Question

A 200 pF parallel-plate capacitor has ±35 nC charges on its plates. The plates are separated by 0.3 mm. Find:

a) the area of each plate;
cm2

b) the potential difference between the plates;
V

c) the electric field between the plates;
N/C

Answer 1

Here ,

d = 0.3 mm

C = 200 pF

a) let the area of plates is A

C = A * epsilon/d

200 *10^-12 = 8.854 *10^-12 * A/(.3 *10^-3)

solving for A

A = 6.77 *10^-3 m^2

A = 67.7 cm^2

area of plate is 67.7 cm^2

b)

as Q = C * V

35 *10^-9 = 200 *10^-12 * V

V = 175 V

the potential difference between the plates is 175 V

c)

electric field = potential difference/d

electric field = 175/(.3 *10^-3)

electric field = 5.83 *10^5 V