A parallel-plate capacitor in air has a plate separation of 1.33 cm and a plate
ID: 2289072 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.33 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 265 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 = nJ
Explanation / Answer
C = A*epsilon/d
= 25*10^-4*8.854*10^-12/0.0133
= 1.66*10^-12
= 1.66 pF
a) Q = C*V
= 1.66*265
= 440 pC
after immersion charge remains same.
Q' = Q = 440 pC
b) C' = k*C
= 80*1.66* pF
= 132.8 pF
V' = V/80
= 265/80
= 3.31 volts
c) Uair = 0.5*C*v^2 = 0.5*1.66*10^-12*265^2 = 5.82*10^-8 J
Uwater = 0.5*C'*v'^2 = 0.5*132.8*10^-12*3.31^2 = 0.073*10^-8 J
delta U = Uwater - Uair = -5.747*10^-8 J = -57.57 nJ