A parallel plate capacitor has a capacitance of 500pf and a charge of magnitude 0.25uc on each plate. What is the potential difference between the plate

Answer 1

The capacitance (C) of a parallel plate capacitor is given by the formula:

C = ε0 * A / d

where:
C = capacitance (500 pF = 500 * 10^(-12) F)
ε0 = permittivity of free space (8.85 * 10^(-12) F/m)
A = area of the plates
d = distance between the plates

From the formula, we can rearrange it to find the area:

A = C * d / ε0

Given that the charge on each plate is 0.25 μC (0.25 * 10^(-6) C) and the equation for potential difference (V) is:

V = Q / C

where:
Q = charge on the plate (0.25 * 10^(-6) C)
C = capacitance (500 pF = 500 * 10^(-12) F)

Substitute the values of Q and C into the equation:

V = (0.25 * 10^(-6)) / (500 * 10^(-12))
V = 0.5 V

Therefore, the potential difference between the plates is 0.5 V.