MAHARASHTRA STATE BOARD CLASS XI AND XII PHYSICS CHAPTER 8 ELECTROSTATICS Introduction. Coulomb's law Calculating the value of an electric field Superposition principle Electric potential Deriving electric field from potential Capacitance Principle of the capacitor Dielectrics Polarization, and electric dipole moment Applications of capacitors.

1. Std : 12th Year : 2022-23
Subject : PHYSICS
Chapter 8 : Electrostatics
CLASSXII
MAHARASHTRA STATE BOARD

2. Can yourecall ?
1. What are conservative force?
Conservative force is any force necessary to move a particle from one point
to another, independent of this particle’s path. Therefore, the work done by
conservative force only relies on the start and end-point, and not on the path
taken.
2. What is potential energy?
Potential energy is the energy difference between the energy of an object in a
given position and its energy at a reference position.
3. What is Gauss’ law and what is a Gaussian surface?
According to Gauss's law, the flux of the electric field through any closed
surface, also called a Gaussian surface, is equal to the net charge enclosed
divided by the permittivity of free space : This equation holds for charges of
either sign, because we define the area vector of a closed surface to point
outward.

3. Electric Flux: Electric flux through a given area held inside an electric field is the
measure of the total number of electric lines of force passing normally
through that area.
Statement of Gauss’ law: The total flux through a closed surface is
𝟏
𝜺𝟎
times the net charge enclosed by the closed surface

4. Application of Gauss' Law:
1. Electric Field Intensity due to Uniformly Charged Spherical Shell or
Hollow Sphere:

5. 2. Electric Field Intensity due to an Infinitely Long Straight
Charged Wire:

6. The direction of the electric field E is directed outward if 𝝀 is positive and inward if is 𝝀
negative.
3. Electric Field due to a Charged Infinite Plane Sheet:
The flux passing through the curved surface is zero as the electric field is
tangential to this surface.

8. 8.3 Electric Potential and Potential Energy:
work done against an electrostatic force = increase in the potential energy of the system.
Expression for potential energy:

9. Units of potential energy:
(i) SI unit= joule (J)
(ii) Another convenient unit of energy is electron volt (eV),
(iii) Other related units are:

10. Concept of Potential:

11. Relation between electric field and electric potential:
The work done = dW = - Fdx = -Edx.
Zero potential:
Once the zero point of the potential is set, then every potential is measured with
respect to that reference.
(i) In case of a point charge or localised collection of charges, the zero point is set
at infinity.
(ii) For electrical circuits the earth is usually taken to be at zero potential.
Thus, the potential at a point A in an electric field is the amount of work done
to bring a unit positive charge from infinity to point A.

12. 8.4 Electric Potential due to a Point Charge, a Dipole and a System of Charges:
(a) Electric potential due to a point charge:
A positively charged
particle produces a
positive electric
potential and a
negatively charged
particle produces a
negative electric
potential.

13. (b) Electric potential due to an electric dipole:

15. Case (i) Potential at an axial point
Case (ii) Potential at an equatorial point

16. (c ) Electrostatics potential due to a system of charges:

17. 8.5 Equipartition Surfaces

18. 8.6 Electrical Energy of Two Point Charges and of a Dipole in an
Electrostatic Field:
a) Potential energy of a system of 2 point charges:
In bringing the first charge 𝑞1 to position 𝐴 ( 𝑟1 ),
no work is done i.e., 𝑊1 = 0.
𝑉1 =
1
4 𝜋 𝜀0
𝑞1
𝑟1
When we bring charge 𝑞2 from infinity to
𝐵 ( 𝑟2 ) at a distance 𝑟12, from 𝑞1, work done is
𝑊2 = (potential at B due to charge 𝑞1) × 𝑞2
=
𝑞1
4 𝜋 𝜀0 𝑟12
× 𝑞2, (𝑤ℎ𝑒𝑟𝑒 𝐴𝐵 = 𝑟12)
∴ 𝑈 =
1
4 𝜋 𝜀0
𝑞1𝑞2
𝑟12

19. b) Potential energy for a system of N point charges:
In bringing a charge 𝑞3 from ∞ to C ( 𝑟3 ) work has to be done against electrostatic
forces of both 𝑞1 and 𝑞2.
∴ 𝑊3 = ( potential at C due to 𝑞1 and 𝑞2) × 𝑞3
𝑈 =
1
4 𝜋 𝜀0
𝑞1
𝑟13
+
𝑞2
𝑟23
× 𝑞3
𝑈 =
1
4 𝜋 𝜀0
𝑞1𝑞3
𝑟13
+
𝑞2𝑞3
𝑟23
Similarly, 𝑊4 =
1
4 𝜋 𝜀0
𝑞1𝑞4
𝑟14
+
𝑞2𝑞4
𝑟24
+
𝑞3𝑞4
𝑟34
𝑈 =
1
4 𝜋 𝜀0
𝑎𝑙𝑙 𝑝𝑎𝑖𝑟𝑠
𝑞𝑗𝑞𝑘
𝑟𝑗𝑘
c) Potential energy of a single charge in an external field:
∴ Work done in bringing a charge q, from ∞ to the given point in the external
field is q V ( 𝒓 ).
∴ PE of a system of a single charge q at 𝒓 in an external field is given by,
PE = q V ( 𝒓 )

20. d) Potential energy of a system of two charges in an external field:
Work done = 𝑞1 𝑉 ( 𝑟1 )
∴ 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑜𝑛 𝑞2 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑖𝑒𝑙𝑑 = 𝑞2 𝑉 ( 𝑟2 ) and
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑜𝑛 𝑞2 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑡ℎ𝑒 𝑓𝑖𝑒𝑙𝑑 𝑑𝑢𝑒 𝑡𝑜 𝑞1 =
𝑞1𝑞2
4 𝜋 𝜀0𝑟12
∴ 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑖𝑛 𝑏𝑟𝑖𝑛𝑔𝑖𝑛𝑔 𝑞2 𝑡𝑜 𝑟2 = 𝑞2𝑉 𝑟2 +
𝑞1𝑞2
4 𝜋 𝜀0𝑟12
Potential energy of the system = Total work done in assembling the configuration
= 𝑞1𝑉 𝑟1 + 𝑞2𝑉 𝑟2 +
𝑞1𝑞2
4 𝜋 𝜀0𝑟12
e) Potential energy of a dipole in an external field:
𝜏 = 𝑝 × 𝐸 𝑜𝑟 𝜏 = 𝑝 𝐸 𝑠𝑖𝑛 𝜃
𝑊 = 𝜃0
𝜃
𝜏𝑒𝑥𝑡 𝜃 𝑑𝜃 = 𝜃0
𝜃
𝑝 𝐸 𝑠𝑖𝑛 𝜃 𝑑𝜃
𝑊 = 𝑝 𝐸 − 𝑐𝑜𝑠 𝜃 𝜃0
𝜃
𝑊 = 𝑝 𝐸 𝑐𝑜𝑠 𝜃0 − 𝑐𝑜𝑠 𝜃
We can choose 𝑈 𝜃0 = 0,
∴ 𝑈 𝜃 − 𝑈 𝜃0 = 𝑝 𝐸 (𝑐𝑜𝑠 𝜃0 − 𝑐𝑜𝑠 𝜃)

21. a) If initially the dipole is perpendicular to the field 𝐸 i.e., 𝜃0 =
𝜋
2
then
𝑈 𝜃 = 𝑝 𝐸 cos
𝜋
2
− cos 𝜃
𝑈 𝜃 = −𝑝 𝐸 cos 𝜃 = − 𝑝 . 𝐸
b) If initially the dipole is parallel to the field 𝐸 then 𝜃0 = 0
𝑈 𝜃 = 𝑝 𝐸 cos 0 − cos 𝜃
𝑈 𝜃 = 𝑝 𝐸 1 − cos 𝜃
Dielectrics and Electric Polarisation:
Dielectrics are insulating materials or non- conducting substances which can be
polarised through small localised displacement of charges.
e.g. glass, wax, water, wood , mica, rubber, stone, plastic etc.
Polar dielectrics Non-Polar dielectrics:
Fig. A polar molecule
Fig. Examples of Polar
molecules HCI, 𝐻20.
Fig.
(a) Nonpolar molecule.
Examples of
Nonpolar molecules
(b) 𝐻2,
(c) 𝐶𝑂2.

22. Polarization of a non-polar dielectric in
an external electric field:
Polarization of a polar dielectric in an
external electric field:
The dipole moment per unit volume is called
polarization and is denoted by 𝑃.
For linear isotropic dielectrics 𝑃 = 𝜒𝑒 𝐸.
Each molecule becomes a tiny dipole
having a dipole moment.
Reduction of electric field due to polarization of a dielectric:

23. Capacitors and Capacitance, Combination of Capacitors in Series and
Parallel:
The potential difference V is the work done to carry a unit positive test charge from
the conductor 2 conductor 1 against the field.
∴ 𝐶 =
𝑄
𝑉
The SI unit of capacitance is farad (F).
Dimensional formula is [ 𝑀−1𝐿−2𝑇4𝐴2]
1 farad = 1 coulomb/1volt
1 μF = 10− 6 𝐹, 1 nf = 10− 9 𝐹, 1 pF = 10− 12 𝐹
Uses of Capacitors
Principle of a capacitor:
Capacity is, 𝐶1 =
𝑄
𝑉
The capacity 𝐶2 =
𝑄
𝑉−𝑉1
∴ 𝐶2 > 𝐶1
If the conductors are plane then it is called parallel
plate capacitor.

24. Combination of Capacitors:
(a) Capacitors in series:
Potential difference across the series combination of capacitor is V volt,
where 𝑉 = 𝑉1 + 𝑉2 + 𝑉3
∴ 𝑉 =
𝑄
𝐶1
+
𝑄
𝐶2
+
𝑄
𝐶3
Let 𝐶𝑆 represent the equivalent capacitance, then 𝑉 =
𝑄
𝐶𝑆
∴
𝑄
𝐶𝑆
=
𝑄
𝐶1
+
𝑄
𝐶2
+
𝑄
𝐶3
∴
1
𝐶𝑆
=
1
𝐶1
+
1
𝐶2
+
1
𝐶3
(for 3 capacitors in series)
∴
1
𝐶𝑒𝑞
=
1
𝐶1
+
1
𝐶2
+. . . . . . . . . +
1
𝐶𝑛
If all capacitors are equal then,
1
𝐶𝑒𝑞
=
𝑛
𝐶
𝑜𝑟 𝐶𝑒𝑞 =
𝐶
𝑛

25. (b) Capacitors in Parallel:
∴ 𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑄 𝑐𝑎𝑛 𝑏𝑒, 𝑄 = 𝑄1 + 𝑄2 + 𝑄3
= 𝐶1𝑉 + 𝐶2𝑉 + 𝐶3𝑉
Let 𝐶𝑝 be the equivalent capacitance of the combination then 𝑄 = 𝐶𝑃𝑉
∴ 𝐶𝑃𝑉 = 𝐶1𝑉 + 𝐶2𝑉 + 𝐶3𝑉
∴ 𝐶𝑃 = 𝐶1 + 𝐶2 + 𝐶3
𝐶𝑃 = 𝐶1 + 𝐶2+. . . . . . . . +𝐶𝑛
If all capacitors are equal then 𝐶𝑒𝑞 = 𝑛 𝐶
Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between
the Plates:
a) Capacitance of a parallel plate capacitor without a dielectric:
𝑬 =
𝝈
𝟐 𝜺𝟎
−
𝝈
𝟐 𝜺𝟎
= 𝟎
𝑬 =
𝝈
𝟐 𝜺𝟎
+
𝝈
𝟐 𝜺𝟎
=
𝝈
𝜺𝟎
=
𝑸
𝑨𝜺𝟎
, 𝑬 =
𝑽
𝒅
𝒐𝒓 𝑽 = 𝑬𝒅
𝑽 =
𝑸
𝑨𝜺𝟎
𝒅 , 𝑪 =
𝑸
𝑽
=
𝑸
𝑸𝒅
𝑨𝜺𝟎
=
𝑨𝜺𝟎
𝒅

26. b) Capacitance of a parallel plate capacitor with a dielectric slab between the plates:
The capacitance of the capacitor is given by, 𝐶0 =
𝐴𝜀0
𝑑
Then the potential difference between the plates is, 𝑉0 = 𝐸0𝑑, Where 𝐸0 =
𝜎
𝜀0
=
𝑄
𝐴𝜀0
,
And σ is the surface charge density on the plates.
The induced field is, 𝐸𝑃 =
𝜎𝑝
𝜀0
=
𝑄𝑝
𝐴𝜀0
𝜎𝑝 =
𝑄𝑝
𝐴
The net field (E) inside the dielectric reduces to 𝐸0 − 𝐸𝑝.
Hence, 𝐸 = 𝐸0 − 𝐸𝑝 =
𝐸0
𝑘
∵
𝐸0
𝐸0−𝐸𝑝
= 𝑘
Where k is a constant called the dielectric constant.
∴ 𝐸 =
𝑄
𝐴𝜀0𝐾
𝑜𝑟 𝑄 = 𝐴𝐾𝜀0𝐸
The potential difference between the capacitor plates is, 𝑉 = 𝐸0 𝑑 − 𝑡 + 𝐸(𝑡)
𝑉 = 𝐸0 𝑑 − 𝑡 +
𝑡
𝑘
∵ 𝐸 =
𝐸0
𝑘
𝑉 =
𝑄
𝐴𝜀0
𝑑 − 𝑡 +
𝑡
𝑘
The capacitance of the capacitor on the introduction of dielectric slab becomes
𝐶 =
𝑄
𝑉
=
𝑄
𝑄
𝐴𝜀0
𝑑−𝑡+
𝑑
𝑘
=
𝐴𝜀0
𝑑−𝑡+
𝑡
𝑘

27. Energy Stored in a Capacitor:
During the process of charging, let q' be the charge on the capacitor and V be the
potential difference between the plates.
Hence 𝐶 =
𝑞′
𝑉
∴ 𝑑𝑊 = 𝑉 𝑑𝑞 =
𝑞′
𝐶
𝑑𝑞
Total work done in transferring the charge,
𝑊 = 𝑑𝑤 = 0
𝑄 𝑞′
𝐶
𝑑𝑞 =
1
𝐶 0
𝑄
𝑞′
𝑑𝑞
𝑊 =
1
𝐶
(𝑞′)2
2 0
𝑄
=
1
2
𝑄2
𝐶
This work done is stored as electrical potential energy U of the capacitor.
∴ 𝑈 =
1
2
𝑄2
𝐶
=
1
2
𝐶 𝑉2 =
1
2
𝑄 𝑉 (∵ 𝑄 = 𝐶 𝑉)

28. Thank You