The capacitor — capacitance, charge, energy

What just happened?

You halved the distance between the plates, and the capacitance doubled. Capacitance and plate separation are inversely proportional.

What is capacitance?

It is a capacitor’s ability to store charge: $C = Q/U$. The same charge at a lower voltage means a larger capacitance. It depends on the plate geometry and the dielectric:

— it grows with the plate area S, drops with the separation d, and filling it with a dielectric (ε_r > 1) increases it. ε₀ = 8.85·10⁻¹² F/m is the permittivity of free space.

Charge and energy

Connected to a voltage U, the capacitor stores a charge $Q = CU$, and in the electric field between the plates it stores energy:

The energy does not sit "in the wire" — it is stored in the electric field in the space between the plates. That is why a charged capacitor can release it suddenly (camera flash, defibrillator).

Invariant: constant charge

Disconnect the source — the charge Q stays trapped. Then from $Q = CU$ it follows that decreasing the capacitance (e.g. pulling the plates apart) increases the voltage U, because the product CU must stay the same.

Why this matters on the exam:

C = ε₀S/d, Q = CU and E = ½CU² solve almost every capacitor problem. The trick: check what stays constant — the voltage U (source connected) or the charge Q (source disconnected) — and only then calculate.

Remember: capacitance grows with the plate area and drops with their separation (C = ε₀S/d), and the capacitor stores energy in the electric field between the plates (E = ½CU²).