Problem 30.

A parallel-plate air capacitor of capacitance of 100 pF has a charge of magnitude 0.1 µC on each plate. The plates are 0.5 mm apart.

(a) What is the potential difference between the plates?

(b) What is the area of each plate?

(c) What is the electric-field magnitude between the plates?

(d) What is the surface charge density on each plate?

Solution

(a) To find the potential difference we need to use the equation:

From this equation we can find:

(b) To find the area of the plates we need to use the expression for the capacitance:

We know d=0.5 mm=0.0005 m, then we can find the area A:

(c) The potential difference is related to the electric field through the following relation:

Then

(d) The surface charge density can be found from the equation: