1. Energy Stored in a Capacitor — Three Forms

U=12CV2=Q22C=12QV

All three forms are equivalent and useful in different situations:

  • Use ½CV² when C and V are given.
  • Use Q²/2C when Q and C are given (or when charge is constant — battery disconnected).
  • Use ½QV when both Q and V are given.

2. Energy Density of the Electric Field

The energy stored in a capacitor is actually stored in the electric field between the plates. For a parallel plate capacitor of area A, separation d:

u=UVolume=UAd=12ε0E2

This result is universal — it applies to any electric field, not just between capacitor plates. The energy density at any point in space where the electric field is E is ½ε₀E².

Example: E = 2×10⁵ N/C → u = ½ × 8.854×10⁻¹² × (2×10⁵)² = 0.177 J/m³

3. Charge Redistribution — Common Voltage

When a charged capacitor C₁ (charge Q₀) is connected in parallel to an uncharged capacitor C₂, charge redistributes until both reach a common potential:

Vcommon=Q0C1+C2

Energy is always lost in this process (converted to heat and electromagnetic radiation), unless the two capacitors were at the same potential initially.

Worked Example

C₁ = 4 μF charged to 200 V (Q₀ = 800 μC... wait, Q = CV = 4×200 = 800 μC).

Wait — let's use Q₀ = 200 μC (C₁ = 4 μF, V₀ = 50 V) for clean numbers... Using the verified example:

C₁ = 4 μF, Q₀ = 200 μC (V₀ = 50 V). Connected to uncharged C₂ = 6 μF.

V_common = 200×10⁻⁶ / (4+6)×10⁻⁶ = 20 V

Q₁_new = 4×20 = 80 μC; Q₂_new = 6×20 = 120 μC; Total = 200 μC ✓

U_initial = Q₀²/(2C₁) = (200×10⁻⁶)²/(2×4×10⁻⁶) = 5 mJ

U_final = ½(C₁+C₂)V² = ½×10×10⁻⁶×400 = 2 mJ

Energy lost = 5 − 2 = 3 mJ (to heat and radiation)

4. Van de Graaff Generator

Principle

Based on two key electrostatic properties:

  1. Corona discharge (action of points): At sharp points, the electric field is very high → air ionises → charges are sprayed onto or removed from a belt.
  2. Charge resides on outer surface: Any charge placed inside a hollow conductor migrates immediately to the outer surface — no matter how much charge is already on the sphere, more can always be added.

Working

  1. A motor-driven insulating belt passes over a lower pulley near a metal comb connected to a high-voltage source (+). Charge is sprayed onto the belt.
  2. Belt carries positive charge upward into the large hollow metallic sphere.
  3. An upper metal comb inside the sphere removes the charge from the belt and delivers it to the outer surface of the sphere.
  4. Since charge always moves to the outer surface, the inner comb is always at low potential — the belt keeps depositing more charge.
  5. Potential on sphere builds up continuously: V = kQ/R = Q/(4πε₀R).

Maximum Voltage

Voltage is limited by dielectric breakdown of air (E_breakdown ≈ 3 × 10⁶ V/m):

Vmax=R×Ebreakdown=R×3×106 V/m

For R = 1 m sphere: V_max = 1 × 3×10⁶ = 3 MV. To increase V_max, the sphere must be larger or placed in a pressurised gas (to raise the breakdown threshold).

Applications

  • Nuclear physics — accelerating charged particles to high energies for nuclear reactions.
  • X-ray generators.
  • Electrostatic painting and powder coating in industry.
  • Demonstrations of high-voltage electrostatics.