Parallel-plate capacitor

Question

A parallel-plate capacitor is connected to a battery and stores 4.7 nC of charge. Then, while the battery remains connected, a sheet of Teflon is inserted between the plates.

By how much does the charge change?

Answer 1

The equation for the charge on the capacitor is

Q = CV.

If the capacitor is not connected to anything the charge is constant. If the capacitor is connected to a battery the voltage is constant. Usually the capacitance is constant but in this case the capacitor changes during the problem. Remember the equation for the capacitance

C = κ ε₀ A/d.

A is the area of the plates, d is the distance between the plates, κ is a constant (dielectric constant) that depends on the material in between the plates and ε₀ is a fundamental constant. This makes the charge on the capacitor at the start of the problem

Q₀ = V ε₀ A/d.

After inserting the Teflon the charge on the plates becomes

Q₁ = V κ ε₀ A/d = κ Q₀.

The change of charge on the plates is given by

Q₀ - Q₁ = κ Q₀ - Q₀ = Q₀(κ - 1).

κ is about 2.02 for Teflon (check the tables in your book or problem set)

Q₀ - Q₁ = 4.7nC (2.02 - 1) = 4.8nC