Question

A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 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.
Cf = F
ΔVf = V
(c) Determine the change in energy of the capacitor.
[ ] nJ

Answer 1

Answer:

a) before immersion

C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F

q = CV = (1.68e-12)(255) = 4.28e-10 C

b) after immersion

q = 4.28e-10 C

Because the capacitor was disconnected before it was immersed, the charge remains the same.

c)*at 20° C

C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F

V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V

e)

U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J

U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J

ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J