Question

A parallel-plate air capacitor with a capacitance of 240pF has a charge of magnitude 0.146uC on each plate. The plates have a separation of 0.367 mm.

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 magnitude of the surface charge density on each plate?

Answer 1

A. The potential difference is given as

V =q/C=0.146uC/240pF

608.33 V

B) The area of each plate is

A =Cd/eo =9.9525*10^-3m ²

C) the electric field magnitude between the plates=V/d

608.33/0.367mm

=1.657x 10^6 N/C

D) The surface-charge density on each plate=q/A

0.146uC/9.9525 x 10^-3m ²

=1.466 x 10^-5 C/m²