A parallelplate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55 uC
What is the potential difference between the plates?
If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled?
How much work is required to double the separation?
4 Answers

1st answer is correct with respect to the potential differences before and after doubling the separation. However, work is equal to the change in potential energy stored in the capacitor.
PE = V^2C/2.
The separation doubles PE because it quadruples V^2 while halving C.
PE1 = V1^2*C1/2 = 3.5339674E03 J
C2 = C1/2
V2 = 2V1
PE2 = V2^2*C2/2 = 4V1^2*C1/4 = V1^2*C = 2PE1 = 7.0679348E03 J
Work = PE2PE1 = PE1 = 3.5339674E03 J

V = Q/C = 2.55 * 10^6/920 * 10^12 = 0.00277 * 10^6 = 2.77 * 10^3 V
V = Q/C = Qd/AEo
if seperation is doubled, potential also doubles!
W = Q(V2 – V1) = QV = 2.55 * 10^6 * 2.77 * 10^3 = 7.0635 * 10^3 J

Q is constant:
C1 = eo A/d1
C2 = eo A/(2*d1) = (C1 / 2)
U1 = Q^2/C1/2
U2 = Q^2/C1
delta_work = U2 – U1 = Q^2/C1/2
delta_work = 0.00353396739 J

(a) By C = εo x A/d =>d = (εo x A)/C =>d = (8.85 x 10^12 x 1.2 x 10^4)/(5.7 x 10^12) =>d = 1.86 x 10^4 m =>d = 0.186 mm (b) By E = 1/2CV^2 =>E = 1/2 x 5.7 x 10^12 x (220)^2 =>E = 1.38 x 10^7 J (c) By C’ = C x k =>C’ = 5.7 x 10^12 x 3.5 =>C’ = 19.95 (pF) By C = Q/V =>Q = V x C’ =>Q = 220 x 19.95 x 10^12 =>Q = 4.39 x 10^9 C