Work done in charging capacitor = ½QV = ½CV². Step-by-step derivation from energy stored in electric field.

Energy Stored in a Capacitor — Derive the Formula U = ½CV²

Question

Derive the expression for the energy stored in a capacitor of capacitance $C$ charged to a potential $V$ . Show that $U = \frac{1}{2}CV^2$ .

Solution — Step by Step

We don’t charge a capacitor for free. Moving each small charge $dq$ from the negative plate to the positive plate requires work — because we’re pushing positive charge against an already-existing electric field. That work gets stored as potential energy.

So we need to find the total work done to move charge from $q = 0$ to $q = Q$ .

At any intermediate stage, suppose charge $q$ has already been transferred. The potential difference across the plates at that instant is:

V' = \frac{q}{C}

This is not the final $V$ — it’s the potential while charging is still in progress.

To transfer a small additional charge $dq$ against potential $V'$ , the work done is:

dW = V' \, dq = \frac{q}{C} \, dq

This is just the definition of work: $W = q \cdot V$ , applied at the microscopic level.

The total work done to charge the capacitor from 0 to full charge $Q$ :

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

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

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

All three are correct. In problems, use whichever form has the known quantities.

Why This Works

When $q = 0$ , no work is needed to transfer the first bit of charge — there’s no opposing field yet. But as charge builds up, the potential increases, and each successive $dq$ costs more work. The integral captures this accumulating resistance exactly.

The factor of $\frac{1}{2}$ is the key here — and students often find it surprising. The average potential during charging is $\frac{V}{2}$ (it goes from 0 to $V$ linearly), so the total work is $Q \times \frac{V}{2} = \frac{1}{2}QV$ . The $\frac{1}{2}$ is not arbitrary; it reflects that you never charged the capacitor at the full final potential.

This energy is stored in the electric field between the plates, not in the charges themselves. NCERT Class 12 and JEE both ask you to connect this to the concept of energy density: $u = \frac{1}{2}\varepsilon_0 E^2$ J/m³.

Alternative Method

Using energy density of the electric field:

The electric field between plates is $E = \frac{\sigma}{\varepsilon_0} = \frac{V}{d}$ . The energy stored per unit volume is:

u = \frac{1}{2}\varepsilon_0 E^2

Multiply by volume $Ad$ of the space between plates:

U = \frac{1}{2}\varepsilon_0 E^2 \cdot Ad = \frac{1}{2}\varepsilon_0 \left(\frac{V}{d}\right)^2 \cdot Ad = \frac{1}{2}\varepsilon_0 \frac{V^2 A}{d}

Since $C = \frac{\varepsilon_0 A}{d}$ , this gives $U = \frac{1}{2}CV^2$ . Same result, different path.

JEE sometimes asks for energy stored in a dielectric-filled capacitor. The formula stays $\frac{1}{2}CV^2$ — you just update $C = \frac{K\varepsilon_0 A}{d}$ where $K$ is the dielectric constant. The energy increases when a dielectric is inserted at constant charge (battery disconnected).

Common Mistake

Using $U = QV$ instead of $U = \frac{1}{2}QV$ .

Students confuse this with the work done on a charge $Q$ moving through a fixed potential $V$ , where $W = QV$ is correct. But during capacitor charging, the potential is not fixed — it grows from 0 to $V$ . The average is $\frac{V}{2}$ , hence the factor of $\frac{1}{2}$ . Writing $U = QV$ gives exactly double the correct answer and is a classic trap in MCQs.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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