Lesson 12 of 13 · 8 min
Energy stored in a capacitor
NCERT §2.15
Meera charges one 10 μF capacitor to 12 V, then connects it across a small bulb and the bulb flashes. Where was that energy, and how much was there?
The lesson in notes
In short
Charging moves charge from the negative plate to the positive plate, which is at the higher potential, so external work is done at every step.
When the charge is Q′ the voltage is Q′/C; moving a further δQ′ takes (Q′/C)δQ′. Adding these from 0 to Q gives W = Q²/(2C).
Stored energy U = Q²/(2C) = ½CV² = ½QV. It is ½QV, not QV, because the voltage builds up from zero while the charge is moved.
The energy can be pictured as stored in the field between the plates. For a parallel plate capacitor U = ½ε₀E² × Ad, where Ad is the volume of the gap.
Energy density of an electric field u = ½ε₀E² (joules per cubic metre). This holds for any field, not only a capacitor's.
Worked value: 900 pF charged to 100 V holds Q = 9 × 10⁻⁸ C and U = 4.5 × 10⁻⁶ J.
Join that charged capacitor to an identical uncharged one: the charge shares equally, V halves to 50 V, and the total energy drops to 2.25 × 10⁻⁶ J. Charge is conserved; the missing half is lost as heat and electromagnetic radiation while charge flows.