝑥 2 | − ln | − 𝑎 + √𝑎2 + 𝑥 2 |
−𝑎
2𝑎
2𝑎
𝑘𝑄
𝑎 + √𝑎2 + 𝑥 2
ln |
|
2𝑎
√𝑎2 + 𝑥 2 − 𝑎
24
Example F: A conducting shell has a total charge of Q and radius R.
a) Find the electric potential outside the shell
b) Find the electric potential inside the shell.
In cases where Gauss’s Law can be readily used, the potential can be found by first
founding the field using Gauss’s Law, and then finding potential by 𝑉 = − ∫ 𝐸 ∙ 𝑑𝑟 from
infinity to the point where potential is being calculated.
a) Step 1: Find the field 𝑄𝑒𝑛𝑐
𝑄
→ 𝐸(4𝜋𝑟 2 ) =
𝜀0
𝜀0
𝑄
𝑘𝑄
→𝐸=
=
4𝜋𝑟 2 𝜀0 𝑟 2
Step 2: Find potential by integrating from infinity to r:
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝑟
𝑉=−∫
∞
→𝑽=
𝑘𝑄
𝐾𝑄 𝑟 𝑘𝑄
𝑑𝑟 = − (−
)| =
−0
2
𝑟
𝑟 ∞
𝑟
𝒌𝑸
𝒓
b) Step 1: Find the field with Gauss’s Law∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
𝑄𝑒𝑛𝑐
→ 𝑄𝑒𝑛𝑐 = 0
𝜀0
→𝐸=0
Step 2: Find potential by integrating from infinity to r:
Coming from infinity, there are two regions that need to be consider, infinity to the surface of the sphere, and
then going in the surface of the sphere to r. The two regions have different fields to integrate:
𝑅
𝑟
𝑘𝑄
𝑉 = − ∫ 2 𝑑𝑟 − ∫ 0 𝑑𝑟
𝑟
∞
→ 𝑉 = − (−
𝑅
𝐾𝑄 𝑅 𝐾𝑄
)| =
𝑟 ∞
𝑅
The potential inside the conducting shell is always
𝐾𝑄
𝑅
. The potential inside of a conductor is constant since the
field inside a conductor is zero.
25
Example G: Find the electric potential inside a uniformly charged insulating sphere of
radius R and total charge Q.
Field Inside:
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝑄𝑒𝑛𝑐
𝜀0
𝑄𝑟
→𝐸=
4𝜋𝜀0 𝑅3
→ 𝐸(4𝜋𝑟2 ) =
𝑄 4 3
4 3 3𝜋𝑟
𝜋𝑅
3
𝜀0
Field Outside:
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
→𝐸=
𝑄
4𝜋𝑟 2 𝜀0
𝑄𝑒𝑛𝑐
𝑄
→ 𝐸(4𝜋𝑟 2 ) =
𝜀0
𝜀0
Potential:
𝑅
𝑟
𝑘𝑄
𝑄𝑟
𝑉 = − ∫ 2 𝑑𝑟 − ∫
𝑑𝑟
𝑟
4𝜋𝜀0 𝑅3
∞
𝑅
𝐾𝑄 𝑅
𝑄𝑟 2
𝑄
𝑟 𝐾𝑄
2
2
→ 𝑉 = − (−
)| − (
)| =
−
3
3 (𝑟 − 𝑅 )
𝑅
𝑟 ∞
𝑅
8𝜋𝜀0 𝑅
8𝜋𝜀0 𝑅
𝑘𝑄 𝑄𝑅 2 − 𝑄𝑟 2
→𝑉=
+
𝑅
8𝜋 𝜀0 𝑅3
Rate your understanding: Electric Potential
0
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I have low potential.
(╯︵╰,)
I understand only a few
aspects of electric
fields.
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I can explain most
aspects of electric fields.
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involve electric fields
with no errors.
4
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electric fields.
Tremendous Potential.
26
Unit 8 – Electric Potential & Capacitance
AP Physics C: Electricity & Magnetism
8.2 Capacitors (Chapter 26)
Focus Question: How does a capacitor store electric potential energy?
▪
A capactior is formed by two conductors seperated by an insulator. When a capacitor is connected to a
battery, it quickly becomes charged, with the two plates taking on opposite
charges. The net charge on a capacitor is zero since the plates have equal but
opposite charges.
▪
Function of a capacitor – Stores electrical energy in the electric field between
two plates.
▪
An electric field exists between the two conductors in a capacitor. The field and electric potential
difference are proportional to the charge on the conductors.
𝑪=
𝑸
𝑽
Q – charge on each plate
V – potential difference
units: Farads (F)
*A capactior always have zero net charge as the plates have equal but opposite charges. The charge in
the equation for capacitance is the charge of each plate.
▪
Capacitor symbol:
Parallel Plate Capacitors
▪
Gauss’s Law can be used to find the electric field between the plates of the capacitor:
𝑄
𝜎𝐴
⃑⃑⃑⃑⃑ = 𝑒𝑛𝑐 → 𝐸𝐴 =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝜖0
𝜀0
𝝈
→𝑬=
𝜺𝟎
Relating E to other variabes: 𝑉 = 𝐸𝑑 →
→𝐶=
𝑄
𝜎
𝜺𝟎
𝑑
→𝐶=
𝑄𝜺𝟎
𝑄
𝐴
𝑑
→ 𝑪 = 𝜺𝟎
𝑄
𝐶
𝑄
= 𝐸𝑑 → 𝐶 = 𝐸𝑑
𝑨
𝒅
Example A: The area of each plate in an air gap capacitor is A m2. The capacitance of the capacitor is C F and
the electric field between the plates is E V/m.
a) What is the potential difference across the plates?
b) How much charge is stored on each plate?
𝑨
a) 𝑪 = 𝜺𝟎 𝒅 → 𝑑 =
b) 𝑄 = 𝐶𝑉 → 𝐶
𝜺𝟎 𝑨
𝑬𝜺𝟎 𝑨
𝑪
𝐶
→ 𝑉 = 𝐸𝑑 = 𝐸
𝜺𝟎 𝑨
𝐶
→𝑽=
𝑬𝜺𝟎 𝑨
𝑪
→ 𝑸 = 𝑬𝜺𝟎 𝑨
27
Energy of a Charged Capacitor
▪
▪
A charged capacitor is used to store electric energy. The amount of energy stored is equal to the work
done to charge the capacitor.
When charging a capacitor, charge is removed from one plate and added to the other. Work is required
to transfer the charge:
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.
𝑞
𝑑𝑊 = 𝑉𝑑𝑞 → 𝑑𝑊 = 𝑑𝑞
𝐶
𝑄
𝑊=∫
0
𝑞
1 𝑄2
𝑑𝑞 =
𝐶
2 𝐶
The work done to charge the plate is equal to the energy stored on the plate”
1 𝑄2
𝑈=
2 𝐶
1
Using Q=CV, the above equation can also be used to derive the more common form: 𝑈 = 2 𝐶𝑉 2
▪
Energy per unit volume:
𝑈
Energy density = 𝑉𝑜𝑢𝑚𝑒 =
1
𝐶𝑉 2
2
𝐴𝑑
=
1 𝑨
𝜺 (𝑬𝒅)𝟐
2 𝟎𝒅
𝐴𝑑
1
= 2 𝜺𝟎 𝑬𝟐
Example B: A capacitor is constructed of two large, identical parallel metal plates seperated by a small
distance 𝑑. A battery fully charges the capacitor and is then disconnected.
a) While the battery is still connected, plate seperation is doubled. Describe the changes, if any, of the
voltage across the capactior, the electric field between the plates, and the energy stored in the
capactior.
Voltage – Since the battery is still connected, the voltage will not change.
𝑨
Electric Field – The field will decrease. The capacitance is given by: 𝑪 = 𝜺𝟎 𝒅, so when the plates
get further apart, the capacitance decreases, which decreases the charge on the plates. Since the
charge on the plate decreases, the electric field decreases.
1
Energy stored – By 𝑈 = 2 𝐶𝑉 2 , the energy stored on the plate decreases. Though the voltage stays
the same, capacitance decreases. Less charge stored leads to less stored energy.
b) The capacitor is disconnected from the battery, and the plate seperation is doubled. Describe the
changes, if any, of the voltage across the capactior, the electric field between the plates, and the
energy stored in the capactior.
Voltage – The voltage will increase. The charge on each plate will stay the same, and the capacitor
will decrease, so the voltage to increase since V=Q/C.
Electric Field – The charge on each plate will stay the same. Since field is only based on charge
density on the plates, the field stays the same.
1 𝑄2
Energy stored – By 𝑈 = 2
𝐶
, the energy increases in this case. The work done to separate the plates
is transferred to the energy stored in the plates.
28
Capacitors in Series and Parallel
Capacitors in Parallel
The capacitors are in parallel, so they have the same voltages, 𝑉1 = 𝑉2 = 𝑉.
The charges add, 𝑄𝑒𝑞 = 𝑄1 + 𝑄2 .
𝐶𝑒𝑞 𝑉 = 𝑄1 𝑉 + 𝑄2 𝑉
Capacitors in parallel act like a single capacitor whose capacitance is the sum of the capacitances.
𝐶𝑒𝑞 = 𝐶1 + 𝐶2
Capacitors in Series
In series, a negative charge flows to C1 and a positive charge flows to C2. The region
between the capacitors will remain neutral. In series, each capacitor stores the same amount
of charge.
𝑄
𝑄
𝑄
1
2
The voltages add: 𝑉𝑒𝑞 = 𝑉1 + 𝑉2 → 𝐶 = 𝐶 + 𝐶
𝑒𝑞
Finding the equivalent capacitance of capacitors in series works the same as finding the equivalent resistance of
resistors in parallel.
1
1
1
=
+
𝐶𝑒𝑞 𝐶1 𝐶 2
Example C: Ponder the circuit on the right:
a) What is the equivalent capacitance of the network?
b) What is the potential difference across the 3 𝜇F capacitor?
c) How much charge is stored in the 6 𝜇F capacitor?
To find the equivalent capacitance, the equivalent circuit needs to be found:
a) 10 𝜇𝐹
29
b) In step 2 of the finding the equivalent circle, the charge on the 2 𝜇𝐹 is the same
as the charge on both the 3 𝜇𝐹 and 6 𝜇𝐹 capacitors since they are in series.
𝑄 = 𝐶𝑉 = (2𝑥10−6 𝐹)(180 𝑉) = 3.6𝑥10−4 𝐶
The 3 𝜇𝐹 capacitor thus has a charge of 3.6𝑥10−4 𝐶 stored:
𝑄 3.6𝑥10−4 𝐶
𝑉= =
= 𝟏𝟐𝟎 𝑽
𝐶
3𝑥10−6 𝐹
c) Since the 3 𝜇𝐹 capacitor has 120 V across it, the 6 𝜇𝐹 capacitor has 60 V across it since they must add to 180
V.
1
1
𝑈 = 𝐶𝑉 2 = (6𝑥10−6 𝐹)(60 𝑉)2 = .011 𝐽
2
2
Example D: Find the capacitance of two concentric conducting cylinders of length
1.0 m. The inner cylinder has radius 1.0 mm and the outer cyilnder has radius 2.0
mm. The capacitors have equal but opposite charges of linear charge density 𝜆.
Cylindrical capatiors are much more common than parallel plate capacitors. To find
the capacitance of a cylindrical capacitor using C=Q/V, the voltage needs to be found based on the field.
Step 1: Found field in the region between the plates using Gauss’s Law.
A cylinder is drawn between the plates to find the field:
⃑⃑⃑⃑⃑ =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝐸(2𝜋𝑟𝐿) =
𝑄𝑒𝑛𝑐
𝜺𝟎
𝜆𝐿
𝜺𝟎
𝜆
2𝜋𝑟𝜺𝟎
→𝐸=
Step 2: Find the potential using 𝑉 = − ∫ 𝐸 ∙ 𝑑𝑟. The potential will be found based on going from the inner to
outer plate.
𝑅2
𝑉=−∫
𝑅1
→𝑉=
𝜆
𝜆
𝜆
𝜆
𝑅2
𝑅
= −(
ln|𝑟|)| 2 = −
(ln|𝑅2 | − ln|𝑅1 |) = −
ln | |
𝑅
2𝜋𝑟𝜺𝟎
2𝜋𝜺𝟎
2𝜋𝜺𝟎
2𝜋𝜺𝟎
𝑅1
1
𝜆
𝑅1
ln | |
2𝜋𝜺𝟎
𝑅2
Step 3: Find the capacitance using C=Q/C
𝐶=
𝜆𝐿
𝜆
𝑅
2𝜋𝜺𝟎
→𝑪=
ln |𝑅1 |
2
𝟐𝝅𝜺𝟎 𝑳
𝑹
𝐥𝐧 |𝑹𝟏 |
𝟐
30
Example E: A spherical capacitor consists of an inner metal sphere of radius r1 supported by
an insulting stand at the hollow metal sphere of radius r2 as shown. The inner sphere has
charge +Q and the outer sphere has charge -Q. Derive an expression for the capacitance.
Step 1: Found field in the region between the surfaces using Gauss’s Law.
A Gaussian sphere is used:
𝑄𝑒𝑛𝑐
∅ = ∮ 𝐸⃑ ∙ ⃑⃑⃑⃑⃑
𝑑𝐴 =
𝜺𝟎
𝑄
𝑄
→ 𝐸(4𝜋𝑟 2 ) =
→𝐸=
𝜀0
4𝜋𝑟 2 𝜀0
Step 2: Find the potential using 𝑉 = − ∫ 𝐸 ∙ 𝑑𝑟.
𝑟2
𝑟2
𝑄
𝑄
𝑑𝑟
𝑉=−∫
=−
∫ 2
2
4𝜋𝑟 𝜀0
4𝜋𝜀0 𝑟
𝑟1
𝑟1
𝑄 1 𝑟2
𝑄
1 1
𝑄 1 1
→ 𝑉 == −
( )|𝑟 = −
( − )=
( − )
4𝜋𝜀0 𝑟 1
4𝜋𝜀0 𝑟2 𝑟1
4𝜋𝜀0 𝑟1 𝑟2
𝑄 𝑟2 − 𝑟1
→𝑉=
(
)
4𝜋𝜀0 𝑟1 𝑟2
Step 3: Find the capacitance using C=Q/C
𝐶=
𝑄
=
𝐶
→𝐶=
𝑄
𝑄
𝑟2 −𝑟1
(
4𝜋𝜀0 𝑟1 𝑟2
𝑟1 𝑟2
)
=
𝑟1 𝑟2 4𝜋𝜀0
𝑟2 − 𝑟1
𝑘(𝑟2 − 𝑟1 )
Rate your understanding: Capacitors
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for more physics.
I understand some
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errors.
3
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errors.
4
I can explain and teach
capacitance.
31
Unit 8 – Electric Potential & Capacitance
AP Physics C: Electricity & Magnetism
8.3 Dielectrics (Chapter 26)
Focus Question: What effect does adding a dielectric have to a capacitor?
▪
Most capacitors have a solid, nonconducting material between the plates. This material is called a
dielectric.
▪
Function of a dielectric:
➢ Keeps plates close, but not in contact
➢ Can tolerate greater field without dielectric breakdown
*Air is more suspect to dielectric breakdown, in which the insulator ionizes and charge
can flow between the plates.
➢ Increases capacitance.
▪
The potential difference between two charge plates decreases when a dielectric is used, but the charge
on each plate stays the same. Electric field decreases when a dielectric is inserted, due to small induced
charges appearing on the surfaces of the dielectric.
The dielectric constant, 𝜅:
▪
𝜿=
𝑪
𝑪𝟎
C – Capacitance with dielectric
C0 – Capacitance without dielectric
*𝜅 ≥ 1
Field Decreases: 𝜅 =
𝑉0
𝑉
=
𝐸0 𝑑
𝐸𝑑
=
𝐸0
𝐸
Permittivity increases: 𝜀 = 𝜅𝜀0
Voltage:
Case 1: If the battery is no longer connected, voltage decreases 𝜅 =
𝑄/𝑉
𝑄/𝑉0
=
𝑉0
𝑉
*in this case, energy decreases due to the work done to insert dielectric.
Case 2: If the capacitor is still connected to the battery, voltage remains constant
*In this case, energy increases since capacitor increases and voltage is constant.
Example A: A 275 F capacitor will be manufactured using a dielectric having a permittivity of 4.000 and
circular plates having a diameter of 2.50 cm. What should the plate separation (and the thickness of the
dielectric) be? Is it likely that this large a capacity could be constructed using parallel plate architecture?
𝜅𝜀0 𝐴
𝜅𝜀0 𝐴
𝐶=
→𝑑=
𝑑
𝐶
4(8.85𝑥10−12
𝐶2
𝑁𝑚2
.025
)(𝜋 (
2
𝑚) )
2
= 6.3𝑥10−11 𝑚
275𝑥10−6 𝐹
It would be extremely difficult to construct a plate separation this small.
𝑑=
32
▪
Electric field in a dielectric using Gauss’ Law:
*induced charges appear on the surface of the dielectric.
𝜎: 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑝𝑙𝑎𝑡𝑒𝑠
𝜎𝑖 : 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
(𝜎 − 𝜎𝑖 )𝐴
𝑄
⃑⃑⃑⃑⃑ = 𝑒𝑛𝑐 → 𝐸𝐴 =
∅ = ∮ 𝐸⃑ ∙ 𝑑𝐴
𝜖0
𝜀0
𝜎 − 𝜎𝑖
→𝐸=
𝜀0
𝜎
*Since the permittivity increases by a factor of 𝜅, E is also 𝐸 = 𝜅𝜀
0
Relationship between induced dielectric charge and dielectric constant:
𝜎 − 𝜎𝑖
𝜎
=
𝜀0
𝜅𝜀0
→𝜿=
𝜅=
▪
𝝈
𝝈 − 𝝈𝒊
𝐸0
𝐸
Energy density in dielectric
1
1
𝐴
𝜅𝐶𝑉 2 2 𝜅𝜀0 𝑑 𝐸 2 𝑑 2 1
𝑈
2
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =
=
=
= 𝜅𝜀0 𝐸 2
𝑉𝑜𝑙𝑢𝑚𝑒
𝐴𝑑
𝐴𝑑
2
Example B: A parallel plate capacitor of area A and plate separation L is filled with a removable dielectric slab
of dielectric slab constant 𝜅. The capacitor is given a charge Q with the slab removed, is disconnected from the
battery and then the slab is inserted.
a) Find the potential difference with and without the dielectric slab.
b) Find the electric field between the plates with and without the slab.
c) Find the induced charge on the slab.
33
Example C: A capacitor having no dielectric has a capacitance of C. It is charged up to
V by momentarily attaching it to a battery, and then disconnecting it.
a) What is the energy stored in the capacitor at this voltage?
b) What is the charge on the capacitor?
c) A dielectric having a permittivity of 20 is now carefully inserted between the plates of the capacitor.
What is its new capacitance?
d) What is the charge on the new capacitor?
e) What is the energy stored in the new capacitor?
f) Explain the discrepancy between (e) and (a).
1
a) 𝑈 = 2 𝐶𝑉 2
b) 𝑄 = 𝐶𝑉
c) 𝜅 =
𝐶′
𝐶
→ 𝐶 ′ = 𝜅𝐶
d) 𝑄 = 𝐶 ′ 𝑉 = 𝜅𝐶𝑉
𝑄
*Inserting a dielectric increases the charge by a factor of kappa: 𝜅 = 𝑄
0
1 𝑄2
e) 𝑈 = 2
𝐶
1 (𝜅𝐶𝑉)2
𝑈=2
𝜅𝐶
The charge and the capacitance both increase by a factor of the dielectric constant:
1
= 2 𝜅𝐶𝑉 2
f) The energy increased by a factor of the dielectric constant even though the circuit was not connected
to power. The work done to insert the dielectric was converted to energy stored in the capacitor.
Rate your understanding: Dielectrics
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1
Die, electrics.
I understand some
aspects of capacitors.
2
I can describe
capacitance with few
errors.
3
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capacitance with no
errors.
4
I can explain and teach
capacitance.
34
AP Physics C
Electricity & Magnetism
Unit 9: Circuits
Section 9.1 – DC Circuits Review……………………………….………………….……….…36
Section 9.2 – Current, Resistance and Power……………………………………….…….……44
Section 9.3 – RC Circuits…………………………………………………..……….………..…48
35
AP Physics C: Electricity & Magnetism
9.1 Notes – DC Circuits Review (Chapter 29)
Unit 9 - Circuits
Focus Question: What causes a magnetic field?
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 energy with them.
Definition of Current
▪
▪
▪
Current – The flow of charge through a conductor. In conductors, electrons move freely. When electrons
move, charge moves around. Negative charge is due to excess electrons. Positive charge is due to a
deficit of electrons.
Current flows in a conductor when there is a potential different from one end of a conductor to another.
The rate of current flow is defined as 𝐶𝑢𝑟𝑟𝑒𝑛𝑡
Current =
𝑐ℎ𝑎𝑟𝑔𝑒
I=
𝑡𝑖𝑚𝑒
𝑄
𝑡
*units: Amperes (A)
Direction of Current
▪
▪
▪
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.
▪
Charge flows around a circuit repeatedly. Electrostatics will never move conventional current from a
negative charge (lower potential) to positive charge (high potential).
*After leaving a source, current circulates to a load. The load consumes energy but not charge.
36
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.
Ohm’s Law
Ohm’s Law relates the difference in potential(known as voltage) from one side of a load to another (V), the
resistance of the load (R) and current ( I ).
𝑂ℎ𝑚′ 𝑠 𝐿𝑎𝑤 - ∆𝑽 = 𝑰𝑹
Series vs. Parallel
▪
When two elements are in series, they are connected end to end so that there is only one way for current
to flow. When resistors are in series, current has to pass through both resistors.
*Req = Equivalent resistance.
Current is the same in series. Voltage is shared between elements in series; elements with more
resistance drop more voltage. Adding a resistor in series with another resistor will increase the
equivalent resistance and decrease the current.
▪
Elements that are in parallel are on different paths, so current is divided between the elements.
However, the potential in each path is the same. In parallel, reciprocals of the resistances add up.
37
Voltage is the same in parallel. Current is split between the elements. More current will go down the
path with less resistance. Adding a resistor in parallel with another resistor will decrease the equivalent
resistance and increase the current in the circuit.
▪
In the circuit below, R1 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.
Apply Kirchoff’s Voltage Law to series circuits.
Kirchhoff’s Voltage (Loop) Law
In a closed loop, the vector sum of the potential differences across each
element is 0.
▪
▪
Current flows from – to + in the source and + to – through the rest of the
circuit. V across the source is positive and negative across the loads.
Kirchhoff’s Voltage Law is consequence of conservation of energy.
Example C: Appreciate the circuit on the right.
a) Find the current in the circuit.
b) Find the voltage drop across each resistor.
All 3 resistors are in series, so they all have the same current. The
current is found by dividing the total voltage by the equivalent
resistance. Once this current is found, it can be used to find the
voltage across each resistor using Ohm’s Law.
*The voltages (3V+18V+9V) add up the battery voltage of 30 V, satisfying Kirchoff’s Loop Law.
38
Circuit Analysis (Current leaves the long side of a
battery) – The battery supplies 30 V to the current
leaving it. When the current gets to the 6Ω resistor, it
uses up 18V, leaving it with 12 V. Current is never
used up so it still have 3 A. The current then loses 9V
in the 3Ω resistor, leaving it with 3V. The 1Ω uses up
the remaining voltage, leaving the current with no
potential.
*Whenever current goes back into the battery, it will
not have any voltage left. All voltage must be used up
in a circuit*
When the current goes back through the battery, it will gain the voltage back and the circuit will repeat.
Batteries do eventually run out of potential over time. (a 9V will eventually have 8V, then 7V,etc.)
*Voltage is potential difference. Therefore, the battery has positive voltage (since it adds voltage) and the
resistors have negative voltage (since they take volts away).
Apply Kirchoff’s Current Law to parallel circuits.
Kirchhoff’s Current (Node) Law
The sum of the current entering a node is equal to the sum of the
current exiting a node.
▪
Kirchhoff’s Current Law is consequence of conservation of charge.
Example D: The circuit to the right has 3 elements in parallel.
a) Find the current that leaves the battery.
b) Find the current in each resistor.
All 3 resistors are in parallel, so they all have the same potential (voltage). The total current can be found by
dividing the battery voltage by the equivalent current. To find the current in each resistor, divide the battery
voltage by each resistance since each resistor gets the full battery voltage.
39
The currents (10A+16A+8A) add up to the total circuit current, satisfying Kirchoff’s Current Law.
Circuit Analysis – Current leaves the battery with 34 A and carries 80
V. Whenever current splits, each branch gets all the volts that the
current was carrying before splitting (the rate of flow, current will be
different). When the currents meet up again, they once again combine,
so there is still 34 A going back to the battery with 0 V (since all 3
branches used up their voltages in the resistors). Again, the battery
will supply 80 V to the current again.
Measure internal resistance.
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.
Example A: Solve for each current in the circuit shown.
There are 3 unknowns, so 3 equations need to be set up using
Kirchoff’s laws using loops or nodes in the circuit.
Loop 1: 0 = +2 − 𝐼1 + 5𝐼2 − 15 − 3𝐼1
→ −4𝐼1 + 5𝐼2 = 13
*Since the loop goes “backwards” through I2, the conventional signs
were reversed in the set-up for that section.
Loops 2: 0 = +12 − 2𝐼3 + 5𝐼2 − 15
→ 5𝐼2 − 2𝐼3 = 3
Node: 𝐼1 + 𝐼2 + 𝐼3 = 0
Solving the equations using a solver: 𝑰𝟏 = −𝟐 𝑨, 𝑰𝟐 = 𝟏 𝑨, 𝑰𝟑 = 𝟏 𝑨
*I1 being negative means current flows opposite of the defined direction.
40
Example B: Solve the circuit shown.
To solve a circuit (find the voltage and current
across each element), 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. You cannot do this in one
step, so you need to combine things step by step.
Step 1
-The 2Ω and 8Ω resistors combine in series (2+8)
-The 12Ω, 15Ω, and 10Ω resistors combine in parallel
(1/(1/12+1/15+1/10))
Step 2
-The 40Ω and 10Ω resistors combine in parallel (1/(1/40+1/10))
-The 4Ω and 6Ω resistors combine in series (4+6)
Step 3
-The 7Ω and 8Ω resistors combine in series (7+8)
Step 4
-The 15Ω and 10Ω resistors combine in parallel (1/(1/15+1/10))
41
Step 5 (Equivalent Circuit)
-The 3Ω and 6Ω resistors combine in series
-The equivalent current is 9Ω
-The battery current is 5 A.
Much like Bilbo Baggins, we have to go there and back again, so we now go backwards through the
steps to the original circuit. If items are in series when going backwards, they have the same current. If
they are in parallel, they have the same voltage drop. We can then use ohm’s law to find the other
quantity for each element.
Back to Step 4
-The 3Ω and 6Ω were in series so they both have the same current, 5A.
*The voltages are found using Ohm’s Law (V=IR). The calculated
voltages (30+15) add up to 45 V, satisfying Kirchoff’s Loop Law.
Back to Step 3
-The 6Ω resistor splits in parallel, so the two resistors have the same
voltage, 30 V.
*The current is found using Ohm’s Law (I=V/R). The calculate currents
(2+3) add up to 5A, satisfying Kirchoff’s Node Law.
Back to Step 2
-The 15Ω resistor splits in series, so its current is the same for the 7Ω
and 8Ω resistors it splits into.
Back to Step 1
-The 8Ω resistor splits in parallel, so the two resistors have the
same voltage, 16V, that the 8Ω resistor had.
- The 10Ω resistor splits in series, so its current is the same for
the 4Ω and 6Ω resistors it splits into.
42
Back to Original Circuit (Answer)
-The 4Ω resistor splits in parallel, the resistors it splits into all have 12 V.
-The 10Ω resistor splits in series, so its current is kept for call elements.
Example B: Appreciate the circuit on the right.
a) Calculate current in bulb A with the switch closed.
b) Which lightbulb is the brightest? Justify your answer.
c) Switch S is opened. Describe what happens to the
brightness of each bulb.
a) The current through A is solved by finding the
equivalent circuit and working backwards.
b) A is the brightest bulb since it gets all the battery current. B and C have to split the current.
c) If the switch is open:
o A gets dimmer. The equivalent resistance goes up, so the total current in the circuit goes down.
o B gets brighter. Even though the current in the circuit decreases, B now gets all the current instead of
splitting it with C.
o C goes out since it is no longer part of the circuit.
43
AP Physics C: Electricity & Magnetism
9.2 Current, Resistance, and Power (Chapter 28)
Unit 9 - Circuits
Focus Question: How does current move through a conductor?
▪
Current is any motion of charge within a conductor. It can be expressed at the amount of charge flowing
through a cross section of a conductor per unit time:
𝒅𝑸
𝑰=
𝒅𝒕
Example A: How much charge passes through a conductor in from t = 0 s to = 5.0 s if the current is given by
𝐼(𝑡) = 0.5sin(0.2𝑡)?
𝐼=
5
𝑑𝑄
→ 𝑄 = ∫ 𝐼𝑑𝑡 = ∫ . 5 sin(.2𝑡)
𝑑𝑡
0
5
→ 𝑄 = (−2.5 cos(. 2𝑡)| = 1.15 𝐶
0
Drift Velocity
▪
Current can be expressed in terms of drift velocity.
𝐼=
𝑄
𝑡
=
𝑛𝑒𝐴𝑑
𝑑/𝑣
→ 𝑰 = 𝒗𝑨𝒏𝒆
n – charge volume density
v – electron drift velocity
Example B: The cross-sectional area of a wire is 2.5𝑥10−6 𝑚2 . The density of free electrons is 𝑛 = 5.0𝑥1028
electrons/m3. A charge of 60 C passes through the wire in 30 minutes. What is the drift velocity of the electrons
in the wire? How long would it take the electrons to go from a light switch to a bulb 20 m away?
Electron drift velocity is very slow even though electrons move extremely fast. This is because electrons move
very erratically, so they mostly move around randomly, but they move on average in the direction of current
very slowly. As shown in the example, it takes a long time for a single electron to get anywhere, but the entire
circuit starts moving at the same time once a circuit is closed.
44
Resistivity
▪
Resistivity – Resistivity is a property of the material. in metals, it is the ratio of electric field to current
density is constant:
𝝆=
𝑬
𝑱
J – current density
*A perfect conductor has zero resistivity. A perfect insulator has infinite resistivity.
*Resistivity is the reciprocal of conductivity. Conductivity = J/E
▪
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
𝜌𝑑𝐿
𝑹=
*𝑑𝑅 = 𝐴
𝑨
R - resistance
L – length of conductor
A – cross sectional area of conductor
Example C: A copper wire (𝜌 = 1.72𝑥10−8 Ωm) has a cross section of area 5.0 mm2 and carries a current of
5.0 A. The density of free electrons in the wire is 1.0𝑥1029 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠/𝑚3 .
a) How many electrons pass through cross section of the wire per unit time?
b) What is the current density in the wire?
c) What is the drift velocity of the electrons?
d) If the wire is 10 cm long then what is its resistance?
e) What is the electric field in the wire?
45
Example D: A coaxial cable consists of two concentric cylindrical conductors. The region between the
conductors is filled with a plastic material. Current leakage through
the plastic, in the radial direction, is unwanted. The radius of the
inner conductor is a a= 0.5 cm, the radius of the outer conductor is
b = 1.0 cm, and the length is L = 10 cm. The resistivity of the plastic
is 1.0𝑥1013 Ω𝑚.
𝑅=
𝜌𝐿
𝜌𝑑𝑟
𝜌𝑑𝑟
→ 𝑑𝑅 =
=
𝐴
𝐴
2𝜋𝑟𝐿
𝑏
𝑏
𝜌
𝜌
𝑑𝑟
→𝑅=∫
𝑑𝑟 =
∫
2𝜋𝑟𝐿
2𝜋𝐿 𝑟
𝑎
𝑅=
𝑎
𝜌
𝜌
𝑏
1.0𝑥1013 Ω𝑚
. 01 𝑚
𝑏
ln|𝑟| =
ln | | =
ln |
| = 𝟏. 𝟏𝒙𝟏𝟎𝟏𝟑 Ω
𝑎 2𝜋𝐿
2𝜋𝐿
𝑎
2𝜋(.10 𝑚)
. 005 𝑚
A high resisitance is desirable going across the coaxial cable to prevent current leakage.
▪
Resistivity and Temperature
∆𝝆 = 𝝆𝟎 𝜶∆𝑻
→ 𝝆 = 𝝆𝟎 (𝟏 + 𝜶∆𝑻)
𝛼 −temperature coefficient of resistivity, fractional change of resistivity per degree
∆𝜌/𝜌0 – fractional change in resistivity
Since resistance varies directly with resistivity, the following is also true: ∆𝑹 = 𝑹𝟎 𝜶∆𝑻
*Resistivity of a semiconductor decreases rapidly with temperature.
Example E: Another copper wire (𝛼 = .00393 /℃) has a resistance of . 01 Ω at 20℃. What is it resistance at
100℃?
1
∆𝑅 = 𝑅0 𝛼∆𝑇 → 𝑅 = 𝑅0 (1 + 𝛼∆𝑇) = (.01 Ω)(1 + (. 00393 ℃) (80 ℃) =. 𝟎𝟏𝟑 Ω
Power
▪
A charge moves through an element as shown. The electric field does work
on the charge as it passes through:
𝑾 = 𝑽𝒒
▪
Power
𝑷𝒐𝒘𝒆𝒓 =
𝑾𝒐𝒓𝒌 𝑽𝒒
𝒒
=
= 𝑽( )
𝒕𝒊𝒎𝒆
𝒕
𝒕
▪ 𝑷 = 𝑽𝑰
46
▪
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
Example F: Describe why a lightbulb will be less bright over time.
Over time, a lightbulb will get hotter, which will increase the resistance of the bulb. As the resistance increase,
the current will decrease. Less current will mean less power and therefore less brightness in the bulb.
Example G: A 90 W stereo is connected to a 120 V source. If energy cost $.10 per kilowatt-hour, find how
much it will cost to leave the stereo on for two days straight playing nothing but “Africa” by Toto on repeat.
Rate your understanding: More Circuits
0
1
I have a high
resistance to physics.
I can solve DC circuits
with help.
2
I can solve DC circuits
with only a few errors.
3
I can solve DC circuits
with no errors.
4
Unlimited power!
47
AP Physics C: Electricity & Magnetism
9.3 RC Circuits (Chapter 29)
Unit 9 - Circuits
Focus Question: How does a capacitor work in a circuit?
▪ 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 the resistor to the battery.
▪ Charge flows in the capacitor until the voltage in the capacitor
equals V, then no more current flows in the circuit.
Charging capacitor – The charge on the capacitor is initially zero.
o Voltage across the resistor: 𝑉𝑅 (𝑡) starts at the V (battery voltage) and decays to zero.
o Current, dq/dt, starts at V/𝑅 and decays to 0.
o Voltage across the capacitor: 𝑉𝐶 (𝑡) starts 0 (acts like a wire) and ends at 𝑉 (acts like a short circuit)
By Kirchoff’s Law:
𝑉 − 𝑉𝑅 − 𝑉𝑐 = 0
𝑞
𝑑𝑞
𝑞
𝑉 − 𝐼𝑅 − = 0 → 𝑉 =
𝑅+ =0
𝐶
𝑑𝑡
𝐶
𝑑𝑞
𝑑𝑞
→ 𝐶𝑉 =
𝑅𝐶 + 𝑞 → − 𝑅𝐶 = 𝑞 − 𝐶𝑉
𝑑𝑡
𝑑𝑡
𝑑𝑞
1
Separating the variables: 𝑞−𝐶𝑉 = −𝑑𝑡 𝑅𝐶
The charge on the capacitor goes from 0 to the charge at some time, Q(t).
𝑄(𝑡)
𝑡
𝑑𝑞
1
∫
=∫ −
𝑑𝑡
𝑞 − 𝐶𝑉
𝑅𝐶
0
0
𝑡
𝑄(𝑡)
ln|𝑞 − 𝐶𝑉|
=−
0
𝑅𝐶
𝑡
𝑄(𝑡) − 𝐶𝑉
𝑡
ln|𝑄(𝑡) − 𝐶𝑉| − ln|−𝐶𝑉| = −
→ ln |
|=−
𝑅𝐶
−𝐶𝑉
𝑅𝐶
𝑡
𝑡
𝑄(𝑡) − 𝐶𝑉
= 𝑒 −𝑅𝐶 → 𝑄(𝑡) = 𝐶𝑉 (1 − 𝑒 −𝑅𝐶 )
−𝐶𝑉
Dividing the result by the capacitance, the voltage across the capacitance as a function of time can be found:
𝒕
𝑽𝑪 (𝒕) = 𝑽 (𝟏 − 𝒆−𝑹𝑪 )
𝑡
This is also commonly written as 𝑉𝐶 (𝑡) = 𝑉 (1 − 𝑒 −𝜏 ), where 𝜏 is the time constant, where 𝝉 = 𝑹𝑪.
The voltage in the resistor acts opposite the capacitor as it drops to zero by the opposite function:
𝒕
𝑽𝑹 (𝒕) = 𝑽𝒆−𝑹𝑪
𝑽
𝒕
The current also drops to zero: 𝑰(𝒕) = 𝑹 𝒆−𝑹𝑪
48
Graphs for to charging circuit:
Example A: Power is supplied to the circuit on the right
a) What steady current flows in the circuit on the right through the 4
ohm resistor? Justify your answer?
b) What is the charge stored in the capacitor?
a) At steady state, current simply stops flowing through the branch with the capacitor, so the circuit will
be a simple loop through the two resistors.
120 𝑉
𝐼 = 4 Ω+8Ω = 𝟏𝟎 𝑨
b) The capacitor is in parallel with the 8 ohm resistor, so it will only have the voltage of that resistor.
8Ω
The voltage of the 8 ohm capacitor is 120 𝑉 (4 Ω+8 Ω) = 80 𝑉
1
1
𝑈 = 2 𝐶𝑉 2 = 2 (6𝑥10−6 𝐹)(80 𝑉)2 =. 𝟎𝟐 𝑱
Discharging Circuit – Suppose the switch in the RC circuit is left closed for a
long time and then opened.
o 𝑉𝑅 (𝑡): starts at the 𝑉𝐶 and decays to zero.
o Current, dq/dt, starts at 𝑉𝐶 /𝑅 and decays to 0.
o 𝑉𝐶 (𝑡): starts at 𝑉𝐶 and decays to zero.
By Kirchoff’s Law:
𝑉𝑐 + 𝑉𝑅 = 0
𝑞 𝑑𝑞
𝑞
𝑑𝑞
+
𝑅=0→
=−
𝐶 𝑑𝑡
𝑅𝐶
𝑑𝑡
Separating the variables:
𝑑𝑞
𝑞
1
= − 𝑅𝐶 𝑑𝑡
The charge on the capacitor starts at its maximum value, Q, and goes to Q(t):
→ ln|𝑞|
𝑄(𝑡) 𝑑𝑞
∫𝑄
𝑞
𝑡
1
= ∫0 − 𝑅𝐶 𝑑𝑡
𝑡
𝑡
𝑡
𝑡
𝑄(𝑡)
𝑡
𝑄(𝑡)
𝑄(𝑡)
=−
→ ln|𝑄(𝑡)| − ln|𝑄| = −
→ ln |
|=−
→
= 𝑒 −𝑅𝐶 → 𝑄(𝑡) = 𝑄𝑒 −𝑅𝐶
𝑄
𝑅𝐶
𝑅𝐶
𝑄
𝑅𝐶
𝑄
49
𝒕
The voltage across the capacitor is then 𝑽𝑪 (𝒕) = 𝑽𝒆−𝑹𝑪
𝒕
The capacitor acts as the battery for the resistor, so 𝑽𝑹 (𝒕) = 𝑽𝒆−𝑹𝑪
𝑽
𝒕
The current also drops asymptotically to zero: 𝑰(𝒕) = 𝑹 𝒆−𝑹𝑪
Graphs for to discharging circuit:
Example B: A circuit with a 9 V battery of negligible internal
resistance is set up with 2 resistors in series with a 10 𝜇𝐹 capacitor.
Two switches exist in the circuit.
a) S1 is closed at time t = 0 s. S2 is kept open.
i. Calculate the time constant.
ii. Calculate the time it would take the capacitor to reach
a potential differenct of 6 V
b) Once the capactior is fully charged, S2 is closed. S1 is opened.
i. Calculate the new constant.
ii. Find the current in the 1 kΩ as a function of time.
a) i. The time constant is based only on the resistors directly in series with the capacitor, in this, both resistors
are in series with the resistor.
𝜏 = 𝑅𝐶 = (3000 Ω + 1000 Ω)(10𝑥10−10 𝐹) = 4𝑥10−6 Ω𝐹
ii. 𝑉𝑐 (𝑡) = 𝑉(1 − 𝑒 −𝑡/𝜏)
→ 6 𝑉 = (12 𝑉) (1 − 𝑒
→ 0.5 = 𝑒
−
𝑡
4𝑥10−6
−
𝑡
4𝑥10−6 Ω𝐹
→ ln|. 5| = −
)
𝑡
→ 𝑡 = − (4𝑥10−6 ) ln|. 5| = 2.8𝑥10−6 𝑠
4𝑥10−6
b) i. Now, only the 1 kΩ resistor is in series with the capacitor.
𝜏 = 𝑅𝐶 = (1000 Ω)(10𝑥10−10 𝐹) = 1𝑥10−6 Ω𝐹
ii. Even though only the kΩ resistor is used for the time constant, the starting current is based on the steadystate current when both resistors were in series:
𝑉
𝑡
9𝑉
𝐼(𝑡) = 𝑅 𝑒 −𝜏 → 𝐼(𝑡) = 4000 Ω 𝑒
−
𝑡
1𝑥10−6
𝐼(𝑡) = .00225𝑒 −1000000𝑡
50
Example C: Here’s a physics problem. V = 100 volts; C1 = 12
microfarads; C2 = 24 microfarads; R = 10 ohms. Initially, Cl and
C2 are uncharged, and all switches are open.
a) First, switch S1 is closed. Determine the charge on Cl when
equilibrium is reached.
b) Next S1 is opened and afterward S2 is closed. Determine the
charge on C1 when equilibrium is again reached.
c) For the equilibrium condition of part (b), determine the
voltage across C1.
d) How much energy is lost in the wires connecting the capacitors while steady is being reached?
a) The total charge once 𝐶1 is charged is 𝑄 = 𝐶1 𝜀 = (12𝑥10−6 )(100 𝑉) = 𝟏. 𝟐 𝒎𝑪
b) At equilibrium the voltage on each is equal.
𝑄1 𝑄2
𝑄1
𝑄2
=
→
=
𝐶1 𝐶2 12 𝜇𝐹 24 𝜇𝐹
→ 𝑄2 = 2𝑄1
The 1.2 mC charge is split between the capacitors:
𝑄1 + 𝑄2 = 1.2𝑥10−3 𝐶
Solving the system: 𝑸𝟏 = 𝟒𝒙𝟏𝟎−𝟒 𝑪, 𝑸𝟐 = 𝟖𝒙𝟏𝟎−𝟒 𝑪
𝑉1 = 𝑉2 →
c) 𝑉 =
𝑄1
𝐶1
4𝑥10−4 𝐶
= 12𝑥10−6 𝐹 = 33 𝑉
1
d) The energy initially stored in C1 is: 𝑈 = 2 (12𝑥10−6 )(100 𝑉)2 = .06 J
1
The energy in C1 after steady state: 𝑈1 = (12𝑥10−6 )(33 𝑉)2 = .013 J
2
1
The energy in C2 after steady state: 𝑈2 = 2 (24𝑥10−6 )(33 𝑉)2 = .013 J
Energy lost = . 06 𝐽 − (.013 𝐽 + .013 𝐽)
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AP Physics C
Electricity & Magnetism
Unit 10: Magnetism
Section 10.1 – Magnetic Force on a Moving Charge………………………..………….………53
Section 10.2 – Magnetic Force on a Conductor………………………………….……………..57
Section 10.3 – Biot-Savart Law………………………………….…………………………..…60
Section 10.4 – Ampere’s Law………………………………….…………………………...…..63
52
AP Physics C: Electricity & Magnetism
10.1 Magnetic Force on a Moving Charge (Chapter 29)
Unit 10 - Magnetism
Focus Question: What causes a magnetic field?
▪
Laws of Magnetism - Magnets have 2 poles. Like poles repel, opposite poles attract.
▪
Cause of Magnetism - A charge in motion creates a magnetic field in the space around it.
*Electric fields describe interaction between charges at rest. Magnetic fields describe interactions between
charges in motion.
▪
▪
▪
▪
Ferro-magnet – naturally occurring magnet caused by clusters of atoms with electrons spinning in the
same direction.
Magnetic Field – a region of space surrounding a magnet in which magnetic forces are applied to other
magnets.
All magnets are dipoles, they have a north and a south.
Like poles repel; opposites attract - Magnetic field lines form closed loops.
By convention, field lines exit north and enter south.
*The earth is a magnet. There is a south pole at the north pole, so the north pole of a bar magnet is
attracted to the earth’s geographic north.
Solve problems involving a charge particle moving in an electric field.
▪
▪
▪
▪
A moving charge sets up a magnetic field around it.
The magnetic field exerts a force on a moving charge in a field.
Field direction: Direction of the magnetic force is perpendicular to both the
magnetic field and the direction of the moving charge; the magnetic force does no
work on a moving charge.
The right hand rule is used for positive particles (left hand for negative)
▪
Force:
53
Example A: A uniform magnetic field 𝐵 = 0.20 𝑇 points in the positive y-direction. If 𝑞 = 3.5𝑥10−6 𝐶 and the
speed of the charge is 8.0𝑥107 𝑚/𝑠 (making a 45o degree angle in the -x, +z as shown) then what is the
magnitude and direction of the force on the charge?
⃑⃗
𝐹 = 𝑞𝑣⃗ 𝑥 𝐵
𝑣⃗ = (− (8𝑥107
𝑚
𝑚
) cos 45°) 𝑖̂ + ((8𝑥107 ) sin 45°) 𝑘̂
𝑠
𝑠
𝐵 = (.2 𝑇)𝑗̂
𝑖̂
𝑗̂
𝑘̂
√2
√2
⃑⃗ = (3.5𝑥10−6 𝐶)|
𝐹 = 𝑞𝑣⃗ 𝑥 𝐵
−8𝑥107
0 8𝑥107 |
2
2
0
0.2
0
̂) 𝑵
𝑭 = (−𝟒𝟎 𝒊̂ − 𝟒𝟎𝒌
Example B: Determine the direction of the unknown variable for an electron using the coordinate axis given.
a)
b)
c)
Since an electron is negatively solved, each of these can be found using the left-hand rule:
a) The magnetic force is out of the page (+z direction).
b) The magnetic field is into the page (-z direction).
c) The velocity is the right (+x direction).
Magnetic Force and circular motion
▪
Since force is always perpendicular to velocity, the magnetic force does no work and the speed of the
charged particle is constant. The direction of velocity changes, but not the magnitude.
The magnetic force keeps the charged particle in the circle, supplying the
centripetal force .
𝐹𝑀𝐴𝐺𝑁𝐸𝑇𝐼𝐶 = 𝐹𝐶𝐸𝑁𝑇𝑅𝐼𝑃𝐸𝑇𝐴𝐿
𝑚𝑣 2
𝐵𝑞𝑣 =
𝑟
𝑚𝑣
→ 𝐵𝑞 =
𝑟
54
Example C: A proton moves in a circular orbit of radius 0.5 m perpendicular to a uniform magnetic field of
magnetic 0.3 T. Find the period of the proton’s orbit. *the mass of a proton = 1.67𝑥10−27 𝑘𝑔
𝐹𝐵 = 𝐹𝑐 → 𝐵𝑞𝑣 =
𝑇=
𝑚𝑣 2
𝐵𝑞𝑟
→𝑣=
𝑟
𝑚
2𝜋𝑟 2𝜋𝑟 2𝜋𝑚
2𝜋(1.67𝑥10−27 𝑘𝑔
= 𝐵𝑞𝑟 =
→𝑇=
= 𝟐. 𝟐𝒙𝟏𝟎−𝟕 𝒔
𝑣
𝐵𝑞
(0.3 𝑇)(1.6𝑥10−19 𝐶)
𝑚
The Lorentz Force
▪
A charge moving in the presence of both a magnetic and electric field experiences both an electric and
magnetic force:
Lorentz Force – total electric and magnetic forces on a moving charge:
⃑𝑭 = ⃑𝑭𝑬 + ⃑𝑭𝑩
⃑𝑭 = 𝒒𝑬
⃑ + 𝒒𝒗
⃑⃑
⃑ 𝒙𝑩
Example D: A charge 𝑞 moves with velocity 𝑣 in a uniform magnetic
field 𝐵.
a) Specify the directions of the magnetic and electric forces.
b) What is the magnitude of E, the electric field, that must exist
so that the charge passes un-defelected between the plates?
a) The electric force is down since the electric field is down. A positive charge experiences a force in the
direction of the electron field.
By the right hand rule, the magnetic force is up.
b) ∑ 𝐹𝑦 = 0 = 𝐹𝐵 − 𝐹𝐸 → 𝐵𝑞𝑣 = 𝐸𝑞 → 𝑬 = 𝑩𝒗
Example E: Behold the system below. A positive charge 𝑞 with mass 𝑚 is accelerated from rest across a
potential difference V in region 2. It passes undeflected in region 2 due to an electric field. In region 3, no
electric field acts. A magnetic field of 𝐵 acts in regions 2 and 3.
a) With what speed does the particle enter region 2?
b) What is the magnitude and direction of the electric field in
region 2?
c) What is the radius of the circular path in region 3?
55
o In region 1, a potential difference does work on the charge to accelerate it from rest.
o In region 2, the particle travels undeflected due to an electric field simultaneously with the magnetic
field.
o In region 3, the particle travels in a circular motion due to the uniform magnetic field in a semicircle
until it hits a detector. The radius of the circular path causes the circular radii to be different for different
elements.
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AP Physics C: Electricity & Magnetism
10.2 Magnetic Force on a Conductor (Chapter 29)
Unit 10 - Magnetism
Focus Question: How do magnetic fields form near conductors?
▪
▪
▪
When a current-carrying conductor lies in a magnetic field, the
field exerts magnetic forces on the moving charges in the
conductor. The forces are transmitted to the conductor and the
conductor experiences a force along its length.
Direction of force: Given by right hand rule. Thumbs points in
the direction of current, fingers curl in the direction of the
magnetic field.
Magnitude of force:
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
⃑⃗𝒙𝑩
⃑⃑⃗
𝑭 = 𝑰𝑳
𝑭 = 𝑩𝑰𝑳 𝐬𝐢𝐧 𝜽
Apply 𝐵𝐼𝑙 .
Example A: A wire that is 8 m long carries a 5 A current from left to
right in the plane of this page. A 1.5 T magnitude field points up and to
the right in the plane of this page at a 60° as shown. What is the
magnitude and direction of the force on the wire?
𝐹 = 𝐵𝐼𝐿 sin 𝜃 = (1.5 𝑇)(5 𝐴)(8 𝑚) sin 60° = 𝟓𝟐 𝑵
By the right hand rule, the force is out of the page.
⃑⃗ = (24 𝜇𝑇)𝑖̂ − (8 𝜇𝑇)𝑘̂. A 1.0 m long
Example B: The Earth’s magnetic field at a certain points is given by: 𝐵
wire carries a current of 500 A in the negative y-direction. What is the force on the wire?
𝑖̂
⃑⃗ 𝑥 𝐵
⃑⃗ = (500 𝐴) |
𝐹 = 𝐼𝐿
0
24𝑥10−6
𝑗̂
𝑘̂
−1
0 |
0 8𝑥10−6
−1
0
0
𝐹 = (500 𝐴)(𝑖̂ |
−6 | − 𝑗̂ |
0 8𝑥10
24𝑥10−6
0
0
̂
−6 | + 𝑘 |
8𝑥10
24𝑥10−6
−1
|
0
𝐹 = (−.004𝑖̂ + .012𝑘̂ )𝑁
57
𝜇𝐼
*The field due to a current carrying wire is 𝐵 = 2𝜋𝑟, where is r is the distance to the wire. This will be used in C
and D and derived in section 10.3
Example C: A pair of long parallel wires are separated by a small
distance, r. If the two currents shown are established in the wires, what is
the force per unit length they exert on each other?
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 = 𝐵2 𝐼1 𝐿 →
𝑭 𝝁𝑰𝟏 𝑰𝟐
→ =
𝑳
𝟐𝝅𝒓
𝐹
𝐿
𝜇𝐼
= (2𝜋𝑟2 ) 𝐼1
Example D: A pair of long parallel wires are separated by a small
distance, r. If the two currents shown are established in the wires, what
is the force per unit length they exert on each other?
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.
𝐹 = 𝐵2 𝐼1 𝐿 →
𝐹
𝜇𝐼2
𝑭 𝝁𝑰𝟏 𝑰𝟐
=(
) 𝐼1 → =
𝐿
2𝜋𝑟
𝑳
𝟐𝝅𝒓
▪ Wires in the same direction attract each other. * “If they flow together, they go together”.
▪ Wires flowing in opposite directions repel each other.
58
Torque on a Current Loop.
▪ A magnetic couple is a pair of equal but opposite forces that are antiparallel separated from each other.
▪ Torque due to a couple:
𝜏 = (𝑓𝑜𝑟𝑐𝑒)𝑥(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑙𝑖𝑛𝑒𝑠 𝑜𝑓 𝑎𝑐𝑡𝑖𝑜𝑛)
𝜏 = 𝐹𝑏 sin 𝛼 = 𝐵𝐼𝑎 sin 𝛼
𝜏 = 𝐵𝐼𝐴 sin 𝛼
*For multiple loops: 𝜏 = 𝑛𝐵𝐼𝐴 sin 𝛼
Example E: A rectangular loop of wire that carries current 𝐼 is placed in
a uniform magnetic field 𝐵 as shown. What torque does it experience?
The current goes in opposite direction above and below the center line as
shown, causing opposing forces, allowing the loop to rotate as shown
since both torques are in the same direction.
𝑙1
𝑙1
∑ 𝜏 = 𝜏𝑜𝑛 𝑡𝑜𝑝 𝑤𝑖𝑟𝑒 + 𝜏𝑜𝑛 𝑏𝑜𝑡𝑡𝑜𝑚 𝑤𝑖𝑟𝑒 = (𝐵𝐼𝑙2 ) ( ) + (𝐵𝐼𝑙2 ) ( )
2
2
→ 𝜏 = 𝐵𝐼𝑙1 𝑙2
This is consistent with 𝜏 = 𝐵𝐼𝐴 since 𝑙1 𝑙2 is the area of the loop.
Example F: A square coil with 12 turns lies in the xy plane. Each side has a length of s = 4.0 cm and the sides
are parallel to the coordinate axes. A uniform magnetic field points in the y-direction, 𝐵 = (2𝑗̂) 𝑇. Calculate the
current needed to produce a net torque about the x-axis of 1.92x10-3 Nm.
𝜏 = 𝑛𝐵𝐼𝐴 → 𝐼 =
▪
𝜏
. 00192 𝑁𝑚
=
= .05 𝐴
𝑛𝐵𝐴 12(2 𝑇)(. 04𝑚)2
Torque on a loop of current:
𝐴 − 𝑎𝑟𝑒𝑎 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑙𝑜𝑜𝑝
𝜇 − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑚𝑜𝑚𝑒𝑛𝑡, describes how loop reacts to torque
𝜇 = 𝐼𝐴
𝜏 =𝜇𝑥𝐵
Rate your understanding: Magnetic Forces
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AP Physics C: Electricity & Magnetism
10.3 Biot-Savart Law (Chapter 30)
Unit 10 - Magnetism
Focus Question: How can the magnetic field due to a current-carrying wire?
▪
Magnetic Field Formula – B is perpendicular to the plane containing 𝑣 and 𝑟 (the vector joining the
source point to the field point):
⃑𝑩
⃑ =
⃑ 𝒙𝒓̂
𝝁𝟎 𝒒𝒗
𝟒𝝅 𝒓𝟐
𝜇0 = magnetic permittivity of free space
𝜇0 = 4𝜋𝑥10−7 Tm/A
𝑟
𝑟̂ = |𝑟|:unit vector in the direction of r.
*B is parallel to 𝑣𝑥𝑟̂
▪
▪
The Biot-Savart Law is method used to find the magnetic field due some amount of length of a currentcarrying wire.
The magnetic field produced at any field point by the current in a conductor is the vector sum of the
fields due to all the moving charges in the conductor. The Biot-Savart Law uses this principle with many
tiny elements of length dL:
Volume of the segment: AdL
# 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒𝑠
Charge in segment: 𝑑𝑞 = (𝑣𝑜𝑙𝑢𝑚𝑒) ( 𝑣𝑜𝑙𝑢𝑚𝑒 ) (𝑞) = 𝐴𝑛𝑞𝐷𝐿
𝜇0 𝑑𝑄𝑣 sin 𝜃
𝑑𝐵 =
4𝜋𝑟 2
(𝐼 = 𝐴𝑛𝑣𝑞)
𝜇0 𝐼𝑑𝑙 sin 𝜃
𝑑𝐵 =
4𝜋𝑟 2
The Biot-Savart Law:
⃑⃑⃑⃑⃑⃑ =
𝒅𝑩
𝝁𝟎 𝑰
⃑⃑⃑⃑ 𝒙 𝒓̂)
(𝒅𝒍
𝟒𝝅𝒓𝟐
*or*
𝝁 𝑰
⃑⃑⃑⃑⃑⃑ = 𝟎 (𝒅𝒍
⃑⃑⃑⃑ 𝒙 𝒓
⃑)
𝒅𝑩
𝟒𝝅𝒓𝟑
*The magnetic field produced by current in a conductor is the vector sum of all moving charges in the
conductor.
*d𝑙⃑ is a vector in the same direction as the current in the conductor.
*𝑟⃗ is the vector from the wire to the point where field is being calculated.
*Field lines produced by the current element are circles in planes perpendicular to dl. The direction of the field
is given by the right-hand rule.
60
Example A: Derive the magnetic field due to the current carrying loop at the center of the loop.
⃑⃑⃑⃗ .
The current carrying-loop is broken up in many tiny current elements, 𝑑𝑙
Every ⃑⃑⃑⃗
𝑑𝑙 is perpendicular to its corresponding 𝑟⃗ vector, so ⃑⃑⃑
𝑑𝑙 𝑥 𝑟̂ is
simply equal to |𝑑𝑙||𝑟̂ | sin 90° = 𝑑𝑙. This is true whenever the two
vectors are perpendicular.
𝑑𝐵 =
𝜇0 𝐼
𝜇 𝐼
𝜇 𝐼
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = ∫ 0 (𝑑𝑙
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = 0 ∫ 𝑑𝑙
(𝑑𝑙
4𝜋𝑅 2
4𝜋𝑅 2
4𝜋𝑅 2
The integral of ∫ 𝑑𝑙 is simply the circumference of the circle.
𝐵=
𝜇0 𝐼
𝝁𝟎 𝑰
(2𝜋𝑅) → 𝑩 =
2
4𝜋𝑅
𝟐𝑹
Example B: Use the Biot-Savart law to calculate the magnetic field at point P in the two cases below.
a)
b)
a) For the first straight segment, the angle between ⃑⃑⃑⃗
𝑑𝑙 and 𝑟̂ is 0°,
⃑⃑⃑
so 𝑑𝑙 𝑥 𝑟̂ |𝑑𝑙||𝑟̂ | sin 0° = 0, so there is no field.
⃑⃑⃑⃗ and 𝑟̂ is always
For the curved segment, the angle between 𝑑𝑙
⃑⃑⃑ 𝑥 𝑟̂ |𝑑𝑙||𝑟̂ | sin 90° = 𝑑𝑙, so ∫ 𝑑𝑙
⃑⃑⃑ 𝑥 𝑟̂ will simply be each to
90°, 𝑑𝑙
the distance through the semi-circle.
For the segment straight segment, the angle between ⃑⃑⃑⃗
𝑑𝑙 and 𝑟̂ is
⃑⃑⃑
0°, so 𝑑𝑙 𝑥 𝑟̂ |𝑑𝑙||𝑟̂ | sin 180° = 0.
Only the curved portion contributes to the electric field, so the Biot-Savart Law will be applied for this segment
only:
𝜇0 𝐼
𝜇 𝐼
𝜇 𝐼
𝜇 𝐼
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = ∫ 0 (𝑑𝑙
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = 0 ∫ 𝑑𝑙 = 0 (𝜋𝑅)
𝑑𝐵 =
(𝑑𝑙
4𝜋𝑅 2
4𝜋𝑅 2
4𝜋𝑅 2
4𝜋𝑅 2
→𝑩=
𝝁𝟎 𝑰
𝟒𝑹
b) Just as in part a), the field is zero in the straight segment, so the Biot-Savart
Law is applied only for the quarter circle segment:
𝜇0 𝐼
𝜇 𝐼
𝜇 𝐼
𝜇 𝐼 𝜋𝑅
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = ∫ 0 (𝑑𝑙
⃑⃑⃑⃗ 𝑥 𝑟̂ ) → 𝐵 = 0 ∫ 𝑑𝑙 = 0 ( )
𝑑𝐵 =
(𝑑𝑙
4𝜋𝑅 2
4𝜋𝑅 2
4𝜋𝑅 2
4𝜋𝑅 2 2
→𝑩=
𝝁𝟎 𝑰
𝟖𝑹
61
Example C: Consider a wire carrying a current I that extends along the y-axis from (0,-a)
to (0,a) . Find the magnetic field at point P, x away from the center of the wire along the xaxis.
⃑⃑⃑⃗ 𝑥 𝑟̂ ) = 𝑑𝑦 sin 𝜃,
Each current-carrying element is along the y-axis, so 𝑑𝑙 = 𝑑𝑦 and (𝑑𝑙
⃑⃑⃑ and 𝑟̂ .
where 𝜃 is the angle between 𝑑𝑙
𝑑𝐵 =
𝜇0 𝐼
⃑⃑⃑⃗ 𝑥 𝑟̂ )
(𝑑𝑙
4𝜋𝑅 2
𝑎
𝑎
𝜇0 𝐼𝑑𝑦 sin 𝜃
𝜇0 𝐼𝑑𝑦 sin 𝜃
𝐵=∫
=
2
∫
4𝜋𝑟 2
4𝜋𝑟 2
−𝑎
0
*𝑟 = √𝑥 2 + 𝑦 2 and sin 𝜃 =
𝑎
𝐵 = 2∫
0
→𝐵=
▪
𝑥
√𝑥 2 +𝑦 2
𝜇𝐼𝑑𝑦
𝑥
2𝜇0 𝐼𝑥 𝑎
𝑑𝑦
→
𝐵
=
∫
2
2
4𝜋(𝑥 + 𝑦 ) √𝑥 2 + 𝑦 2
4𝜋 0 (𝑥 2 + 𝑦 2 )32
𝜇𝐼𝑥
𝑦
𝝁𝟎 𝑰
𝒂
𝑎
(
)| → 𝑩 =
2𝜋 𝑥 2 √𝑥 2 + 𝑦 2 0
𝟐𝝅𝒙 √𝒂𝟐 + 𝒙𝟐
B for an infinitely long conductor can be found by taking the limit as 𝑎 → ∞
𝐵 = lim
𝜇0 𝐼
𝑎→∞ 2𝜋𝑥
𝑎
→𝑩=
√𝑎2 +𝑥 2
𝝁𝟎 𝑰
𝟐𝝅𝒙
where x is the distance to the wire
Example D: Find the magnetic field in the center of the rectangular loop.
The rectangle consists of 4 wires of finite length, with two pairs of wires
with identical fields. This problem can be simplified with using the result for
a wire of finite length from C.
𝐻
𝜇0 𝐼
𝐵=2
2
+2
𝐿
(
2𝜋 2 √ 𝐻 2
𝐿 2
( ) +( )
2
2
)
𝐿
𝜇0 𝐼
2
𝐻
2𝜋 2 √ 𝐻 2
𝐿 2
(
)
+
(
)
(
2
2 )
→𝐵=
2𝜇0 𝐼
𝜋𝐿𝐻
The field is into the page using the right hand rule with any of the segments of wire.
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62
AP Physics C: Electricity & Magnetism
10.4 Ampere’s Law (Chapter 30)
Unit 10 - Magnetism
Focus Question: What is Ampere’s Law used for?
▪
▪
Ampere’s Law plays the same role for magnetic fields that Gauss’s Law plays for electric fields:
➢ Gauss’s Law for electric fields relates the integral of the perpendicular component of the
electrical field over a closed surface to the total charge enclosed by the surface.
➢ Ampere’s Law for magnetic fields relates the integral of the tangential component of a magnetic
field at points on a closed curve to the current passing through the area bounded by the curve.
Consider a long straight wire carrying current I. Choose a circle perpendicular to the wire and center on
the wire as the path:
∮ 𝐵 ∙ 𝑑𝑙 is a line integral, which is evaluated by dividing the path in a bunch of 𝑑𝑙’s. The line
integral adds up 𝐵 ∙ 𝑑𝑙 for all the dl’s along the length.
𝜇 𝐼
Everywhere on the path, the magnitude of B is constant and tangent to the circle. B is equal to 𝑜
2𝜋𝑟
at all points, where is the distance to the wire.
𝜇0 𝐼
𝜇0 𝐼
∮ 𝐵 ∙ 𝑑𝑙 = 𝐵 ∮ 𝑑𝑙 = 2𝜋𝑟 ∮ 𝑑𝑙 = 2𝜋𝑟 (2𝜋𝑟)
→ 𝐵 = 𝜇0 𝐼
▪
Ampere’s Law: Ampere’s Law is used to find the magnetic field at some distance from a currentcarrying wire:
∮ 𝑩 ∙ 𝒅𝒍 = 𝝁𝟎 𝑰𝒆𝒏𝒄
*Chose an integration path that is a closed loop that passes through the point where you want to calculate the
magnetic field. If the integration path is a circle, then ∮ 𝑑𝑙 = 2𝜋𝑟.
Gauss’s Law used an imaginary surface that enclosed the charge the electric field due to that charge. Ampere’s
Law involves drawing an imaginary loop, or path, around the current that the magnetic field is being calculated
for. As with Gauss’s law, the Amperian path chosen needs to have appropriate symmetry.
Example A: Field around a long straight conductor.
An Amperian circle or radius is drawn around the conducting wire. The circle encloses all the
current of the wire:
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
→ 𝐵 ∮ 𝑑𝑙 = 𝜇0 𝐼 → 𝐵(2𝜋𝑟) = 𝜇0 𝐼
→𝑩=
𝝁𝟎 𝑰
𝟐𝝅𝒓
FB is constant and tangent to the circle at all points, so ∮ 𝐵 ∙ 𝑑𝑙 simply equal to B times the length of the
Amperian path, which is common for most cases in which Ampere’s Law is applied in Physics C.
63
Example B: Field inside a long straight conductor.
The Amperian path is again a circle. Since the field is to be found inside the conductor, the
path is chosen with r < R. The path only encloses part of the current, so the current enclosed
will be: 𝐼𝑒𝑛𝑐 = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 𝑎𝑟𝑒𝑎 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 → 𝐼𝑒𝑛𝑐 = 𝐽𝐴
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐 → 𝐵(2𝜋𝑟) = 𝜇0 𝐽𝐴𝑒𝑛𝑐
→ 𝐵(2𝜋𝑟) =
𝜇0 𝐼
𝜇0 𝑰𝒓
(𝜋𝑟 2 ) → 𝑩 =
2
𝜋𝑅
𝟐𝝅𝑹𝟐
Graph for the field due to a current-carrying as a function of distance from the center of the wire:
Example C: Field inside a solenoid.
▪
A solenoid is a helical winding of wire around a cylinder. The turns are closely spaced so that each can
be considered as a circle. When the solenoid is long compared to its cross-sectional area, the field near
the center can be considered uniform. The external field is small. Ampere’s Law for a solenoid is used to
find the internal field.
An Amperian circle would have no symmetry here, so the
Amperian path is a rectangle as shown. For the long part of
the rectangle running through the center of the solenoid,
∮ 𝐵 ∙ 𝑑𝑙 = 𝐵 ∮ 𝑑𝑙 since B is parallel to the field and constant
inside. For the long section outside the solenoid, the field is
negligible. For the two widths of the rectangle, B is
perpendicular to the field, so ∮ 𝐵 ∙ 𝑑𝑙 = 0.
Ampere’s Law is applied only for the long segment inside the
solenoid:
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
The current enclosed is NI, where N is the number of loops in the solenoid. Each loop of wire carries the
current, so each loop enclosed will enclose the current.
→ 𝐵 ∫ 𝑑𝑙 = 𝜇0 𝑁𝐼 → 𝐵𝐿 = 𝜇0 𝑁𝐼
𝝁 𝑵𝑰
→ 𝑩 = 𝟎𝑳
The field for a solenoid is also written as 𝐵 = 𝜇0 𝑛𝐼, where n is the number of loops per unit length: n=N/L.
64
A toroid, or toroidal solenoid is formed by winding wire around a ring. It is essential a
solenoid bent into a circle so both ends meet. A toroid has all of its field within the region
between the wires. Since current flows in opposite directions on the inner and outer parts
of the region, the total current enclosed outside of the toroid is zero.
Example D: Field inside a toroid.
Find the field in the following regions:
a) r < a
b) r > b
c) a < r < b
For all parts, the Amperian path will be a circle in order to have the appropriate symmetry with the circular
toroid.
a) The current enclosed by the path is zero.
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐 → 𝐼𝑒𝑛𝑐 = 0
→𝐵=0
b) As with the solenoid, the toroid encloses the current on the inner part N times, where N is the number of
loops.
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐 → 𝐵 ∮ 𝑑𝑙 = 𝜇0 𝑁𝐼
→ 𝐵(2𝜋𝑟) = 𝜇0 𝑁𝐼
𝜇0 𝑁𝐼
→𝐵=
2𝜋𝑟
c) As shown on the diagram, the current flows into the page on the inner part of the toroid, and out of the page
on the outer surface, so every loop contains current flowing into and out of the page, making the total current
enclosed by the Amperian loop zero.
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐 → 𝐼𝑒𝑛𝑐 = 0
→𝐵=0
65
Example E: Coaxial Cable
▪
A coaxial conductor consists of a cylindrical conductor or radius R1 that carries current I into the page
and an outer cylindrical conductor (of inner radius R2 and outer radius R3) that carries current I out of
the page.
Find the field for:
a) R1< r < R2
Only the current on the inner conductor is enclosed:
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
→ 𝐵 ∮ 𝑑𝑙 = 𝜇0 𝐼 → 𝐵(2𝜋𝑟) = 𝜇0 𝐼
→𝑩=
𝝁𝟎 𝑰
𝟐𝝅𝒓
b) R > R3
The currents flow in opposite direction.
∮ 𝐵 ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
→ 𝐵 ∮ 𝑑𝑙 = 𝜇0 (𝐼 − 𝐼)
→𝑩=𝟎
*Since the field is kept within the wire, coaxial cables are effective for carrying weak signals that would not
interact well with outside signals.
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AP Physics C
Electricity & Magnetism
Unit 11: Electromagnetism
Section 11.1 – Magnetic Flux…………………………….……….………………….…...……68
Section 11.2 – Faraday’s Law………………………………………………………………..…70
Section 11.3 – Motors & Inductance……………………….……….………………..…………75
Section 11.4 – Inductance…………………………….………………….…………….……….76
Section 11.5 – Maxwell’s Equations………………………………………………….………..83
67
Unit 11 – Electromagnetic Induction
AP Physics C: Electricity & Magnetism
11.1 Magnetic FLUX! (Chapter 30)
Focus Question: What is magnetic flux?
▪
Magnetic Flux – 𝜙, Amount of magnetic field that goes through a surface.
⃑ ∙ ⃑⃑⃑⃑⃑
𝜙 = ∫𝐵
𝑑𝐴 = 𝐵𝐴 cos 𝜃
Units: 1 weber (Wb) = 1 T𝑚2
𝑑𝜙
*Field vs. flux - 𝑑𝜙 = 𝐵⊥ 𝑑𝐴 → 𝐵⊥ = 𝑑𝐴 . The magnetic field equals the flux per unit area, so the
magnetic field is sometimes called “flux density”.
▪
Gauss’s Law for Magnetism concerns magnetic flux through a closed surface:
Gauss’s Law for Magnetism is true since there are no magnetic monopoles.
Example A: A uniform magnetic field |𝐵| = 0.80 𝑇 acts in the positive y-direction.
a) Calculate the flux through surface ABDE.
b) Calculate the flux through surface ABCF.
a) The flux and dA are in opposite directions since the flux to in the
+y to the right, and dA is out of the surface, which is to the left:
𝜙 = 𝐵𝐴 cos 180° = −(. 8 𝑇)(3.5 𝑚)(4 𝑚) = −11.2 𝑊𝑏
b) dA points directly out of the surface as shown. The field is in the positive ydirection:
3
𝜙 = 𝐵𝐴 cos 𝜃 = (.8 𝑇)((3.5 𝑚)(5 𝑚) = 8.4 𝑊𝑏
5
Example B: A uniform 2.0 T magnetic field passes through a circular loop with a radius of 0.2 m that has 4
turns. Calculate the flux through the loop.
𝜙 = 𝐵𝐴 = (2 𝑇)(𝜋(. 2 𝑚)2 )𝑥4 = 1 𝑊𝑏
68
Example C: An infinitely long conducting wire carries a current of I
and is a distance of 𝑎 from a rectangular surface as shown. Derive an
expression for the magnetic flux through the rectangle.
The field is not constant through the loop, so it is broken into many
dA’s along its length in the x-direction since field varies in x. Each dA
has a thickness of dx and an area of hdx. .
𝜇 𝐼
0
The field in each dA is constant vertically and is equal to 2𝜋𝑟
, where r is
the distance to the wire.
𝑎+𝑤
𝑎+𝑤
𝑎
𝑎
𝜇0 𝐼
𝜇0 𝐼ℎ
𝑑𝑥
𝜙 = ∫ 𝐵 ∙ 𝑑𝐴 = ∫
ℎ𝑑𝑥 =
∫
2𝜋𝑥
2𝜋
𝑥
→
𝜇𝑜 𝐼ℎ
𝑎 + 𝑤 𝜇0 𝐼ℎ
ln |𝑥|
=
(ln|𝑎 + 𝑤| − ln|𝑎|)
𝑎
2𝜋
2𝜋
𝝓=
𝝁𝟎 𝑰𝒉
𝒂+𝒘
𝐥𝐧 |
|
𝟐𝝅
𝒂
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AP Physics C: Electricity & Magnetism
11.2 Faraday’s Law (Chapter 31)
Unit 11 - Electromagnetism
Focus Question: What causes an induced emf?
▪
When the magnetic field through a loop of wire changes, a current is induced in the loop of wire. In the
diagram below, the electrons in the wire experience a force, and the resulting movement induces current
in the wire, even though the wire is not connected to a power source.
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:
𝐹𝐵 = 𝐹𝐸
𝑞𝐵𝑣 = 𝐸𝑞 → 𝐵𝑣 =
𝜀
𝑙
→ 𝜺 = 𝑩𝒗𝒍
*induced emf in a bar of length moving at speed in a magnetic field of B
If the bar is part of a completed circuit loop as shown, then the moving bar acts as a battery due to the induced
emf in the bar.
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
𝜙
𝑡
= 𝐵𝑣𝑙 since velocity is constant. This is equal to the emf.
Example A: A bar of lenth .20 m is moved to the right at a speed of v =4.0 m/s in a
field of magntidue B = 2.0 T as shown. The resistor has a value of 𝑅 = 2.0 Ω.
a) Find the current in the circuit.
b) Calculate the force required to keep the rod moving.
𝜀
a) 𝐼 = 𝑅 =
𝐵𝑙𝑣
𝑅
=
𝑚
𝑠
(2 𝑇)(.2 𝑚)(4 )
2Ω
= 0.8 𝐴
b) 𝐹 = 𝐵𝐼𝐿 = (2 𝑇)(. 8 𝐴)(. 2 𝑚) = .32 𝑇
70
▪
Faraday’s Law with Calculus – The induced emf due to a changing magnetic field is equal to the rate of
change of the magnetic flux through the surface bounding by the circuit:
𝜺=
𝜺=
−𝑑𝜙
𝒅𝒕
−𝑑
⃑⃑ ∙ ⃑⃑⃑⃑⃑
∫𝑩
𝒅𝑨 𝐜𝐨𝐬 𝜽
𝒅𝒕
*The minus sign indicates the magnetic force produced by the induced current opposes the force that
causes it.
▪
Ways to change flux:
o Change the magnitude of the magnitude.
o Change the size of the area the flux is passing through.
o Change the angle made by the field 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 B: 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.
71
Example C: A magnetic field has a strength given by 𝐵(𝑡) = 2𝑡 2 − 2𝑡 + 2 and acts out of the plane of this
page. A circular loop of wire of radius R = 2.0 m within the plane of this page has the
field pass through it.
a) Find the induced emf in this wire as a function of time.
b) Find the direction of the current in the wire at t = 2.0 s.
𝑑𝜙
𝑑𝐵
a) 𝜀 = − 𝑑𝑡 = −𝐴 𝑑𝑡 (A is constant and B changes with t)
→ ε = −(π)(2 )2 (4𝑡 − 2)
→ 𝜀(𝑡) = 16𝜋𝑡 2 − 8𝜋
𝑑𝑏
b) At t = 2, 𝑑𝑡 = 4(2) − 2 = 6. B is increasing since dB/dt is positive. Thus the flux is increasing out of
the page, so the change in flux is out of the page. The induced current creates a field into the page to the
induced current is clockwise.
▪ Flux Linkage
*Each loop produces its own emf, and the emfs from each loop add to the total emf.
Example D: A metal wire coil of radius 0.5 m has 1000 loops, forming a solenoid. The solenoid is placed in a
500 mT magnetic field. The field drops to 0 T in 10 s. What is the induced emf in the coil?
𝜀=−
𝑁𝑑𝜙
𝑑𝐵
0 − (500𝑥10−3 𝑇) 25𝜋
= −𝑁𝐴
= −(1000)(𝜋)(0.5𝑚)2
=
𝑉
𝑑𝑡
𝑑𝑡
10 𝑠
2
Example E: A square loop (s = 0.25 m) is placed around a solenoid of 1000 loops with a diameter of .10 m and
length of 0.2 m. The loop’s plane is perpendicular to the axis of the solenoid. The current in the Solenoid is
𝑡
given by 𝐼 = 100 (1 − 𝑒 −5 ). Calculate the induced emf in the loop as a function of time.
Using the field of a solenoid derived in 10.4: 𝜙 = 𝐵𝐴 =
→𝐵=
𝜇0 𝑁𝐼(𝑡)
𝐿
𝜋𝑟 2
𝑡
𝜇0 𝑁𝜋𝑟 2
100(1 − 𝑒 −5 )
𝐿
(4𝜋𝑥10
𝑡
𝑑𝐵
𝜇 𝑁𝜋𝑟 2
2 0
𝜀=−
=𝑠
20𝑒 −5 = (. 25 𝑚)2
𝑑𝑡
𝐿
−7 𝑇𝑚 (1000)𝜋(.
)
05
𝐴
.2 𝑚
𝑚)2
𝑡
(20)𝑒 −5
𝑡
→ 𝜀 = .001𝑒 −5
72
Example F: A conducting bar of mass M and length L is connected to two long
vertical conducting rails. The two rails are connected by a resistor of resistance R at
the top. The entire apparatus is in a magnetic field of magnitude B directed into this
page. The bar is released from rest and slides without friction down the rails.
a) What is the direction of current in the resistor?
b) Is the magnitude of the net magnetic field above the bar at point C
greater than, less than, or equal to the magnitude of the net magnetic
field before the bar is released?
c) Determine an expression for the terminal velocity of the bar.
d) Derive an expression for the power dissipating in the resistor when the bar is at terminal velocity.
e) Derive an expression for the velocity of the bar as a function of time until it reaches terminal
velocity.
a) Due to the downward motion due to gravity of the bar, the magnetic field into the
page will push positive charge in the bar to the right due to the right-hand rule,
causing a clockwise-current. The current in the resistor is to the left.
b) The induced current in the bar creates a magnetic field out of the page on top of
the bar and into the page below the bar by the right hand-rule. Above the bar, this
field goes against the existing field, so the net magnetic field in C decreases.
c) Due to the right hand rule again, the current to the right causes an upward
magnetic force. With a downward force of gravity the forces on the bar are:
∑ 𝐹 = 𝑚𝑔 = 𝑚𝑔 − 𝐹𝐵
→ 𝑚𝑎 = 𝑚𝑔 − 𝐵𝐼𝐿
The current is based on the induced emf in the bar:
𝜀
𝐵𝐿𝑣
→ 𝑚𝑎 = 𝑚𝑔 − 𝐵 ( ) 𝐿 = 𝑚𝑔 − 𝐵 (
)𝐿
𝑅
𝑅
𝐵 2 𝐿2 𝑣
→ 𝑚𝑎 = 𝑚𝑔 −
𝑅
At terminal velocity, the acceleration is zero:
𝐵 2 𝐿2 𝑣𝑇
𝐵 2 𝐿2 𝑣𝑇
0 = 𝑚𝑔 −
→ 𝑚𝑔 =
𝑅
𝑅
𝒎𝑹𝒈
→ 𝒗𝑻 = 𝟐 𝟐
𝑩 𝑳
𝜀 2
d) Power = 𝐼 2 𝑅 = (𝑅) 𝑅 =
𝐵 2 𝐿2 𝑣𝑇2
→𝑃=
=
𝑅
𝑚2 𝑅𝑔2
→𝑃= 2 2
𝐵 𝐿
(𝐵𝐿𝑣𝑇 )2
𝑅2
2
𝑚𝑅𝑔
𝐵 2 𝐿2 (𝐵2 𝐿2 )
𝑅
𝑅
𝑚2 𝑅 2 𝑔2
= 2 2
𝐵 𝐿𝑅
73
e) From c) 𝑚𝑎 = 𝑚𝑔 −
𝐵2 𝐿2 𝑣
𝑅
𝑑𝑣
𝐵 2 𝐿2 𝑣
→𝑚
= 𝑚𝑔 −
𝑑𝑡
𝑅
This is a free-fall terminal velocity question like in AP Physics C: Mechanics, just with a different drag force.
𝑅
To get v positive and without and cofficients (to make integration easy), multiply both sides by − 𝐵2 𝐿2
→−
𝑚𝑅 𝑑𝑣
𝑚𝑅𝑔
=𝑣− 2 2
2
2
𝐵 𝐿 𝑑𝑡
𝐵 𝐿
Separate the variables:
𝐵 2 𝐿2
𝑚𝑅𝑔 = − 𝑚𝑅 𝑑𝑡
𝑣−
𝑑𝑣
𝐵2 𝐿2
The velocity starts at zero and goes to some value at time, v(t).
𝑣(𝑡)
𝑡
𝐵 2 𝐿2
∫
𝑚𝑅𝑔 = ∫ − 𝑚𝑅 𝑑𝑡
𝑣−
𝑑𝑣
𝐵2 𝐿2
0
0
𝑚𝑅𝑔
𝑣(𝑡) − 𝐵2 𝐿2
𝑚𝑅𝑔 𝑣(𝑡)
𝐵 2 𝐿2
𝑚𝑅𝑔
𝑚𝑅𝑔
𝐵 2 𝐿2
𝐵 2 𝐿2
→ ln |𝑣 − 2 2 |
=−
𝑡 → ln |𝑣(𝑡) − 2 2 | − ln |− 2 2 | = −
𝑡 → ln |
|=−
𝑡
𝑚𝑅𝑔
𝐵 𝐿
0
𝑚𝑅
𝐵 𝐿
𝐵 𝐿
𝑚𝑅
𝑚𝑅
−
𝐵2 𝐿2
𝑚𝑅𝑔
→
𝑣(𝑡) − 𝐵2 𝐿2
𝑚𝑅𝑔
− 𝐵2 𝐿2
→ 𝒗(𝒕) =
=𝑒
−
𝐵2 𝐿2
𝑡
𝑚𝑅
𝑩𝟐 𝑳 𝟐
𝒎𝑹𝒈
−
𝒕
𝒎𝑹 )
(𝟏
−
𝒆
𝑩𝟐 𝑳𝟐
𝑚𝑅𝑔
By this equation, the velocity starts at 0 and increases asymptotically to 𝐵2 𝐿2, which is the terminal velocity
from c).
Rate your understanding: Faraday’s Law
0
1
I know Lenz’s Law.
I understand parts of
Faraday’s Law.
2
3
4
I can solve for induced
emf with minor errors.
I can solve for induced
emf with no errors.
I can explain and teach
Faraday and Lenz’s law.
74
AP Physics C: Electricity & Magnetism
11.3 Motors & Generators (Chapter 31)
Unit 11 - Electromagnetism
Focus Question: How is Faraday’s law applied in the modern world?
An AC generator converts mechanical energy into electrical energy.
Suppose a wire loop is made to rotate as shown on the right by some
mechanical source:
▪
As the loop rotates, the magnetic flux through the loop changes
with time:
𝜙 = 𝐵𝐴 cos(𝜔𝑡)
*𝜔 = 2𝜋𝑓 (angular frequency)
▪
The motional emf can be thus be found:
𝜀=
−𝑵∆Φ
∆𝒕
=
𝑵𝒅(𝐵𝐴 cos t)
𝒅𝒕
 = NBA sin t
▪
The emf is at a maximum when the plane of the loop is parallel to the magnetic field:
𝜀𝑃𝐸𝐴𝐾 = 𝑵𝑩𝑨
Example A: Generator - A 250-turn coil of wire has dimensions 2.5 cm by 3.2 cm. It is rotating at a
frequency of 60 Hz in a constant magnetic field having a strength of 1.5 T. What is its peak voltage 0?
𝜀0 = 𝜀𝑃𝐸𝐴𝐾 = 𝑁𝐵𝐴 = (2𝜋)(60 𝐻𝑧)(250)(1.5 𝑇)(. 025 𝑚)(. 032 𝑚) = 𝟏𝟏𝟑 𝑽
Example B: Motor - A motor contains a coil with a total resistance of 10 Ω and is supplied by a voltage of 120
V. When the motor is running at its maximum speed, the back emf is 70 V.
a) Find the current in the coil at the instant the motor is turned on.
b) Find the current in the coil when the motor has reached maximum speed.
a) When the motor is turned on, the available voltage is one 120 V – 70 V = 50 V, since the back emf
prevents the emf from jumping to the full supply value. This is because coils prevent current from
instantly jumping to their steady-state values. Back emf is due to inductance, which is the topic of 10.4
50 𝑉
𝐼 = 10 Ω = 5 𝐴
b) 𝐼 =
120 𝑉
10 Ω
= 12 𝐴
Rate your understanding: Motors & Generators
0
1
2
I understand parts of
Faraday’s Law.
I can solve for induced
emf with only
occasional errors.
3
4
I can solve for induced
emf with no errors.
I can explain and teach
Faraday and Lenz’s law.
75
AP Physics C: Electricity & Magnetism
11.4 Inductance (Chapter 32)
Unit 11 - Electromagnetism
Focus Question: What happens current runs through an inductor?
Inductance
▪
Inductor – A component that stores energy temporarily in a magnetic field. Inductors resist changes in
current.
*usually created by coiling wire into many loops. More loops = stronger magnetic field
Schematic symbol:
▪
In a circuit with an inductor, current doesn’t automatically jump to its
𝜀
max value, 𝐼 = . As current increases, magnetic flux through the
𝑅
inductor increases with time, inducing an emf in the inductor; opposite
the direction of the current.
Self-Inductance – The circuit itself produces a changing magnetic flux
and thus induces an emf due its own current.
▪
𝜺𝑳 = −𝑵
▪
𝒅𝝓
𝒅𝒕
Inductance - Property of inductor; ratio of magnetic flux to current flow.
𝑳=𝑵
𝝓
𝑰
𝜺
𝑳 = − 𝒅𝑰 → 𝜺𝑳 = −𝑳
𝒅𝒕
𝒅𝑰
𝒅𝒕
𝑑𝑣
L, inductance, is basically “electrical inertia”. The second equation above is comparable to 𝐹 = 𝑚 𝑑𝑡 .
units of inductance: Henry (H); 1 H = 1 Vs/A
*inductance can be thought of as resistance to change in current, not to current itself.
Example A: Consider the solenoid shown with current flowing in the
indicated direction.
a) Find the direction of the self-induced emf when the current is
increasing.
b) Find the direction of the self-induced emf when the current is
decreasing.
The current flows from right to left.
a) The current is to the left and increasing, the change in current is the left, inducing an emf to resist this change
to the right. The solenoid acts as an inductor, which does not let current suddenly change, so the current will
slowly start to flow to the left rather than jumping immediately when power is connect.
b) The current is to the left, but decreasing, so the change in current is the right. The solenoid will resist this
change by inducing an emf with a current to the left.
76
Example B: Calculate the self-inductance of a solenoid of radius R and length L with N loops.
𝜇 𝑁𝐼
The magnetic field in a solenoid is given by 𝐵 = 0𝐿 as derived in 10.4.
𝜇0 𝑁𝐼
2
𝑁𝜙 𝑁𝐵𝐴 𝑁 ( 𝐿 ) (𝜋𝑅 )
𝐿=
=
=
𝐼
𝐼
𝐼
𝑁 2 𝜇0 𝜋𝑅 2
→𝐿=
𝐿
*The current cancels out. Inductance is a property of the inductor and is independent of current.
▪
Energy stored in an inductor –
Consider the time when current in an inductor increases from 0 to I:
𝑑𝑖
𝟏
𝑊 = ∫ 𝑃(𝑡)𝑑𝑡 = ∫ 𝜀𝐼𝑑𝑡 = ∫ 𝐿 𝑖𝑑𝑡 = ∫ 𝐿𝑖𝑑𝑖 = 𝑳𝑰𝟐
𝑑𝑡
𝟐
*the energy is stored in the magnetic field of the inductor
Example C: A toroid solenoid with N turns has mean radius R and a cross-sectional area of A. A current of I
flows through the toroid. What is the energy stored in this toroid?
𝜇0 𝑁𝐼
The magnetic field in a toroid is given by 𝐵 = 2𝜋𝑟
as derived in 10.4.
𝜇0 𝑁𝐼
𝑁𝜙 𝑁𝐵𝐴 𝑁 ( 2𝜋𝑅 ) 𝐴 𝜇𝑁 2 𝐴
𝐿=
=
=
=
𝐼
𝐼
𝐼
2𝜋𝑅
𝑈=
1 2 1 𝜇0 𝑁 2 𝐴 2
𝐿𝐼 = (
)𝐼
2
2 2𝜋𝑅
𝜇0 𝑁 2 𝐴𝐼 2
→𝑈=
4𝜋𝑅
LR Circuits
▪
▪
RL Circuit – Contains a resistor and inductor in parallel.
As the current increases towards its maximum value
(when switch is closed), an emf is induced in the inductor:
The inductor prevents the current from immediately jumping to
𝜀/𝑅. Instead the current will build to this value over time.
o Voltage across the resistor the resistor, 𝑉𝑅 (𝑡), will start at
zero, but will build to 𝜀 over time.
o Voltage across the inductor, 𝑉𝐿 (𝑡), will start at the battery
value, 𝜀, giving a back emf to prevent current. At steady state, this emf will drop to zero, and the
inductor will act as normal current-carrying wire, but it will have energy stored in its magnetic field.
o Current, 𝐼(𝑡), will start at zero since the inductor doesn’t let current immediately jump when the battery
is connected. Over time, it will build to 𝜀/𝑅 at steady state.
77
By Kirchoff’s Law:
𝜀 − 𝑉𝑅 − 𝑉𝐿 = 0
→ 𝜀 − 𝐼𝑅 − 𝐿
𝑑𝐼
𝑑𝐼
= 0 → 𝜀 − 𝐼𝑅 = 𝐿
𝑑𝑡
𝑑𝑡
Multiplied both sides by -1/R to get I by itself to make integration easier:
𝜀
𝐿 𝑑𝐼
=−
𝑅
𝑅 𝑑𝑡
𝐼−
Separate the variables:
𝑑𝐼
𝜀
𝐼−𝑅
=−
𝑅
𝑑𝑡
𝐿
The current starts at 0 and goes to I at some time, t, I(t):
𝐼(𝑡)
∫
0
𝑡
𝑑𝐼
𝜀
𝐼−𝑅
= ∫−
0
𝑅
𝑑𝑡
𝐿
𝜀
𝐼(𝑡) − 𝑅
𝜀
𝑅
𝜀
𝜀
𝑅
𝑅
→ ln |𝐼 − | 𝐼(𝑡) = − 𝑡 → ln |𝐼(𝑡) − | − ln |− | = − 𝑡 →= ln |
𝜀 |=− 𝑡
𝑅 0
𝐿
𝑅
𝑅
𝐿
𝐿
−𝑅
𝜀
→
𝐼(𝑡) − 𝑅
𝜀
−𝑅
𝑅
= 𝑒 −𝐿𝑡
𝑹
𝜺
o 𝑰(𝒕) = 𝑹 (𝟏 − 𝒆− 𝑳 𝒕 )
𝜺
𝒕
𝑰(𝒕) = 𝑹 (𝟏 − 𝒆−𝝉 )
𝐿
o The time constant for an LR is 𝜏 = 𝑅.
𝒕
o Using ohm’s law for the resistor, the voltage across the resistor is: 𝑽𝑹 (𝒕) = 𝜺(𝟏 − 𝒆−𝝉 )
o When the voltage is not dropped in the resistor, it is dropped in terms of back emf in the inductor:
𝒕
𝑽𝑳 (𝒕) = 𝜺𝒆−𝝉
o Charging circuit graphs:
78
▪
Discharging Circuit – Suppose switch 1 is open and switch 2 is
closed at t = 0 after steady state conditions have been reached.
When the circuit is charged and disconnected, the energy stored in
the inductor’s magnetic field acts a battery for the resistor. The
current will continue flowing in the direction it was going (since
the inductor resists the current decreasing) until the inductor runs
out of emf.
By Kirchoff’s Law:
𝑉𝑅 + 𝑉𝐿 = 0
𝑑𝐼
𝑑𝐼
𝐿 𝑑𝐼
−𝐼𝑅 − 𝐿 𝑑𝑡 = 0 → 𝐼𝑅 = −𝐿 𝑑𝑡 → 𝐼 = − 𝑅 𝑑𝑡
→
𝑑𝐼
𝐼
𝑅
= − 𝐿 𝑑𝑡
𝜀
The current starts at 𝑅 and decreases to some value, I(t) after time t:
𝐼(𝑡) 𝑑𝐼
∫𝜀
𝑅
𝐼
𝑡
𝑅
= ∫0 − 𝐿 𝑑𝑡
𝐼(𝑡)
𝑅
𝜀
𝑅
𝐼(𝑡)
𝑅
→ ln|𝐼| 𝜀 = − 𝐿 𝑡 → ln|𝐼(𝑡)| − ln |𝑅| = − 𝐿 𝑡 → ln | 𝜀 | = − 𝐿 𝑡
𝑅
𝑅
→
𝐼(𝑡)
𝜀/𝑅
𝑅
=− 𝑡
𝐿
𝜺
𝑹
𝜺
o 𝑰(𝒕) = 𝑹 (𝟏 − 𝒆− 𝑳 𝒕 )
𝒕
𝑰(𝒕) = 𝑹 (𝟏 − 𝒆−𝝉 )
𝒕
o By ohm’s law: 𝑽𝑹 (𝒕) = 𝜺𝒆−𝝉
𝒕
o The inductor acts as the battery, so its emf drains with the circuit: 𝑽𝑳 (𝒕) = 𝜺𝒆−𝝉
o Disharging circuit graphs:
79
Example D: Appreciate the circuit given on the right.
Assume V = 100 V, R = 20 Ω and L = 4.0 H.
a) What is the current through resistor R right when the switch is
closed?
b) What is the voltage across the inductor just after the switch is
closed?
c) Find the potential across the inductor as a function of time.
d) At what rate is the energy stored in the inductor increasing at time
t=2 s.0?
a) The current is zero initially, as the inductor will not allow the current to immediately jump to a value right
when the power in connected.
b) The back emf is originally 100 V against the direction of the current.
𝐿
4𝐻
c) The time constant of the inductor is 𝜏 = 𝑅 = 20 Ω = 0.2
𝑡
𝑉𝐿 (𝑡) = 100𝑒 −.2 → 𝑽𝑳 (𝒕) = 𝟏𝟎𝟎𝒆−𝟓𝒕
d) The rate of change of energy is power.
𝑑𝑈
𝑑 1
𝑑𝐼
𝑃=
= ( 𝐿𝐼 2 ) = 𝐿𝐼
𝑑𝑡 𝑑𝑡 2
𝑑𝑡
100 𝑉
The current as a function of time is given by: 𝐼(𝑡) = 20 Ω (1 − 𝑒 −5𝑡 ) = 5(1 − 𝑒 −5𝑡 )
𝑑𝐼
= 25𝑒 −5𝑡
𝑑𝑡
𝑑𝐼
→ 𝑃 = 𝐿𝐼 𝑑𝑡 = (4𝐻)(5(1 − 𝑒 −5(2) )(25𝑒 −5(2) ) = .02 𝑊
Example E: A L = 5.0 Henry inductor carrying an initial current of 100 A is
discharged through a resistor of R = 1.0 Ω.
a) Find the voltage drop across the resistor as a function of time.
b) Find the time needed for the current to reach 10 A.
𝑅
1Ω
a) 𝑉𝑅 (𝑡) = 𝐼𝑅𝑒 − 𝐿 𝑡 → 𝑉𝑅 (𝑡) = (100 𝐴)(1 Ω)𝑒 −5 𝐻𝑡
→ 𝑉𝑅 (𝑡) = 100𝑒 −.2𝑡
𝑅
1
b) 𝐼(𝑡) = 𝐼0 𝑒 − 𝐿 𝑡 → 𝐼(𝑡) = 100𝑒 −5𝑡
→ 10 = 100𝑒 −.2𝑡 → .1 = 𝑒 −.2𝑡 → ln|. 1| = −.2𝑡
→𝑡=
ln|. 1|
= 11.5 𝑠
−.2
80
Example F: The circuit on the right is allowed to reach a steady state. At
time t = 0, the switch is closed.
a) What is the potential across the inductor at t = 0?
b) At what rate is the current increasing in the inductor at t = 0?
c) How much current flows through the inductor after the switch has
been closed for a long time?
a) The inductor is in parallel with the 40 Ω resistor, so it only
produces a back emf to match the voltage across the 40 Ω resistor
at steady state.
40
𝑉𝐿 = 20+40 (120 𝑉)80 𝑉
b) 𝜀 = −𝐿
→
𝑑𝐼
𝑑𝑡
→
𝑑𝐼
𝑑𝑡
=−
𝜀
𝐿
𝑑𝐼 80 𝑉
𝐴
=
= 32
𝑑𝑡 2.5 𝐻
𝑠
c) At steady-state, the inductor acts a wire with no resistance. As a result, the 40 Ω is “skipped” by the
current, so current flows from the battery, through the 20 Ω resistor and then through the inductor, so
current flows through a circuit with only 20 Ω of resistance. The current is then:
𝐼=
120 𝑉
20 Ω
=6𝐴
The LC Circuit
▪
LC Circuit – Contains a charged capacitor and inductor (with little resistance).
If a fully charged capacitor is connected to an inductor, it will discharge through the
inductor. When the capacitor runs out of charge to discharge through the inductor, the
inductor will produce an induced current to stop the current from decreasing, sending
charge to the other plate of the conductor. The opposite plate eventually becomes fully
charged, and then discharges again through the inductor, which does its thing again. The LC
oscillates as a result.
81
By Kirchoff’s Law: 𝑉𝐶 + 𝑉𝐿 = 0
𝑞
𝑑𝑖
→ +𝐿 =0
𝐶
𝑑𝑡
𝑑𝑞 𝑑𝑖
𝑑2 𝑞
Since 𝑖 = 𝑑𝑡 , 𝑑𝑡 = 𝑑𝑡 2
𝑞
𝑑2𝑞
→ +𝐿 2 =0
𝐶
𝑑𝑡
𝑑2𝑞
1
→ 2+
𝑞=0
𝑑𝑡
𝐿𝐶
The energy above matches the equation for simple harmonic motion from AP Physics C: Mechanics:
𝑑2 𝑥
( 𝑑𝑡 2 + 𝜔2 𝑥 = 0)
1
In this case, the variable is the charge over time and the angular frequency is 𝜔 = √𝐿𝐶.
o The period of an RC circuit is 𝑇 =
2𝜋
𝜔
=
2𝜋
1
𝐿𝐶
→ 𝑻 = 𝟐𝝅√𝑳𝑪
√
1
1
o The total energy in an LC circuit is 𝐸 = 2 𝐿𝐼 2 + 2 𝐶𝑉𝐶2
o From the general solution to the differential equation, the charge over time can be given by:
1
𝑄(𝑡) = 𝑄0 cos(𝜔𝑡) → 𝑄(𝑡) = 𝑄0 cos(
𝑡)
√𝐿𝐶
Rate your understanding: Inductance
0
1
L
I understand some
concepts in inductance,
but can’t solve complex
problems.
2
I understand
inductance and can
solve most problems
involving it.
3
I can solve problems in
inductance without any
significant errors.
4
I can explain and teach
inductance.
82
Unit 11 – Electromagnetic Induction
AP Physics C: Electricity & Magnetism
11.5 Maxwell’s Equations (Chapter 34)
Focus Question: What is the purpose of Maxwell’s Equations?
Maxwell’s equations are four equations that unite electricity and magnetism and can be used (along with
Lorentz force) as the basis of all phenomena in these areas.
➢ Gauss’s Law – The total flux through any closed surface equals the net charge inside that surface
divided by the permittivity of free space.
∮ 𝑬 ∙ 𝒅𝑨 =
𝑸
𝜺𝟎
∗Charges create electric fields.
➢ Gauss’s Law for Magnetism – The net magnetic flux through a closed surface is zero.
∮ 𝑩 ∙ 𝒅𝑨 = 𝟎
*Isolated monopoles cannot exist.
➢ Ampere’s Law – The line integral of the magnetic field around any closed path is determined by the net
current and the rate of change of electric flux through any surface bounded by that path.
∮ 𝑩 ∙ 𝒅𝒍 = 𝝁𝟎 𝑰 + 𝜺𝟎 𝝁𝟎
𝒅𝝓
𝒅𝒕
*An electric current and a changing electric field create a magnetic field.
➢ Faraday’s Law of Induction – The line integral of the electric field around any closed path equals the
rate of change of magnetic flux through any closed surface bounded by that path.
∮ 𝑬 ∙ 𝒅𝒓 = 𝜺 = −
𝒅𝝓
𝒅𝒕
*A changing magnetic field creates an electric field.
Rate your understanding: Maxwell’s Equations
0
1
2
I ran. I ran so far away
from physics.
Somebody once told
E&M was gonna roll me.
I’ll understand this in a
day or twoooooooo!!!
3
4
We’re no strangers to
physics. You know the
rules and so do I.
It’s going to take a lot to
drag me away from you.
83