Learn first: Electric field and potential
A capacitor is a device that stores charge and the energy of an electric field. You find one in a camera flash, a defibrillator, the tuned circuit of a radio, computer memory, motor-starting circuits and on almost every circuit board. This topic applies what we know about electric fields to practical problems and prepares the ground for “Electromagnetic oscillations”.
When an isolated conductor is given a charge, its potential rises in proportion to the charge. The constant of proportionality shows how well the conductor stores charge:
C = q/U, unit farad (F): 1 F = 1 C/V
Capacitance does not depend on the charge or the voltage, only on the size, shape and arrangement of the conductors and on the surrounding medium. Doubling the charge doubles the voltage, and the ratio stays the same.
1 F is a huge unit: a sphere with capacitance 1 F would need a radius of 9·10⁹ m (about 1400 times the Earth’s radius). In practice smaller units are used: 1 μF = 10⁻⁶ F, 1 nF = 10⁻⁹ F, 1 pF = 10⁻¹² F.
A single conductor has a small capacitance. To store a lot of charge we use a capacitor: two conductors (plates) separated by a thin layer of dielectric. When charged, the plates carry equal and opposite charges (+q and −q); the charge of a capacitor means the charge q on one plate. The field is almost entirely between the plates.
A parallel-plate capacitor is two parallel plates a distance d apart:
C = εε₀S/d, ε₀ = 8.85·10⁻¹² F/m
The field inside is uniform: E = U/d.
By dielectric, capacitors may be paper, ceramic, mica or electrolytic; in a variable capacitor one set of plates rotates and changes the overlapping area, which is how a radio is tuned. Every capacitor has a maximum working voltage; above it the dielectric breaks down.
Each plate has an area of 100 cm², the gap is 1 mm and filled with air. Find the capacitance.
If a problem changes the capacitor’s parameters (d, S, ε), first decide its situation:
This also decides how the energy and the field strength change.
The plate separation of a disconnected capacitor is doubled. How do C, U, E and W change?
Parallel: C = C₁ + C₂ + ..., same voltage, charges add
Series: 1/C = 1/C₁ + 1/C₂ + ..., same charge, voltages add
In series the voltage divides in inverse proportion to the capacitances: the smaller capacitor gets the larger voltage. So capacitors are put in series to withstand a high voltage and in parallel to get a large capacitance.
A dielectric of permittivity ε completely fills an air capacitor of capacitance C, and a capacitor Cₓ is connected in series with it; the total capacitance is C/2. Find Cₓ.
While charging, the source moves charge from one plate to the other and does work; since the voltage rises gradually from zero to U, the average voltage is U/2:
W = qU/2 = CU²/2 = q²/(2C)
The energy is stored in the electric field between the plates. Pulling apart the plates of a disconnected capacitor requires work, so the energy grows; for a connected capacitor the source exchanges energy as well.
Putting C = εε₀S/d and U = Ed into W = CU²/2 for a parallel-plate capacitor gives the energy per unit volume:
Energy density of the field: w = W/(Sd) = εε₀E²/2
This formula holds for any electric field, not only in a capacitor: the energy is in the field itself, not in the charges.
When a charged capacitor is connected to an uncharged one, charge is conserved, but some energy is lost as heat in the wires and as radiation.
A 10 μF capacitor is charged to 200 V. Find the charge and the energy.
A 2 μF capacitor charged to 100 V is connected in parallel with an identical uncharged one. How do the voltage and energy change?