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UY1: Energy Stored In Capacitors

Derive capacitor energy formulas and electric-field energy density with clear units and physical interpretation.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

This page gives the UY1 working model/result for Energy Stored In Capacitors. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Charging a capacitor requires work, and that work is stored as electric potential energy in the field.

1) At a glance

  • Prerequisites: Q = CΔ V, basic integration
  • Outcomes: derive U = Q²/(2C) = (1/2)C(Δ V)² = (1/2)QΔ V
  • Key results: U = Q²/2C = (1/2)C(Δ V)² = (1/2)QΔ V and u = (1/2)ε₀E² (vacuum)
  • Field view: energy density in vacuum is u = (1/2)ε₀E²
  • Common trap: confusing battery work QΔ V with stored energy (1/2)QΔ V

Motivation / intuition

Charging builds the field up gradually: early on the voltage is small, later it is larger, so the “average voltage during charging” is half the final value. That is why the stored energy comes out with a factor 1/2.

2) Setup

Let q be instantaneous charge during charging from 0 to final Q.

At each stage, V(q) = q/C Differential work done by external agent: dW = V dq = (q/C)dq

3) Core derivation/explanation

Integrate from q = 0 to q = Q: U = W = ∫₀^Q(q/C)dq = Q²/2C Using Q = CΔ V: U = (1/2)C(Δ V)² = (1/2)QΔ V

For ideal parallel plates in vacuum: C = ε₀A/d, E = (Δ V)/d Then U = (1/2)ε₀E²(Ad) Since Ad is field volume, energy density is u≡U/volume = (1/2)ε₀E²

Units check:

  • [U] = J
  • [u] = J/m³
Quick checks (units + limits/sign)
  • Units: [U] = J and [u] = J/m³ (as shown).
  • Limits/signs: stored energy must be nonnegative; if Q → 0 (or Δ V → 0), then U → 0; for fixed Q, increasing C decreases U (since U = Q²/(2C)).

4) Worked example(s)

A capacitor has C = 4.0 μF and Δ V = 150 V.

Charge: Q = CΔ V = (4.0 × 10⁻⁶)(150) = 6.0 × 10⁻⁴ C Energy: U = 1/2 C(Δ V)² = (1/2)(4.0 × 10⁻⁶)(150)² = 4.5 × 10⁻² J

5) Practice set (with hints + answers)

  1. If voltage doubles with fixed C, how does stored energy change?
  2. Write one expression for U in terms of Q and C only.
  3. For vacuum field E = 3.0 × 10⁵ V/m, find u.

Hints

  1. Use U = 1/2 C(Δ V)².
  2. Integrate or rearrange.
  3. Use u = (1/2)ε₀E².

Answers

  1. It becomes four times larger.
  2. U = Q²/(2C).
  3. u ≈ 0.40 J/m³.

6) Summary + next steps

  • Capacitor energy is the integrated charging work, not simply final QΔ V.
  • Equivalent formulas let you choose the most convenient variables.
  • Energy is stored in the electric field with density u = (1/2)ε₀E² in vacuum.

Next: Capacitance Of Spherical Capacitor Previous: Capacitors In Series And In Parallel Back To Electromagnetism (UY1)