Energy stored in a capacitor — derive U = ½CV² from first principles

Question

Derive the expression for energy stored in a charged capacitor: $U = \frac{1}{2}CV^2$ . Show the two other equivalent forms.

(JEE Main 2022, similar pattern)

Solution — Step by Step

Consider a capacitor being charged from 0 to final charge $Q$ . At some intermediate stage, let the charge on the capacitor be $q$ and the potential difference be $v = q/C$ .

To bring an additional small charge $dq$ to the capacitor, we must do work against the existing potential:

dW = v \cdot dq = \frac{q}{C} \, dq

Total work done in charging from 0 to $Q$ :

W = \int_0^Q \frac{q}{C} \, dq = \frac{1}{C} \cdot \frac{q^2}{2}\Bigg|_0^Q = \frac{Q^2}{2C}

This work is stored as electrostatic potential energy in the capacitor.

Using $Q = CV$ , we get three equivalent expressions:

\boxed{U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV}

Each form is useful depending on what quantities are given in the problem:

$U = \frac{1}{2}CV^2$ — when $C$ and $V$ are known
$U = \frac{Q^2}{2C}$ — when $Q$ and $C$ are known
$U = \frac{1}{2}QV$ — when $Q$ and $V$ are known

Why This Works

The energy comes from the work done by the external agent (battery) in moving charge against the electric field inside the capacitor. As more charge accumulates, the voltage increases, so each subsequent bit of charge requires more work — hence the integral and the factor of $\frac{1}{2}$ .

The energy is stored in the electric field between the plates. The energy density (energy per unit volume) in the field is:

u = \frac{1}{2}\epsilon_0 E^2

This is a fundamental result that applies to any electric field, not just capacitors.

Alternative Method — Using Energy Density

For a parallel plate capacitor with plate area $A$ , separation $d$ , and field $E = V/d$ :

U = u \times \text{volume} = \frac{1}{2}\epsilon_0 E^2 \times Ad = \frac{1}{2}\epsilon_0 \frac{V^2}{d^2} \times Ad = \frac{1}{2}\frac{\epsilon_0 A}{d}V^2 = \frac{1}{2}CV^2

JEE frequently tests what happens to the stored energy when a dielectric is inserted. With the battery connected: $V$ stays constant, $C$ increases, so $U = \frac{1}{2}CV^2$ increases. With the battery disconnected: $Q$ stays constant, $C$ increases, so $U = Q^2/(2C)$ decreases. Use the appropriate formula based on what stays constant.

Common Mistake

Students often write $U = QV$ instead of $U = \frac{1}{2}QV$ . The factor of $\frac{1}{2}$ is essential — it arises because the voltage builds up gradually during charging (it’s not constant at $V$ throughout). The battery delivers energy $QV$ , of which $\frac{1}{2}QV$ is stored in the capacitor and $\frac{1}{2}QV$ is dissipated as heat in the connecting wires.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Energy Density

Common Mistake

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