EM-11 · Electrostatics field

Capacitance

Capacitance is the amount of charge two conductors can hold at the same voltage. Narrowing the plate gap and inserting a dielectric, you confirm for yourself that capacitance is set by shape and material (C = εA/d).

Narrow the gap to raise the capacitance

In a parallel-plate capacitor, the red plate on the left is charged + and the blue plate on the right −, with uniform field lines running between them. Narrowing the gap with the slider raises the capacitance, and filling the gap with a dielectric using the button raises it once more. The multiple on the indicator compares with a vacuum at the reference gap.

Plate gap dd = 1.00

Drag to rotate; the slider sets the plate gap d, and the buttons insert or remove 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

Put two conductors close together, give one + charge and the other − charge, and an electric field and a potential difference V appear between them. The quantity that says how much charge Q they can hold for a given voltage is the capacitance C, and C = Q/V. Its unit is the farad (F). The larger C, the more charge is stored at the same voltage, and such a paired set of conductors is called a capacitor.

ObserveC = QV

Capacitance is the charge held per unit voltage.

Parallel plates, C = εA/d

The simplest capacitor is two parallel plates of area A facing each other across a gap d. If the plates are much wider than the gap, the bending of the field at the edges can be ignored, and the field between the plates is nearly uniform, E = V/d. Solving with Gauss's law gives the capacitance C = εA/d. The wider the plates (A↑), the more charge they take in, and the narrower the gap (d↓), the stronger the field at the same voltage and the more charge they hold. So C is set not by the charge held or the voltage applied but by shape and material.

ChooseC = ε ?

The capacitance of parallel plates is proportional to area and inversely proportional to the gap.

Boosting it with a dielectric

Filling the gap with a dielectric raises the capacitance by a factor of ε_r (EM-10). For the same charge, the dielectric's polarization weakens the field by a factor of ε_r, so the potential difference V drops by that much. In C = Q/V, a smaller V means a larger C; as a formula, putting in ε = ε_r ε₀ gives C = ε_r ε₀ A/d, that is, ε_r times the vacuum value. Real capacitors insert a thin film with high permittivity to raise capacitance. If the film is made too thin, however, the voltage at which the insulation breaks down drops, so there is a trade-off between capacitance and the voltage it can withstand.

Fill inC = ?

Inserting a dielectric multiplies the capacitance by ε_r.

The stored energy

Charging a capacitor means pushing charge from one plate over to the other, and that work is stored as energy. The stored energy is U = ½CV² = ½QV = Q²/(2C). At first the potential difference is small and charge moves easily, but as charge builds up the potential difference grows and moving more gets harder, so averaging gives the factor of one half (½). This energy is actually held in the electric field between the plates, and the next unit looks at it again 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, the field in a narrow gap is stronger and pulls more charge onto the plates. Inserting the dielectric made the capacitance jump once more by a factor of ε_r, because polarization weakens the field and lowers the potential difference for the same charge. What sets capacitance is not the charge held or the voltage but area, gap, and permittivity, that is, shape and material. C = εA/d.

Capacitance C is the amount of charge two conductors hold at the same voltage. C = Q/V. The capacitance of a parallel-plate capacitor is proportional to plate area A, inversely proportional to the gap d, and proportional to the permittivity ε of the filling material. C = εA/d. Inserting a dielectric multiplies C by ε_r. The stored energy is U = ½CV².

What comes next

The energy stored in a capacitor, U = ½CV², is actually held in the electric field that fills the space between the plates. The next unit (EM-12 electrostatic energy) revisits it as the field's energy density u = ½ε E². The view that energy is not attached to charges but spread through the field in space leads on to electromagnetic waves (EM-23 onward).