Electrostatic Potential and Capacitance Class 12 Notes (2026-27) — CBSE
Class 12 Physics Chapter 2 notes: electric potential, equipotential surfaces, potential energy, capacitance, parallel-plate capacitors, dielectrics and energy stored.
Electrostatic Potential and Capacitance — Class 12 Physics Notes
Chapter Snapshot
This chapter builds on Chapter 1: instead of forces and fields, it uses energy and potential to describe electrostatics. You learn electric potential and potential difference, equipotential surfaces, electrostatic potential energy, and then capacitance — how conductors store charge — including the parallel-plate capacitor, dielectrics, combinations, and the energy stored.
Board relevance: together with Chapter 1 this is ~8 marks. The reliable questions are a capacitance/combination numerical and a potential-or-energy derivation. Learn the formulas and the sign of the dipole potential energy.
Key Concepts & Definitions
Electric potential (V) — the work done per unit positive charge to bring a test charge from infinity to a point, without acceleration:
V = W/q (unit: volt, V = J/C; a scalar)
Potential difference VAB = VA − VB = work per unit charge to move it from B to A.
Potential due to a point charge: V = kq/r (k = 1/4πε₀ = 9 × 10⁹). Potential is positive near +q, negative near −q, and adds as a scalar for many charges.
Potential due to a dipole: on the axis V = kp/r² (r ≫ a); on the equatorial line V = 0. (Contrast with field: axial and equatorial fields are non-zero.)
Equipotential surface — a surface of constant potential. Key properties:
- No work is done moving a charge on it (V is constant).
- The electric field is always perpendicular to it.
- Equipotentials never intersect; they are closer where the field is stronger.
Relation between field and potential:
E = −dV/dr
The field points in the direction of decreasing potential; its magnitude equals the potential gradient.
Electrostatic Potential Energy
Of a system of two charges: U = kq₁q₂/r₁₂.
For many charges, sum over all pairs.
Of a dipole in a uniform field: U = −pE cos θ = −p·E.
- θ = 0° (aligned): U = −pE (minimum, stable equilibrium).
- θ = 180°: U = +pE (maximum, unstable).
- Work to rotate the dipole from θ₁ to θ₂: W = pE(cos θ₁ − cos θ₂).
Conductors in electrostatics: inside a conductor E = 0; the whole conductor is an equipotential; charge resides on the surface; field just outside is σ/ε₀ and perpendicular to the surface.
Formulas — Capacitance
Capacitance: C = Q/V (unit farad, F). It depends only on geometry and the medium, not on Q or V.
Configuration Capacitance
Parallel-plate (vacuum) C = ε₀A/d
Parallel-plate (dielectric K) C = Kε₀A/d
Isolated sphere (radius R) C = 4πε₀R
- A dielectric (insulator) increases capacitance by a factor K (dielectric constant) because it reduces the field between the plates.
- Combinations:
Combination Rule Shared
Series 1/C = 1/C₁ + 1/C₂ + … Same charge Q
Parallel C = C₁ + C₂ + … Same voltage V
Energy stored:
U = ½CV² = ½QV = Q²/2C
The energy resides in the electric field; the energy density (energy per unit volume) is u = ½ε₀E².
Worked Examples
Example 1 — Potential of a point charge: Find the potential 0.09 m from a charge of 2 × 10⁻⁹ C.
V = kq/r = (9 × 10⁹)(2 × 10⁻⁹)/0.09 = 18/0.09 = 200 V.
Example 2 — Parallel-plate capacitor: Two plates of area 100 cm² separated by 1 mm in air. Find C (ε₀ = 8.85 × 10⁻¹²).
C = ε₀A/d = (8.85 × 10⁻¹²)(100 × 10⁻⁴)/(10⁻³) = 8.85 × 10⁻¹¹ F (88.5 pF).
Example 3 — Combination: Three capacitors 2, 3, 6 μF are in series. Find the equivalent.
1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 → C = 1 μF. (In parallel they would give 11 μF.)
Example 4 — Energy: A 10 μF capacitor is charged to 100 V. Find the stored energy.
U = ½CV² = ½(10 × 10⁻⁶)(100²) = ½(10 × 10⁻⁶)(10⁴) = 0.05 J.
Important Question Patterns
1. Potential/energy (2–3 marks): potential due to point charges or a dipole; potential energy of a charge configuration; work done moving a charge between two points (W = qΔV).
2. Dipole in a field (3 marks): torque τ = pE sin θ and potential energy U = −pE cos θ; work to rotate.
3. Capacitance derivation (3 marks): derive C = ε₀A/d for a parallel-plate capacitor; effect of a dielectric slab (partly/fully filled).
4. Combinations (3 marks): equivalent capacitance of a series/parallel network; charge and voltage on each capacitor.
5. Energy (2–3 marks): energy stored; energy change when a dielectric is inserted, or when capacitors are connected (redistribution of charge → energy loss).
⚡ Quick Revision
- Potential: V = W/q (scalar, volt); point charge V = kq/r. Dipole: axial V = kp/r², equatorial V = 0.
- Equipotential surface: constant V, field ⟂ to it, no work done along it. E = −dV/dr (field points to lower V).
- Potential energy: two charges U = kq₁q₂/r; dipole in field U = −pE cos θ (min at θ = 0).
- Inside a conductor E = 0; it is an equipotential; charge on surface.
- Capacitance C = Q/V (farad). Parallel plate C = ε₀A/d; dielectric multiplies by K.
- Series: 1/C = Σ1/Cᵢ (same Q). Parallel: C = ΣCᵢ (same V).
- Energy U = ½CV² = ½QV = Q²/2C; energy density u = ½ε₀E².
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