A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 244 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume that distilled water is an insulator.
(a) What is the charge on the plates before immersion?
pC
What is the charge on the plates after immersion?
pC

(b) What is the capacitance after immersion?
pF
What is the potential difference after immersion?
V

(c) What is the change in energy of the capacitor?
?U = Uwater - Uair = 5 nJ

Explanation / Answer

a)
C = kA/d

k=1 for air
is  8.85x10-12F/m
A = .0025m2
d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)
C = kA/d,  for distilled water k is approx. 80
Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: U = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
U = 5.204nJ

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