Electrostatic Potential and Capacitance Class 12 Physics Notes

• Potential is a scalar; the SI unit is the volt (1 V = 1 J/C).
• Potential due to a dipole at a point on its axis:
• $V=\frac{kp}{r^2}$
• .
• Equipotential surfaces: potential is constant, so no work is done moving along them; they are perpendicular to field lines.

• Parallel-plate, spherical and cylindrical capacitors are the common geometries.
• Capacitance depends only on geometry and the dielectric — not on charge.
• Unit: farad (1 F = 1 C/V), commonly microfarads and picofarads.

• Non-polar dielectrics polarise via induced dipoles; polar dielectrics align their permanent dipoles.
• Field inside a dielectric:
• $E=\frac{E_0}{\kappa }$
• .
• Maximum breakdown field limits how much charge a capacitor can hold.

• Series:
• $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\dots$
• — same charge, voltages add.
• Parallel:
• $C=C_1+C_2+\dots$
• — same voltage, charges add.
• In series the smallest capacitor limits the charge; in parallel capacitances add.

Example: Two capacitors $4\mu \text{F}$ and $6\mu \text{F}$ are connected in series. Find the equivalent capacitance.

Solution: $\frac{1}{C}=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}\Rightarrow C=\frac{12}{5}=2.4\mu \text{F}$.

Exercises (NCERT Q2.1–Q2.11)

• 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.
• 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.
• 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?
• 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?
• 2.5) A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10⁻¹² F). 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?
• 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?
• 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.
• 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?
• 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.
• 2.10) A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
• 2.11) A 600pF capacitor is charged by a 200V 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?

Answers at a glance

1. 1
   Q2.1 → Potential is zero at 10 cm from the 5 μC charge (between the charges) and at 40 cm from it on the far side of the positive charge.
2. 2
   Q2.2 → V = 2.7 × 10⁶ V at the centre.
3. 3
   Q2.3 → (a) The plane bisecting AB perpendicularly (V = 0). (b) The field is perpendicular to this plane at every point.
4. 4
   Q2.4 → (a) E = 0. (b) 1.0 × 10⁵ N/C, radially outward. (c) 4.4 × 10⁴ N/C, radially outward.
5. 5
   Q2.5 → C = 96 pF.
6. 6
   Q2.6 → (a) C = 3 pF. (b) 40 V across each capacitor.
7. 7
   Q2.7 → (a) C = 9 pF. (b) Charges: 2 × 10⁻¹⁰ C, 3 × 10⁻¹⁰ C and 4 × 10⁻¹⁰ C respectively.
8. 8
   Q2.8 → C = 17.7 pF; q ≈ 1.8 × 10⁻⁹ C on each plate.
9. 9
   Q2.9 → (a) Supply on: V stays 100 V while the charge rises to ≈ 1.1 × 10⁻⁸ C. (b) Supply off: charge stays fixed and V drops to ≈ 16.7 V.
10. 10
    Q2.10 → U = 1.5 × 10⁻⁸ J.
11. 11
    Q2.11 → Energy lost = 6 × 10⁻⁶ J (half the stored energy disappears into the sharing process).

MCQ Questions

1 Mark Questions

1. 1
   Define the capacitance of a conductor and give its SI unit.
2. 2
   What is the dielectric constant of a medium, and how does it affect the capacitance of a parallel-plate capacitor?
3. 3
   Why is the electric field always perpendicular to an equipotential surface?
4. 4
   When two capacitors are connected in series, what is the charge on each capacitor?

2 Mark Questions

1. 1
   A charge of 2 μC is placed in air. Find the electric potential at a point 1 m away. (Ans: V = kq/r = 1.8 × 10⁴ V)
2. 2
   A 10 μF capacitor is charged to 200 V. Find the energy stored in it. (Ans: U = ½CV² = 0.2 J)
3. 3
   Capacitors of 3 μF and 6 μF are connected in series. Find the equivalent capacitance. (Ans: C = 2 μF)

3 Mark Questions

1. 1
   Derive the expression for the capacitance of a parallel-plate capacitor of plate area A and separation d with air between the plates. (Ans: C = ε₀A/d)
2. 2
   Derive the expression for the energy density of an electric field between the plates of a parallel-plate capacitor. (Ans: u = ½ε₀E²)
3. 3
   Explain, with a diagram, how inserting a dielectric slab between the plates of a capacitor increases its capacitance. (Ans: polarization lowers the net field and hence raises charge capacity at fixed V)

5 Mark Questions

1. 1
   Derive the expression for the energy stored in a capacitor. Hence obtain the energy density and explain how the energy is released when the plates are connected. (Ans: U = ½CV² = Q²/2C; u = ½ε₀E²)
2. 2
   Describe the construction and working of a van de Graaff generator. Explain the role of the sharp points (corona discharge) and why the outer shell is maintained at a high potential.