EM-11 · Electrostatics field

Capacitance

Capacitance is how much charge two conductors hold at a given voltage. Narrow the plate gap and insert a dielectric to see that capacitance is set by geometry and material (C = εA/d).

Narrow the gap to raise capacitance

Use the slider to narrow the plate gap of the parallel-plate capacitor. How does the capacitance change? And how does it change again when you fill the gap with a dielectric?

Plate gap dd = 1.00

Drag to orbit. Slider for the plate gap d, buttons for the dielectric.

This capacitor's capacitance

C = Q/V = εA/d · set by geometry and material

proportional to area A, inverse to gap d, proportional to ε

Capacitance (vs vacuum, d=1) · ×1.0

Small capacitance

What capacitance is

Place two conductors close, put +charge on one and −charge on the other, and a field and a voltage V appear between them. How much charge Q they can hold at a given voltage is the capacitance C: C = Q/V, measured in farads (F). A larger C stores more charge at the same voltage. Such a pair of conductors is a capacitor.

ObserveC = QV

Capacitance is charge stored per volt.

Parallel plates, C = εA/d

The simplest capacitor is two parallel plates of area A facing each other a gap d apart. The field between is nearly uniform, E = V/d. Solving with Gauss’s law gives the capacitance C = εA/d. Wider plates (A↑) take more charge, and a smaller gap (d↓) makes a stronger field at the same voltage, holding more charge. So C is set by geometry, not by the charge held or the voltage.

ChooseC = ε ?

Parallel plates scale with area, inverse with the gap.

Boosting it with a dielectric

Fill the gap with a dielectric and the capacitance grows by ε_r (EM-10). For the same charge, the dielectric’s polarization weakens the field by ε_r, so the voltage V drops by the same factor. In C = Q/V a smaller V means a larger C. With ε = ε_r ε₀ the formula reads C = ε_r ε₀ A/d, that is ε_r times the vacuum value. Real capacitors therefore pack in thin, high-permittivity films to raise capacitance.

Fill inC = ?

A dielectric makes the capacitance ε_r times larger.

The stored energy

Charging a capacitor means pushing charge from one plate to the other, and that work is stored as energy. The stored energy is U = ½CV² = ½QV = Q²/(2C). At first the voltage is small and charge moves easily, but as charge piles up V rises and it gets harder, so the average brings in the factor of one half. This energy in fact lives in the electric field between the plates. The next unit re-sees it as field energy.

On your ownU = ?

The stored energy is ½CV².

Back to the first screen

On the first screen, narrowing the plate gap d raised the capacitance (C ∝ 1/d): at the same voltage a tighter gap makes a stronger field that pulls in more charge. Inserting a dielectric bumped it up by ε_r once more, because polarization weakens the field and drops the voltage for the same charge. What sets capacitance is not the charge held or the voltage, but area, gap, and permittivity — geometry and material. C = εA/d.

The capacitance C is how much charge two conductors hold at a given voltage: C = Q/V. A parallel-plate capacitor is proportional to plate area A, inverse to the gap d, and proportional to the permittivity ε of the filling: C = εA/d. A dielectric makes C ε_r times larger. The stored energy is U = ½CV².

What comes next

The energy U = ½CV² stored in a capacitor in fact resides in the electric field filling the gap. The next unit (EM-12, electrostatic energy) re-sees it as a field energy density u = ½ε E². The view that energy lives not on the charges but spread through the field in space carries forward to electromagnetic waves (EM-23 onward).