Energy Stored in a Capacitor

You’ll be able to

Derive U = Q²/2C = ½CV² by integrating dU = (q/C) dq
Show that the energy resides in the field, with density u = ½ε₀E²
Track energy when a charged capacitor’s geometry is changed at fixed Q or fixed V

Charging costs work — integrate it

To charge a capacitor you ferry charge from one plate to the other against the growing potential difference. When charge q has accumulated, the next increment dq must be lifted through v = q/C, costing dU = v·dq = (q/C)·dq. The very first bit of charge is free (v = 0); the last bit is the most expensive (v = V). Total stored energy is the integral over the whole charging history — and the factor of ½ appears automatically, because the average lift is only V/2.

Worked example

Derive the energy stored in a capacitor charged to final charge Q and voltage V, then evaluate for a 6.0 μF capacitor at 200 V.

1.Sum the work for every increment: U = ∫₀^Q (q/C) dq.
2.Integrate: U = (1/C)·[q²/2] from 0 to Q = Q²/(2C).
3.Substitute Q = CV for the equivalent forms: U = ½CV² = ½QV. All three are the same statement in different variables.
4.Numerically: U = ½(6.0 × 10⁻⁶ F)(200 V)² = ½(6.0 × 10⁻⁶)(4.0 × 10⁴) = 0.12 J.
5.Check with the average-voltage picture: U = ½QV = ½(CV)(V) — moving Q through an average of V/2 ✓.

Answer: U = Q²/2C = ½CV² = ½QV; here 0.12 J

Stored energy (three equivalent forms)

U = Q²/(2C) = ½CV² = ½QV

Choose the form whose variables are held constant in your scenario: Q²/2C when the capacitor is isolated, ½CV² when a battery pins the voltage.

The energy lives in the field

Where is that energy physically? In the field filling the gap. For a parallel-plate capacitor, U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E²·(A·d) — the last factor is exactly the volume between the plates. So the field carries an energy density u = ½ε₀E², in J/m³, and this turns out to be a completely general result: any electric field, anywhere, stores ½ε₀E² per unit volume. (In Unit 5 the magnetic field gets its twin, u = B²/2μ₀.)

Energy density of the electric field

u = ½ε₀E² (energy per unit volume)

General, not just for capacitors. At air’s breakdown field, 3 × 10⁶ V/m, u ≈ 40 J/m³ — why electric fields make poor bulk energy storage.

Worked example

A 4.0 μF parallel-plate capacitor is charged to 50 V, then disconnected from the battery. The plates are pulled apart to double the separation. Find the new voltage and stored energy, and account for the energy change.

1.Disconnected ⇒ the charge is trapped: Q = C₀V₀ = (4.0 × 10⁻⁶)(50) = 2.0 × 10⁻⁴ C, fixed throughout.
2.Doubling d halves the capacitance (C = ε₀A/d): C = 2.0 μF. New voltage: V = Q/C = (2.0 × 10⁻⁴)/(2.0 × 10⁻⁶) = 100 V — it doubles.
3.Initial energy: U₀ = ½C₀V₀² = ½(4.0 × 10⁻⁶)(2500) = 5.0 × 10⁻³ J.
4.Final energy (fixed-Q form): U = Q²/2C = (2.0 × 10⁻⁴)²/(2 × 2.0 × 10⁻⁶) = (4.0 × 10⁻⁸)/(4.0 × 10⁻⁶) = 1.0 × 10⁻² J — it doubles.
5.The extra 5.0 × 10⁻³ J is the work you did pulling the oppositely charged plates apart against their attraction. Field view: E = σ/ε₀ is unchanged, but the field now fills twice the volume, so u·(volume) doubles ✓.

Answer: V = 100 V; U = 1.0 × 10⁻² J. The added 5.0 mJ equals the mechanical work done separating the plates.

Tip

Before touching any energy formula, decide what is held fixed. Battery removed ⇒ Q is constant ⇒ use U = Q²/2C. Battery attached ⇒ V is constant ⇒ use U = ½CV². Using the wrong pair silently assumes the wrong quantity is constant and flips your answer (double vs. half).

Checkpoint

Why is the stored energy ½QV rather than QV?

Checkpoint

A parallel-plate capacitor stays connected to its battery while the plate separation is doubled. The stored energy:

Answer the 2 checkpoints as you read.