Electric Potential
Electric potential: Electric potential at a point in space is amount of energy per unit
charge when placed at that point.
It is also defined as the amount of work done required to bring unit positive (test
charge) charge from infinity to the given point against external electric field slowly.
𝐏
⃗ . ⃗⃗⃗
V= - ∫∞ 𝐄
𝐝𝐥
✓ SI unit of electric potential is J/C = volt
Potential energy: Potential energy of a charge Q placed at a point P where electric
potential is V is the total energy of charge due to electric field is given as
U = QV
Potential difference or Voltage: Potential difference between any two points is the
difference of potential of that two point.
∆𝐯 = V2 – V1
Amount of energy required to move charge Q from point 1 to 2 having electric
potential V1 and V2 respectively is
∆𝑢 = Q (V2 - V1 )
∆𝒖= Q ∆𝐯
Note:
✓ Electron volt is the amount of energy required to move an electron through unit
potential difference.
Electric potential due to point charge
To move unit positive charge slowly
⃗
𝐅ext = - 𝐄
𝐅ext =
1
𝑄
4𝜋𝜀0 𝑟 2
𝒊
Amount of work done required to
bring unit positive charge from ∞ to p due to external force is given as
𝑅
⃗⃗⃗⃗
W = ∫ 𝐅ext. 𝐝𝐫
∞
𝑅
W = ∫∞ 𝐹𝑒𝑥𝑡 𝑑𝑟
𝑅
W = ∫∞
W=
1
𝑄
4𝜋𝜀0 𝑟 2
dr
𝑅 𝑑𝑟
𝑄
∫
4𝜋𝜀0 ∞ 𝑟 2
𝑄
1
W = [4𝜋𝜀 (𝑟 )]𝑅∞
0
Physics Point
Page 1
W=
W=
𝑄
1
4𝜋𝜀0
1
( 𝑅 - ∞)
𝑸
𝟒𝝅𝜺𝟎 𝑹
Electric potential at a point is the amount of energy required to bring unit positive
charge from ∞ to that point.
V=W
V=
𝟏
𝑸
𝟒𝝅𝜺𝟎 𝒓
Note:
✓ Electric Potential is a scalar quantity.
✓ S.I. unit of Electric Potential is J/C or volt(v)
Electric potential due to electric dipole
Net potential at point P is
VP= VP1 + VP2
RP1 =√𝑟 2 + 𝑎2 + 2𝑎𝑟 𝑐𝑜𝑠𝜃
RP2 =√𝑟 2 + 𝑎2 − 2𝑎𝑟 𝑐𝑜𝑠𝜃
VP1
=
VP2 =
−𝑄
1
𝑄
4𝜋𝜀0 𝑅𝑃2
VP1 =
VP2 =
VP =
1
4𝜋𝜀0 𝑅𝑃1
1
−𝑄
4𝜋𝜀0
√𝑟 2 + 𝑎2+ 2 𝑎𝑟 𝑐𝑜𝑠 𝜃
1
𝑄
4𝜋𝜀0 √𝑟 2 + 𝑎2− 2 𝑎𝑟 𝑐𝑜𝑠 𝜃
𝑄
1
{
4𝜋𝜀0 √𝑟 2 +𝑎2 − 2𝑎𝑟 𝑐𝑜𝑠𝜃
−
1
√𝑟 2 +𝑎2 + 2𝑎𝑟 𝑐𝑜𝑠𝜃
If point p is very far r >> a =>
1
1
= 𝑟(1+
=
1
=
1
𝑟
𝑟
2𝑎 cos 𝜃
𝑟
(1−
𝑎
2𝑎
√𝑟 2 (1+( ) 2+ 𝑐𝑜𝑠𝜃)
𝑟
𝑟
=
1
1
2𝑎𝑐𝑜𝑠𝜃
𝑟(1+
)2
𝑟
1
2
𝑟
𝑎 cos 𝜃
≈0
−
)
1 2𝑎 cos 𝜃
(1−2
𝑟
1
=
√𝑟 2 +𝑎2 + 2𝑎𝑟 𝑐𝑜𝑠𝜃
𝑎
)
……(i )
)
𝑟
}
Similarly
1
1
√𝑟 2 +𝑎2 − 2𝑎𝑟 𝑐𝑜𝑠𝜃
=𝑟 ( 1 +
𝑎 cos 𝜃
𝑟
……(ii )
)
From equation ( i ) and ( ii )
VP =
=
=
𝑄
1
{ (1+
𝑎 cos 𝜃
4𝜋𝜀0 𝑟
𝑄
4𝜋𝜀0
𝑄
4𝜋𝜀0
𝑎 cos 𝜃
{(
𝑟2
𝑟2
)−
1
𝑟
𝑎 cos 𝜃
)+(
2𝑎 cos 𝜃
{(
𝑟
𝑟2
(1−
𝑎 cos 𝜃
𝑟
)}
)}
)}
Physics Point
Page 2
VP =
[𝑽 =
𝟐𝑸 𝐚𝐜𝐨𝐬 𝜽 𝟏
𝟒𝝅𝜺𝟎
𝒓𝟐
𝑲𝒑
𝒄𝒐𝒔]
𝒓𝟐
Since,
⃗ . 𝐫 = 2𝑄𝑎𝑟 cos 𝜃
𝒑
⃗ .𝐫
𝒑
VP = 4𝜋𝜀
1
0
𝑟3
⃗ .𝐫
𝒑
VP = 4𝜋𝜀
1
3
0 |𝐫|
Relation between electric field and potential
Electric potential at point 1 and 2 is given as,
1
⃗ . ⃗⃗⃗
V1 = − ∫ 𝐄
dl
∞
2
⃗ . ⃗⃗⃗
V2 = − ∫∞ 𝐄
dl
𝟐
⃗ . ⃗⃗⃗
V2 – V1 = − ∫𝟏 𝐄
𝐝𝐥
Note:
✓ Electric potential is a referential quantity, and its value depends upon where we
have taken a reference point.
𝜕𝑉
𝜕𝑉
𝜕𝑉
✓ Finding Electric field if electric potential is known 𝐸⃗ =
𝑖̂ +
𝑗̂ +
𝑘̂
𝜕𝑥
𝜕𝑦
𝜕𝑧
✓ Negative sign in the expression tells that Electric field is always directed from
high potential to low potential.
Equipotential surface: Those surfaces on which each point is having same electric
potential is called equipotential surface.
Note:
✓ Electric field is always in the direction normal to the equipotential surface.
Equipotential surface due to some standard charge distribution
Due to +ve charge
Due to -ve point charge
Note:
✓ Equipotential surface due to same type of charge repels each other and due to
opposite type repel each other.
✓ Equipotential surface for point charge is concentric sphere.
Physics Point
Page 3
Equipotential surface due to Plane sheet charge
Equipotential surface Plane due to electric dipole
Potential energy of system
Potential energy of the system is the amount of energy required to make the system
1. For two-point charge system
Let we have two charge system Q1 and Q2.
To form the system first bring charge Q1 at
point 1 since there is no any other charge so
potential energy of Q1 is
U1 = 0
Then bring charge Q2at point 2
Potential at point 2 is
V2 =
𝑄1
4π𝜀𝑜 𝑟21
Potential energy of Q2 is
U2 = Q2V2 =
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
Potential energy of the system is
U = U1 + U2 = 0 +
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
=
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
𝑸 𝑸𝟐
U =𝟒𝛑𝜺𝟏
𝒐 𝒓𝟐𝟏
2. For three point charge system
Let we have three charge system Q1, Q2and Q3.
To form the system first bring charge Q1 at point 1
since there is no any other charge so
potential energy of Q1 is
U1 = 0
Then bring charge Q2 at point 2
𝑸𝟏
Potential at point 2 is
V2 =
Potential energy of Q2 is
U2 = Q2V2 =
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
Then bring charge Q3 at point 3
𝑸𝟏
Potential at point 3 is
V3 =
+
Potential energy of Q2 is
U3 = Q3V3 =
𝟒𝛑𝜺𝒐 𝒓𝟑𝟏
𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟑𝟐
𝑸𝟏 𝑸𝟑
𝟒𝛑𝜺𝒐 𝒓𝟑𝟏
+
𝑸𝟑 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟑𝟐
Potential energy of the system is
U = U1 + U2 + U3 = 0 +
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
+
𝑸𝟏 𝑸𝟑
𝟒𝛑𝜺𝒐 𝒓𝟑𝟏
+
𝑸𝟑 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟑𝟐
Physics Point
Page 4
𝐐 𝐐𝟐
U =𝟒𝛑𝛆𝟏
𝐨 𝐫𝟐𝟏
+
𝐐𝟏 𝐐𝟑
𝟒𝛑𝛆𝐨 𝐫𝟑𝟏
+
𝐐𝟑 𝐐𝟐
𝟒𝛑𝛆𝐨 𝐫𝟑𝟐
Potential energy of dipole inside external electric field:
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E
In a uniform electric field, torque on dipole is given by
⃗
⃗ × 𝑬
⃗ = 𝒑
𝝉
The amount of work done by the external torque will be given by
𝜃
𝜃
𝑈 = ∫ 𝜏𝑒𝑥𝑡
. ⃗⃗⃗⃗
𝑑𝜃 = ∫ 𝑝𝐸𝑠𝑖𝑛𝜃𝑑𝜃
𝜃0
𝜃0
𝑈 = −𝑝𝐸(𝑐𝑜𝑠𝜃 − 𝑐𝑜𝑠𝜃0 )
If zero potential reference is taken at θ0 = 900
then potential energy of dipole is 𝑼 = −𝒑𝑬𝒄𝒐𝒔𝜽
⃗
⃗ .𝐄
U = −𝐩
Potential energy of system inside external electric field
It is the total potential energy of the system which consists of
✓ Energy required to form the system
✓ Energy gain by electric field
1. For two-point charge system
Let we have two charge system Q1 and Q2.
To form the system first bring charge Q1 at point 1
Potential energy of Q1 is
V1 = V(due to electric fd.) +V(due to charge)
Since there is no charge initially V1 = V1E
U1 = Q1V1 = Q1V1E
Then bring charge Q2 at point 2
Potential at point 2 is V2 = V2E +
𝑸𝟏
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
Potential energy of Q2 is
U2 = Q2V2 = Q2V2E +
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
Potential energy of the system is
U = U1 + U2 = Q1V1E + Q2V2E +
U = Q1V1E + Q2V2E +
𝑄1 𝑄2
4π𝜀𝑜 𝑟21
𝑸𝟏 𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
2. For dipole
Let we have a dipole of charge Q and length 2a. To form the system first bring
+Q charge at point 1 .
Potential energy at point 1 is
V1 = V(due to electric fd.) +V(due to charge)
Since there is no charge initially V1 = V1E
U1 = QV1 = QV1E
Physics Point
Page 5
Then bring -Q charge at point 2
Potential at point 2 is
𝑸
V2 = V2E +𝟒𝛑𝜺
𝒐 𝒓𝟐𝟏
Potential energy of Q2 is
𝑸𝟐
U2 = - QV2 = - QV2E – 𝟒𝛑𝜺
𝒐 𝒓𝟐𝟏
Potential energy of the system is
𝑄2
U = U1 + U2 = QV1E - QV2E - 4π𝜀
U = QV1E− QV2E −
𝑜 𝑟21
𝑸𝟐
𝟒𝛑𝜺𝒐 𝒓𝟐𝟏
Electrostatics of conductor
✓ Those substances having very large number of free electron or free charge
carriers are called conductor.
✓ Those substances which are not conductor are called Di-Electric.
Conductor inside electric field
➢ When a conductor is placed inside
electric field. Free electron experience
force opposite to the direction of
external electric field, due to which
charge separation occurs.
➢ Which creates an electric field inside conductor in the direction opposite to the
direction to that of external field, that electric field is called induced electric
field.
➢ if the magnitude of induced electric field is less than the external field, the free
electron continues to move in the direction opposite to the external electric
field, due to which magnitude of induced field increases and this process
continues until the net electric field inside conductor doesn’t becomes zero.
Note:
✓ Inside any conductor electric field is zero.
✓ Region inside conductor is equipotential region including the surface point.
✓ Surface of any conductor is always equipotential surface.
✓ Electric field is always perpendicular to the conductor surface.
✓ Inside any conductor the charge never accumulates; it comes to its surface in
very short interval of time.
Dielectric inside External Electric field:
When a dielectric is placed inside electric field, there induces electric dipole which
align themselves in the direction of electric field or align themselves in the direction
of electric field. Due the alignment of dipole there appears charge on the opposite
Physics Point
Page 6
surface, due to which there creates electric field inside dielectric in the direction
opposite to the direction of external electric field which reduces net electric field
inside dielectric.
Polarization
Net dipole moment induces per unit volume inside dielectric when placed inside
external electric field is called polarization.
Let there are N dipole induced in volume ΔV
Dipole moment induced per unit volume
⃗ =
𝐏
∑𝑵
𝒑𝒊
𝒊=𝟏 ⃗⃗⃗⃗
∆𝑽
Polarization depends upon external electric field and for
linear and isotropic dielectric it is given as
⃗𝑷
⃗ = 𝝃𝒆 ⃗𝑬
⃗
Where 𝝃𝒆 is
dielectric.
called
electrical
susceptibility
of
the
Susceptibility: It is the measure of degree of polarization of a dielectric inside
external electric field.
Dielectric Constant (K): It is the number of times electric field inside dielectric
decreases as compared to external electric field.
⃗ 𝒊𝒏 =
𝑬
⃗𝑬𝒆𝒙𝒕
𝑲
Dielectric Breakdown: When external electric field is increased force on bound
electron increases. If Electric field is increased to such an amount that the force on
bound electron is greater than nuclear force then electron will come out from the
atom and number of free electrons inside dielectric increases to very large amount
and the dielectric behaves as conductor, this is called dielectric breakdown.
Dielectric Strength: The minimum electric field at which dielectric breakdown
occurs is called dielectric strength.
Cell and battery
Cell is the smallest unit which stores electrical energy. Battery is the combination of
more than one cell.
It is represented as
Note:
✓ Cell is having two terminal one is having higher potential and another has
lower potential.
✓ Cell maintains a constant potential difference across its terminal.
✓ The high potential is also called positive terminal and low potential is called
negative terminal.
Physics Point
Page 7
Capacitor
When two conductors are separated by some distance, the system is called capacitor.
Capacitance
Capacitance of a capacitor is amount of charge required to create unit potential
difference across its capacitor.
Let, we have given Q charge on a capacitor due to which a
potential difference of V volt is created across it, then
capacitance is given as
𝑪=
𝑸
𝑽
where,
➢ Q = charge on capacitor.
➢ V= voltage across capacitor.
Note:
✓ SI unit of capacitance is C/V also called Farad (F).
✓ The plate which is having positive charge is called positive plate and which is
having negative charge is called negative plate.
Parallel plate capacitor
In parallel plate capacitor there are two conductor plates placed parallel to each
other.
Finding capacitance of a capacitor
Steps:
➢ Ground one plate of the capacitor and call it –ve plate.
➢ Give charge Q to other plate called +ve plate.
➢ Find the electric field between the plates of the capacitor.
➢ Find the potential difference between the plates of the capacitor using
+𝑣𝑒 𝑝𝑙𝑎𝑡𝑒
𝑉 = 𝑉+𝑣𝑒 𝑝𝑙𝑎𝑡𝑒 − 𝑉−𝑣𝑒 𝑝𝑙𝑎𝑡𝑒 =
➢ Find the capacitance using 𝑪 =
∮
𝐸⃗ . ⃗⃗⃗
𝑑𝑙
−𝑣𝑒 𝑝𝑙𝑎𝑡𝑒
𝑸
𝑽
Capacitance of a parallel plate capacitor
Let we have a capacitor of plate area A and they are separated by distance d.
Electric field between the plate.
⃗ = 𝑄 𝑖̂ + 𝑄 𝑖̂
𝐄
2𝜖 𝐴
2𝜖 𝐴
0
0
Physics Point
Page 8
⃗ =
𝐄
𝑄
𝜖0 𝐴
𝑖̂
Voltage across capacitor
V = V2 –V1
2 → →
= -∫1
𝑬
.
0
V = − ∫𝑑
𝒅𝒍
𝑄
𝜖0 𝐴
𝑖̂ . 𝑑𝑥 𝑖̂
0
∫𝑑 𝑑𝑥
𝐴
0
𝑄
V = −𝜖
V=
C=
C=
𝑄
𝜖0 𝐴
𝑄
𝑉
𝑑
=
𝑄
𝑄
𝑑
𝜖0 𝐴
=
𝜖0 𝐴
𝑑
𝝐𝟎 𝑨
𝒅
Note:
✓ Capacitance of a capacitor is independent of charge and voltage of the
capacitor.
✓ Capacitance of a capacitor depends upon geometry of capacitor and medium
between the plates of the capacitor.
Capacitive circuit
Kirchhoff’s Charge Law: At any junction in the circuit net
incoming charge is equal to net outgoing charge.
i.e. charge cannot accumulate at any point in the circuit.
𝑸𝒊𝒏𝒄𝒐𝒎𝒊𝒏𝒈 = 𝑸𝒐𝒖𝒕𝒈𝒐𝒊𝒏𝒈
Q1 = Q2 + Q3
Kirchhoff’s Voltage Law: In any close loop sum of voltage across different circuit
element is equal to zero.
Steps for solving capacitative circuit
➢
➢
➢
➢
➢
Do the charge distribution in the circuit
Apply the sign convention across circuit elements
Apply KCL at each junction in the loop
Apply KVL in each fundamental loop
Solve the above equation to solve the circuit
Energy stored inside parallel plate capacitor
Let at any time t charge on the capacitor is q and the potential difference across the
plates of capacitor is v. if dq amount of charge crosses capacitor the energy stored
inside capacitor is dU
dU = dq V
We have to sum all these small energies given to capacitor by the small charges up to
when the charge on capacitor becomes 0 to Q is
Physics Point
Page 9
𝑄
𝑈 = ∫ 𝑑𝑞𝑉
0
𝑄
𝑈=∫
0
𝑞
𝑑𝑞
𝐶
𝑄
1
𝑈 = ∫ 𝑞𝑑𝑞
𝐶
0
𝑸𝟐
𝑼 = 𝟐𝑪 =
𝟏
𝟐
𝑪𝑽𝟐
Energy Density inside capacitor: It is the energy present per unit volume in the the
capacitor.
𝑄2
Total energy inside the capacitor = 2𝐶
Total volume of the capacitor= Ad
Energy density Ed =
𝑄2
2𝐶𝐴𝑑
=
𝑄2
2𝜀𝑜 𝐴2
=
1
𝜀 𝐸
2 𝑜
2
Note:
✓ Relative permittivity and Dielectric constant of a dielectric are having same
magnitude.
Combination of capacitor
1. Series Combination
𝟏
𝑪𝒆𝒒
=
𝟏
𝑪𝟏
+
𝟏
𝑪𝟐
+.......
+
𝟏
𝑪𝑵
2.Parallel Combination
Ceq = C1 + C2 + . . . . . . +
Capacitance with dielectric inside it
Capacitance without Dielectric
CN
Capacitance with Dielectric
Physics Point
Page 10
Electric field between the plate.
⃗𝑬
⃗o =
𝒊
𝝐𝟎 𝑨
0
= − ∫𝑑
𝑄
𝑉
𝑖̂.𝑑𝑥 𝑖̂
𝐴
0
∫ 𝑑𝑥
𝜖0 𝐴 𝑑
𝑄
𝑄
𝜖0 𝐴
=
𝑄
𝜖0
Voltage across capacitor
V = V2 –V1
0
𝑄
=
𝜖0 𝐴
𝑑
𝑄
= − ∫𝑑
𝐾𝜖0 𝐴
𝑖̂.𝑑𝑥 𝑖̂
0
∫𝑑 𝑑𝑥
𝐴
0
𝑄
= − 𝐾𝜖
=
𝑑
𝑄
𝑑
𝜖0 𝐴
𝐊𝛜𝟎 𝐀
2
= − ∫1 ⃗⃗𝐸. ⃗⃗⃗⃗
𝑑𝑙
2
⃗⃗⃗
= − ∫1 𝐸⃗𝑜 . 𝑑𝑙
= −
𝐊
𝐐
E=
Voltage across capacitor
V = V2 –V1
Co =
𝐄𝐨
E=
𝑸
=
Electric field between the plate.
C=
𝑄
𝐾𝜖0 𝐴
𝑄
𝑉
=
𝑑
𝑄
𝑄
𝑑
𝐾𝜖0 𝐴
=
𝐾𝜖0 𝐴
𝑑
C = K Co
Physics Point
Page 11