Theme: Electricity
Terms/Laws
Definition, Formula and Derivation, Theory/Notes
Electric Charge,𝑄
Coulomb’s Law
𝑄 = 𝑛𝑒 = 𝑛𝑝 = 𝐼𝑡 =
It states that the
electrostatic force, 𝑭
between two separated
point charges 𝑸 and 𝒒 is
directly proportional to the
product of the charges,
𝑸𝒒 and inversely
proportional to the square
of the distance between
them,𝒓𝟐
𝑄𝑞
𝑟2
𝑄𝑞
1
𝐹=𝑘 2
[𝑘 =
]
𝑟
4𝜋𝜀0
Charges in free space,
𝑄𝑞
𝐹=
4𝜋𝜀0 𝑟 2
Charges in insulating medium,
𝑄𝑞
𝐹=
[𝜀 = 𝜀0 𝜀𝑟 ]
4𝜋𝜀0 𝜀𝑟 𝑟 2
𝑄𝑞
𝐹=
4𝜋𝜀𝑟 2
𝐹∝
From Gauss’ law,
𝑄
𝐸=
4𝜋𝜀0 𝑟 2
𝐹 = 𝐸𝑞
𝑄
𝐹=(
)𝑞
4𝜋𝜀0 𝑟 2
𝑄𝑞
𝐹=
4𝜋𝜀𝑟 2
Electric Field,𝐸
A region around a
charged particle where
free positive test charge
will experience an
electric force exerted by
the particle.
𝑊 𝐸
=
𝑉
𝑉
Repulsive Force
- Like charges repel each other with forces that same in magnitude and inopposite
direction.
- Action-reaction pair that obeys Newton’s Third Law of Motion.
Attractive Force
-Unlike charged attract each other with forces that same in magnitude and in
opposite direction.
Electric field produced by :
Positive point charge +Q
Negative point charge -Q
Electric dipole
Two like charges
Point charge and plate
with unlike charges
The effect of an electric field on a metal coated polystyrene ball
Parallel plates with unlike charges
The effects of an electric field on a candle flame
Electric Field
Strength/Intensity,𝐸
It defined as the electric
force acting on a unit
positive charge placed at the
point.
Electric force 𝐹 per unit
positive charge +𝑞 at the
point.
𝐸=
𝐹
𝑞
𝑄𝑞
1
𝑄𝑞
(
)
[𝐹
=
]
4𝜋𝜀0 𝑟 2 𝑞
4𝜋𝜀0 𝑟 2
𝑄
𝐸=
4𝜋𝜀0 𝑟 2
𝐸=
Coulomb’s Law related
Motion of a charge in Uniform Electric Field
For uniform electric field,
𝐸=
𝑉
𝑑
-
A force 𝐹 = 𝐸𝑞 is acting on a charge+𝑞 that causes it accelerates to move
in a straight line in the direction od electric field.
𝐹 = 𝑚𝑎 = 𝐸𝑞
𝑚𝑎 = 𝐸𝑞
𝐸𝑞
𝐸𝑒
(𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 + 𝑞) / 𝑎 = − (𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 − 𝑒)
𝑎=
𝑚
𝑚
Parabolic path of charge in Uniform Electric Field
Gauss’ Law and
Electric Flux, 𝜙
Electric flux 𝜙 is the number
of electric field lines 𝐸 that
intersect/penetrate an area
𝐴.
When A is the normal to E,
𝜙 = 𝐸𝐴
When A is not the normal to E,
𝜙 = (𝐸 cos 𝜃)𝐴
The charge experiences projectile motion.
A constant force pulling the charge downward.
𝑣𝑥 increases , but 𝑣𝑦 remains constant.
Gauss’ law states that the
electric flux 𝜙 through a
closed surface (Gaussian
surface) is given by
Σ𝑄
𝜙=
𝜀0
Electric field produced by :
𝜙=
Σ𝑄
𝜀0
Isolated point charge
where Σ𝑄 is the sum of the
charges enclosed by the
surface and 𝜀0 is the
permittivity of free space.
▪
Electric field strength,E
Electric flux through Gaussian surface,
𝜙 = 𝐸𝐴
𝜙 = 𝐸(4𝜋𝑟 2 ) [𝐴 = 4𝜋𝑟 2 ]
Applying Gauss’ law,
Σ𝑄
𝜙=
𝜀0
Q
𝐸(4𝜋𝑟 2 ) =
[ Σ𝑄 = 𝑄]
𝜀0
𝐸=
𝑄
4𝜋𝜀0 𝑟 2
Isolated charged conducting sphere
▪
▪
▪
Inside the sphere (𝑟 < 𝑅), there is no charge.
Σ𝑄 = 0
If inside the sphere (𝑟 < 𝑅) have charges Σ𝑄 = 𝑄 , the mutual repulsive
force of like charges would push themselves away from each other until
they reach the sphere surface.
Hence, charges only appear on the sphere surface and distributed
uniformly.
➢ Electric field strength inside the sphere (𝑟 < 𝑅)
Electric flux through Gaussian surface,
𝜙 = 𝐸𝐴
𝜙 = 𝐸(4𝜋𝑟 2 ) [𝐴 = 4𝜋𝑟 2 ]
Applying Gauss’ law,
Σ𝑄
𝜙=
𝜀0
0
𝐸(4𝜋𝑟 2 ) =
[ Σ𝑄 = 0]
𝜀0
0
𝐸=
4𝜋𝜀0 𝑟 2
𝐸=0
➢ Electric field strength outside the sphere surface (𝑟 ≥ 𝑅)
Electric flux through Gaussian surface,
𝜙 = 𝐸𝐴
𝜙 = 𝐸(4𝜋𝑟 2 ) [𝐴 = 4𝜋𝑟 2 ]
Applying Gauss’ law,
Σ𝑄
𝜙=
𝜀0
Q
𝐸(4𝜋𝑟 2 ) =
[ Σ𝑄 = 𝑄]
𝜀0
𝑄
𝐸=
4𝜋𝜀0 𝑟 2
On the sphere surface (𝑟 ≥ 𝑅),
𝑄
𝐸=
4𝜋𝜀0 𝑅2
Graph E - r
Uniformly charged plate
Electric Potential,𝑉
and Electric Potential
Energy,𝑈
Electric potential 𝑉 is the
work done 𝑊 to bring a unit
of positive charge +q from
infinity ∞ to a point.
Electric potential energy 𝑈 is
the work done 𝑊 to bring
the positive charge +q from
infinity ∞ to a point.
Electric potential and Electric potential energy due to a point charge +Q
Force exerted on +q by +Q,
𝑄𝑞
4𝜋𝜀0 𝑥 2
Work done to bring +q through a small displacement -dx,
𝑊 = 𝐹𝑠
𝑑𝑊 = 𝐹(−𝑑𝑥)
𝑄𝑞
𝑑𝑊 = −
𝑑𝑥
4𝜋𝜀0 𝑥 2
Work done to bring +q from infinity to a point ,P ,
𝐹=
𝑟
𝑊=∫ 𝑊
𝑟
∞
𝑄𝑞
𝑑𝑥
2
∞ 4𝜋𝜀0 𝑥
𝑄𝑞 𝑟
𝑊 = − [−
]
4𝜋𝜀0 𝑥 ∞
𝑄𝑞
𝑊 = −(−
− 0)
4𝜋𝜀0 𝑟
𝑄𝑞
𝑊=
4𝜋𝜀0 𝑟
𝑊 = −∫
From the definition of electric potential,
𝑉 =
𝑊
𝑄𝑞 1
=(
)
𝑞
4𝜋𝜀0 𝑟 𝑞
𝑄
𝑉=
4𝜋𝜀0 𝑟
Relationship between V and U,
𝑈
𝑞
𝑈 = 𝑞𝑉
𝑄𝑞
𝑈=
4𝜋𝜀0 𝑟
𝑉=
Electric potential due to a charged conducting sphere,
➢ Inside the sphere and on the sphere surface (𝑟 ≤ 𝑅)
𝑄
𝑉=
4𝜋𝜀0 𝑅
➢ Outside the sphere (𝑟 > 𝑅)
𝑄
𝑉=
4𝜋𝜀0 𝑟
Graph V-r
Equipotential surface
A space where all points
charge have the same
electric potential.
𝑄
• 𝑉 = 4𝜋𝜀
0𝑟
at all point.
• No work done W in moving a charge from a point to another.
𝑊=0
Electric Current,I
Rate 𝑡 of flow of charge Q in
a conductor.
Current
𝑄
𝐼=
𝑡
•
•
•
Current is a form of positive charge flow.
Direction of current or direction of positive charge flow is from positive
terminal (+) to negative terminal (−).
Current can flow is due to a potential difference between two points in a
circuit.
Ammeter
•
•
Potential Difference /
Voltage,𝑉
Work done W to bring a unit
of positive charge from one
point to another against the
electric field.
𝑉=
𝑉=
𝑊
𝑞
Measures current in Ampere (A)
It is in series arrangement in a circuit.
Potential difference between two points
𝐸𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑
𝑞
•
•
Electric field is a conservative field.
Potential difference between two points is same via any path.
𝑉1 = 𝑉2 = 𝑉3
•
Velocity of a particle
Work done to move +q (electric potential energy),
𝑊
𝑉=
𝑞
𝑊 = 𝑞𝑉
After being accelerated through a potential difference V,
Gain in KE = Loss in Electric PE
1
𝑚𝑣 2 = 𝑞𝑉
2
2𝑞𝑉
𝑣2 =
𝑚
2𝑞𝑉
𝑣=√
𝑚
Potential difference between anode and cathode
•
Velocity of the electron
Horizontal component (x-axis)
After the electron being accelerated
via a potential difference V ,
Gain in KE = Loss in Electric PE
1
𝑚 𝑣 2 = 𝑒𝑉
2 𝑒 𝑥
2𝑒𝑉
𝑣𝑥 2 =
𝑚𝑒
2𝑒𝑉
𝑣𝑥 = √
𝑚𝑒
Vertical component (y-axis)
Uniform acceleration in the electric
field,
𝐸𝑒
𝑎=
𝑚𝑒
When electron emerges from electric
field,
𝑣𝑦 = 𝑢 + 𝑎𝑡
𝐸𝑒 𝐿
𝑣𝑦 = 0 + ( ) ( )
𝑚𝑒 𝑣𝑥
𝐿𝐸𝑒
𝑣𝑦 =
𝑚𝑒 𝑣𝑥
When electron enter the uniform
electric field, 𝑣𝑥 constant.
•
•
Time taken to transverse the entire
electric field,
𝐿
𝑡=
𝑣𝑥
Resultant velocity, v
𝑣 = √𝑣𝑥 2 + 𝑣𝑦 2
Angle to the x-axis ,𝜃
tan 𝜃 =
𝑣𝑦
𝑣𝑥
Voltmeter
•
•
Measures voltage in Volt (V)
It is in parallel arrangement in a circuit.
Projectile motion- in the
electric filed
Straight line motion – outside
the electric field
Relationship between V and E
➢
➢ V-x and E-x graphs
➢ Uniform electric field in parallel plates
Resistance
Ohm’s Law
Electrical Energy
Electrical Power
Capacitance,𝐶
The ratio of the charge Q on
the capacitor to the
potential difference V across
it.
Capacitor
•
•
Used to store and release electric charges and electrical energy.
It can be arranged in series and parallel.
•
It consists of two parallel plates.
Charged capacitor
•
•
•
•
•
𝐶=
𝑄
𝑉
When a capacitor is connected directly to a battery, electrons from one
plate will be transferred to another.
It causes one plate is deficit of electrons with +Q whereas another is
excess of electrons with -Q.
Two plates have equal amount of charges but different of types.
+𝑄 = −𝑄
The plate with +Q at a higher potential whereas plate with -Q at a lower
potential.
So, there is a potential difference V across the plates.
Graph Q-V
# Capacitance is independent with potential difference across it.
#𝑄 ∝𝑉
Capacitance in isolated charged sphere
𝑄
4𝜋𝜀0 𝑅
𝑄
𝐶=
𝑉
4𝜋𝜀0 𝑅
𝐶 = 𝑄(
)
𝑄
𝐶 = 4𝜋𝜀0 𝑅
𝑉=
Capacitance of Parallel-Plate capacitor
For uniform electric field,
𝜎=
𝑄
𝑉
𝑎𝑛𝑑 𝐸 =
𝐴
𝑑
Using Gauss’ law,
𝐸=
𝜎
𝜀0
•
•
Therefore,
Then,
𝐸=𝐸
𝜎 𝑉
=
𝜀0 𝑑
𝜀0 𝑉
𝜎=
𝑑
𝜎=𝜎
𝑄 𝜀0 𝑉
=
𝐴
𝑑
𝑄 𝜀0 𝐴
=
𝑉
𝑑
𝜀0 𝐴
𝐶=
𝑑
•
[
𝑄
= 𝐶]
𝑉
The electrons on X induce +Q on Y.
The more electrons flow to X , the more +Q is induced by electrons , the
potential difference V increases.
When potential difference across plates = voltage of battery
✓ The flow of electrons from battery stops
✓ The charge on the plates is 𝑄 = 𝐶𝑉
Dielectric
Series Circuit and
Parallel Circuit
Series circuit
Parallel circuit
Diagram
𝐼 = 𝐼1 = 𝐼2 = 𝐼3 = ⋯
𝑉 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯
𝑅 = 𝑅1 + 𝑅2 + 𝑅3 + ⋯
𝑄 = 𝑄1 = 𝑄2 = 𝑄3 = ⋯
Current
Voltage/Potential
difference
Effective
Resistance
Charge
𝐼 = 𝐼1 + 𝐼2 + 𝐼3 + ⋯
𝑉 = 𝑉1 = 𝑉2 = 𝑉3 = ⋯
1
1
1
1
=
+
+
+⋯
𝑅 𝑅1 𝑅2 𝑅3
𝑄 = 𝑄1 + 𝑄2 + 𝑄3 + ⋯
𝑉 = 𝑉1 + 𝑉2 + 𝑉3 + ⋯
𝑄 𝑄
𝑄
𝑉 = + + +⋯
𝐶1 𝐶2 𝐶3
1
1
1
𝑉 = 𝑄( + + + ⋯)
𝐶1 𝐶2 𝐶3
𝑉
1
1
1
= + + +⋯
𝑄 𝐶1 𝐶2 𝐶3
1
1
1
1
= + + +⋯
𝐶 𝐶1 𝐶2 𝐶3
•
When a bulb/ device is not function in the circuit,
the other devices cannot function.
Effective
Capacitance
𝑄 = 𝑄1 + 𝑄2 + 𝑄3 + ⋯
𝑄 = 𝐶1 𝑉 + 𝐶2 𝑉 + 𝐶3 𝑉 + ⋯
𝑄 = 𝑉(𝐶1 + 𝐶2 + 𝐶3 + ⋯ )
𝑄
= 𝐶1 + 𝐶2 + 𝐶3 + ⋯
𝑉
𝐶 = 𝐶1 + 𝐶2 + 𝐶3 + ⋯
Other Features
Applications
•
When a bulb/ device is not function in the circuit, the
other devices still can function.