A parallel-plate capacitor in air has a plate separation of 1.68 cm and a plate
ID: 1553131 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.68 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 260 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before________pC
after _________pC
b) Determine the capacitance and potential difference after immersion.
(c) Determine the change in energy of the capacitor.
nJ
Explanation / Answer
a)
use this formula
C = eo*A/d = (8.84e-12*25e-4)/(1.68e-2) = 1.315*10^-12 F
Q = C*V = eo*A*V/d = 342 pC
before Q = 342 pC
in After Q = 32 pC
b) use in this formula
Cf = k*C = 80*1.315 pF = 105.2*10^-12 F
Vf= V/k = 260/80 = 3.25 v
C) that's du = 0.5*Q*V - 0.5*Q*Vf = 0.5*342e-12*(260-3.25) =
43.9 nJ