Electrostatic Potential and Capacitance Class 12 Notes | CBSE Physics Chapter 2

Electrostatic Potential and Capacitance is Chapter 2 of CBSE Class 12 Physics. This chapter builds on Chapter 1 by introducing the concept of electric potential — a scalar quantity that simplifies many electrostatic calculations. You will learn about potential due to various charge configurations, equipotential surfaces, and capacitors.

This chapter carries 7–8 marks in board exams. Numericals on capacitors (series/parallel combinations), derivation of potential due to a dipole, and the concept of equipotential surfaces are most frequently tested.

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Key Concepts

1. Electrostatic Potential

The electric potential at a point is the work done per unit positive charge in bringing a test charge from infinity to that point (against the electric field).

V = W/q₀ = kQ/r (due to a point charge Q)

Unit: Volt (V) = Joule/Coulomb

Potential is a scalar quantity — it has magnitude but no direction. This makes it easier to work with than electric field.

Potential Due to Various Configurations

Configuration	Potential
Point charge Q at distance r	V = kQ/r
System of charges	V = k(q₁/r₁ + q₂/r₂ + … ) — algebraic sum
Dipole (axial point)	V = kp cos θ/r²
Dipole (equatorial point)	V = 0
Uniformly charged sphere (outside)	V = kQ/r
Uniformly charged sphere (surface)	V = kQ/R

Potential Difference

V_A − V_B = work done per unit charge in moving a test charge from B to A.

Relation with electric field: E = −dV/dr (field points in the direction of decreasing potential)

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2. Equipotential Surfaces

An equipotential surface is a surface where every point has the same electric potential. No work is done in moving a charge along an equipotential surface.

• Electric field lines are always perpendicular to equipotential surfaces
• For a point charge: equipotential surfaces are concentric spheres
• For a uniform field: equipotential surfaces are parallel planes perpendicular to the field
• Two equipotential surfaces never intersect
• Closer equipotential surfaces = stronger electric field

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3. Electrostatic Potential Energy

Potential energy of a system of charges is the work done in assembling the charges from infinity.

For two charges: U = kq₁q₂/r

For three charges: U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)

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4. Conductors and Capacitors

Properties of Conductors in Electrostatics

• Electric field inside a conductor is zero
• Any excess charge resides on the surface
• Electric field at the surface is perpendicular to the surface
• The entire conductor is at the same potential (equipotential body)

Capacitor

A capacitor is a device that stores electric charge and energy. It consists of two conductors separated by an insulator (dielectric).

Capacitance: C = Q/V

Unit: Farad (F). 1 F = 1 C/V (very large — usually μF, nF, or pF are used)

Parallel Plate Capacitor

C = ε₀A/d

• A = area of each plate
• d = separation between plates
• With dielectric (constant κ): C = κε₀A/d — capacitance increases κ times

Combination of Capacitors

Feature	Series	Parallel
Charge	Same on each (Q = Q₁ = Q₂)	Divides (Q = Q₁ + Q₂)
Voltage	Divides (V = V₁ + V₂)	Same across each (V = V₁ = V₂)
Equivalent	1/C = 1/C₁ + 1/C₂ + …	C = C₁ + C₂ + …
Effect	Total C decreases	Total C increases

Energy Stored in a Capacitor

U = ½CV² = ½QV = Q²/(2C)

Energy density (energy per unit volume) in electric field: u = ½ε₀E²

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Important Definitions

Term	Definition
Electric potential	Work done per unit charge in bringing a test charge from infinity to a point
Equipotential surface	Surface where every point has the same potential
Capacitance	Ratio of charge stored to potential difference: C = Q/V
Dielectric	Insulating material placed between capacitor plates that increases capacitance
Dielectric constant (κ)	Factor by which capacitance increases when dielectric is inserted

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Solved Examples

Example 1

Two capacitors of 6 μF and 3 μF are connected in series across a 12 V battery. Find the equivalent capacitance and charge on each.

Answer: 1/C = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. So C = 2 μF.

Charge: Q = CV = 2 × 10⁻⁶ × 12 = 24 μC (same on both in series).

Example 2

A parallel plate capacitor has plates of area 100 cm² separated by 2 mm. Find the capacitance. If a dielectric of κ = 5 is inserted, find the new capacitance.

Answer: C = ε₀A/d = (8.854 × 10⁻¹²)(100 × 10⁻⁴)/(2 × 10⁻³) = 44.27 pF.

With dielectric: C’ = κC = 5 × 44.27 = 221.35 pF.

Example 3

Find the potential at a point 9 cm from a charge of 4 × 10⁻⁷ C.

Answer: V = kQ/r = (9 × 10⁹)(4 × 10⁻⁷)/0.09 = 3600/0.09 = 4 × 10⁴ V = 40 kV

Example 4

A capacitor of 10 μF is charged to 100 V. Find the energy stored.

Answer: U = ½CV² = ½ × 10 × 10⁻⁶ × (100)² = ½ × 10⁻⁴ × 10⁴ = 0.05 J = 50 mJ

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Important Questions for Board Exams

1-Mark Questions

1. What is an equipotential surface?
2. How does the capacitance of a parallel plate capacitor change when a dielectric is inserted?
3. What is the SI unit of capacitance?

3-Mark Questions

1. Derive the expression for potential due to an electric dipole at a general point.
2. Derive the formula for the equivalent capacitance of capacitors in series and parallel.
3. Show that the electric field is always perpendicular to an equipotential surface.

5-Mark Questions

1. Derive the expression for capacitance of a parallel plate capacitor with and without a dielectric medium.
2. What is electrostatic potential energy? Derive the expression for energy stored in a capacitor. Also find the energy density.

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Quick Revision Points

• Potential V = kQ/r (scalar); Field E = kQ/r² (vector); E = −dV/dr
• Equipotential surfaces ⊥ field lines; no work to move charge along them
• PE of two charges: U = kq₁q₂/r
• Capacitance C = Q/V; Parallel plate: C = ε₀A/d; with dielectric: C = κε₀A/d
• Series: 1/C = 1/C₁ + 1/C₂ (same charge, voltage divides)
• Parallel: C = C₁ + C₂ (same voltage, charge divides)
• Energy: U = ½CV² = ½QV = Q²/(2C); Energy density: u = ½ε₀E²
• Inside conductor: E = 0, V = constant, charge on surface only

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Previous Chapter: Chapter 1 — Electric Charges and Fields
Next Chapter: Chapter 3 — Current Electricity