A parallel plate capacitor made of two plates that can carry equal and opposite charge.

Recall the symmetry of an infinite plane dictates that the electric field is always perpendicular to the plane. Since the plate areas of a parallel plate capacitor are much larger than the separation between the plates, we can treat them as infinite planes.

We found the electric field of an infinite plane of charge does not depend on the distance from the plane. The magnitude of the E field is the surface charge density over twice the permittivity constant.

By the principle of superposition, the electric fields of the two plates of a capacitor simply add. Since the charges on the plates have opposite sign, the fields are in opposite directions. The field of the positive plate points outward, while the field of the negative plate points inward toward the plate.

Between the plates, the fields add together to make twice the magnitude of a single plate. The direction of the E field is from positive to negative. Outside the capacitor, the E fields cancel, leaving zero field outside.

Consider the relationship between gravitational potential energy and the work done by gravity on a massive object. If a ball is at rest with a gravitational potential energy U₀ at a height y₀ and is released, it falls to a height y_f and has a gravitational potential energy U_f. If the ball is near Earth's surface, the gravitational force is constant.

Δr is positive in the dot product so we write it as y₀ - y_f. The force is in the same direction as the displacement so cos θ = 1. The change in gravitational potential energy takes its familiar form.



Similarly, we can define the change in electric potential energy as the work done on the charged particle by the electric field. If we have a constant electric field, we don't need to do an integral. We have to be more careful here because charged particles can have two different signs. For example, if we are considering a charged particle moving in a constant electric field, we need to do the dot product to get the correct sign for the change in potential.

Sometimes we only care about the magnitude of the potential difference. We can use the relationship between potential difference and the E field to find the potential difference inside a capacitor. Using the definition of the charge density as Q/A, we can relate the potential difference of a capacitor to its physical dimensions.

Sample problem

Two metal plates each measure 2.0 cm x 2.0 cm, and are spaced 2.0 mm apart. The electric field between the plates is 5.0 x 10⁵ V/m.

Use ε₀ = 8.85 x 10 ^-12 C²/Nm², m_e = 9.11 x 10^-31 kg, q_e = 1.60 x 10^-19 C.

a. What is the voltage across the capacitor?

b. What is the charge on each plate?

c. An electron is launched from the positive plate straight toward the negative plate. It strikes the negative plate at a speed of 2.0 x 10⁷ m/s. What was the initial speed of the electron?

A capacitor with plate separation d is charged by wiring it to a battery as shown. After charging, it has the same potential difference as the battery, ΔV. Then the wires are disconnected and the plates are pulled apart to a distance 2d by insulating handles.

a. Does the charge on the plates change?

b. Does the electric field between the plates change?

c. Does the potential difference across the plates change?