Class 12 Physics Notes – Chapter 2: Electrostatic Potential and Capacitance

1. Electrostatic Potential

The electrostatic potential at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.

Formula: V = W / q

Its SI unit is Joule/Coulomb (J/C) which is also known as Volt (V).

2. Potential due to a Point Charge

The potential V at a distance r from a point charge Q is given by:

Formula: V = 1 / (4πε₀) · (Q / r)

Unlike electrostatic force and electric field which are vector quantities, electrostatic potential is a scalar quantity.

3. Equipotential Surfaces

An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single point charge, the equipotential surfaces are concentric spherical surfaces centered at the charge.

Properties of Equipotential Surfaces:

• No work is done in moving a test charge over an equipotential surface.
• The electric field is always normal (perpendicular) to the equipotential surface at every point.
• Two equipotential surfaces can never intersect each other.

Relation between Field and Potential: E = – ΔV / Δl (Electric field is in the direction in which the potential decreases steepest).

4. Potential Energy of a System of Charges

The potential energy of a system of charges is the work done (by an external agency) in bringing the charges from infinity to their respective locations.

For two charges q₁ and q₂ separated by distance r: U = 1 / (4πε₀) · (q₁q₂ / r)

Potential Energy of a Dipole in an External Field

When an electric dipole of moment p is placed in a uniform electric field E, it experiences a torque. The potential energy of the dipole is:

U = – p · E = – pE cosθ

5. Electrostatics of Conductors

• Inside a conductor, electrostatic field is zero.
• At the surface of a charged conductor, electrostatic field must be normal to the surface at every point.
• The interior of a conductor can have no excess charge in the static situation.
• Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

6. Dielectrics and Polarisation

Dielectrics are non-conducting substances. When placed in an external electric field, they develop an induced net dipole moment. The dipole moment per unit volume is called polarisation (P).

7. Capacitors and Capacitance

A capacitor is a system of two conductors separated by an insulator, used to store electric charge and electrical energy. The capacitance C is the ratio of charge Q to the potential difference V.

Formula: C = Q / V

The SI unit of capacitance is Farad (F). 1 F = 1 Coulomb / Volt.

The Parallel Plate Capacitor

For a parallel plate capacitor consisting of two large plane parallel conducting plates separated by a small distance d, with area A:

C = ε₀A / d

Effect of Dielectric on Capacitance

If the medium between the plates is filled with an insulating substance (dielectric) of dielectric constant K, the capacitance increases K times: C = K · C₀

8. Combination of Capacitors

• Capacitors in Series: 1 / C_s = 1 / C₁ + 1 / C₂ + … + 1 / C_n
• Capacitors in Parallel: C_p = C₁ + C₂ + … + C_n

9. Energy Stored in a Capacitor

The energy stored in a charged capacitor is the work done in charging it. It is given by:

U = 1/2 CV² = Q² / 2C = 1/2 QV

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Multiple Choice Questions (MCQs)

1. The SI unit of electric potential is:
a) Joule
b) Newton/Coulomb
c) Volt
d) Farad
Answer: (c) Volt

2. The shape of the equipotential surface for a single point charge is:
a) Plane
b) Cylinder
c) Concentric spheres
d) Parabola
Answer: (c) Concentric spheres

3. Work done in moving a charge of 2C on an equipotential surface is:
a) 2 Joules
b) 0 Joules
c) 4 Joules
d) Infinity
Answer: (b) 0 Joules

4. The electric field inside a charged spherical conductor is:
a) Maximum
b) Zero
c) Depends on radius
d) Depends on charge
Answer: (b) Zero

5. The relation between electric field (E) and potential (V) is:
a) E = dV/dr
b) E = – dV/dr
c) V = dE/dr
d) V = – dE/dr
Answer: (b) E = – dV/dr

6. What happens to the capacitance of a parallel plate capacitor when a dielectric is inserted between its plates?
a) It decreases
b) It remains same
c) It becomes zero
d) It increases
Answer: (d) It increases

7. Two capacitors of 2 μF and 4 μF are connected in parallel. Their equivalent capacitance is:
a) 1.33 μF
b) 6 μF
c) 8 μF
d) 2 μF
Answer: (b) 6 μF

8. Energy stored in a capacitor of capacitance C charged to potential V is:
a) CV²
b) 1/2 CV²
c) 1/2 Q²V
d) QV
Answer: (b) 1/2 CV²

9. Dielectrics are materials that are:
a) Good conductors
b) Semiconductors
c) Insulators
d) Superconductors
Answer: (c) Insulators

10. The direction of the electric field is always:
a) Parallel to equipotential surface
b) Perpendicular to equipotential surface
c) At 45 degrees to equipotential surface
d) Random
Answer: (b) Perpendicular to equipotential surface

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Textbook Exercise Solutions

Q 2.1: Two charges 5 × 10^-8 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Solution:
Let P be the point on the line joining the charges where potential is zero. Let its distance from the positive charge be x cm. So, distance from negative charge is (16 – x) cm.
V = V₁ + V₂ = 0
⇒ k(5 × 10^-8)/x + k(-3 × 10^-8)/(16 – x) = 0
⇒ 5/x = 3/(16 – x)
⇒ 80 – 5x = 3x ⇒ 8x = 80 ⇒ x = 10 cm.
If the point lies outside the charges on the side of the negative charge, distance is x and (x – 16).
5/x = 3/(x – 16) ⇒ 5x – 80 = 3x ⇒ 2x = 80 ⇒ x = 40 cm.
So, potential is zero at 10 cm and 40 cm from the positive charge.

Q 2.2: A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Solution:
In a regular hexagon, the distance of the centre from each vertex is equal to the side of the hexagon. So, r = 10 cm = 0.1 m.
Number of charges n = 6. Charge q = 5 × 10^-6 C.
Total Potential V = 6 × (1 / 4πε₀) × (q / r)
V = 6 × (9 × 10⁹) × (5 × 10^-6) / 0.1
V = 2.7 × 10⁶ V.

Q 2.3: Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Solution:
(a) The system is an electric dipole. The equipotential surface where V = 0 is the equatorial plane, which is a plane perpendicular to the line AB and passing through its mid-point.
(b) The direction of the electric field is from the positive charge to the negative charge (i.e., from A to B). It is perpendicular (normal) to the equipotential surface at every point.

Q 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7 C distributed uniformly on its surface. What is the electric field:
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?
Solution:
Radius R = 0.12 m, q = 1.6 × 10^-7 C.
(a) Inside the conductor, electric field is always Zero.
(b) Just outside (r = R = 0.12 m): E = k · q / R² = (9 × 10⁹ × 1.6 × 10^-7) / (0.12)² = 10⁵ N/C.
(c) At r = 18 cm = 0.18 m: E = k · q / r² = (9 × 10⁹ × 1.6 × 10^-7) / (0.18)² = 4.44 × 10⁴ N/C.

Q 2.5: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Solution:
Initial capacitance C₀ = ε₀A / d = 8 pF.
New distance d’ = d / 2. Dielectric constant K = 6.
New capacitance C’ = K · ε₀A / d’ = K · ε₀A / (d / 2) = 2K · (ε₀A / d) = 2K · C₀
C’ = 2 × 6 × 8 pF = 96 pF.

Q 2.6: Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Solution:
(a) In series, 1 / C_s = 1 / C₁ + 1 / C₂ + 1 / C₃ = 1/9 + 1/9 + 1/9 = 3/9 = 1/3.
So, Total Capacitance C_s = 3 pF.
(b) Since the capacitors are identical and in series, the voltage divides equally among them.
V₁ = V₂ = V₃ = V / 3 = 120 / 3 = 40 V.

Q 2.7: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
(a) In parallel, C_p = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF.
(b) In parallel, potential V is same across all capacitors (100 V).
Charge Q₁ = C₁ × V = 2 × 10^-12 × 100 = 2 × 10^-10 C.
Charge Q₂ = C₂ × V = 3 × 10^-12 × 100 = 3 × 10^-10 C.
Charge Q₃ = C₃ × V = 4 × 10^-12 × 100 = 4 × 10^-10 C.

Q 2.8: In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Solution:
Area A = 6 × 10^-3 m², d = 3 mm = 3 × 10^-3 m.
Capacitance C = ε₀A / d = (8.85 × 10^-12 × 6 × 10^-3) / (3 × 10^-3) = 17.7 × 10^-12 F = 17.7 pF (approx 18 pF).
Charge Q = C × V = 17.7 × 10^-12 × 100 = 1.77 × 10^-9 C.

Q 2.9: Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
Solution:
New capacitance C’ = K · C = 6 × 17.7 pF = 106.2 pF.
(a) When supply remains connected: Potential V remains same (100 V). Charge increases. New charge Q’ = C’ × V = 106.2 × 10^-12 × 100 = 1.06 × 10^-8 C.
(b) When supply is disconnected: Charge Q remains conserved (1.77 × 10^-9 C). Potential decreases. New potential V’ = Q / C’ = V / K = 100 / 6 = 16.6 V.

Q 2.10: A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Solution:
C = 12 pF = 12 × 10^-12 F, V = 50 V.
Energy U = 1/2 CV² = (1/2) × (12 × 10^-12) × (50)²
U = 6 × 10^-12 × 2500 = 15000 × 10^-12 J = 1.5 × 10^-8 J.

Q 2.11: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Solution:
Initial Energy U_i = 1/2 C₁V₁² = (1/2) × (600 × 10^-12) × (200)² = 1.2 × 10^-5 J.
When connected to uncharged capacitor C₂ (600 pF), the common potential V’ = (C₁V₁ + C₂V₂) / (C₁ + C₂) = (600 × 200 + 0) / (600 + 600) = 100 V.
Final Energy U_f = 1/2 (C₁ + C₂) (V’)² = (1/2) × (1200 × 10^-12) × (100)² = 0.6 × 10^-5 J.
Energy lost = U_i – U_f = 1.2 × 10^-5 – 0.6 × 10^-5 = 0.6 × 10^-5 J (or 6 × 10^-6 J).