A parallel-plate capacitor in air has a plate separation of 1.46 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.46 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 235 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.

(b) Determine the capacitance and potential difference after immersion.

(c) Determine the change in energy of the capacitor.
_______________nJ

Explanation / Answer

C = eoA/d

C = 8.85 x 10^-12 x 25 x 10^-4 / 1.46 x 10^-2

C = 1.52 x 10^-12 F

part a )

Q = CV

Q = 357.2 pC

after immersion

charge remain same because distill water act as a insulator

after = 357.2 pC

part b ) 1.52 x 10^-12 C

after immersion = Cf = k * C

k = 80 dielectricconstant for water

Cf = 1.216 x 10^-10 F

Vf = V/k

Vf = 235/80 = 2.9375 V

part c )

Ui = 1/2 * CV^2 = 4.1971x 10^-8

Uf = 1/2*Cf*Vf^2 = 4.281935 x 10^-10

change in energy = Uf - Ui

dU = 41.54 nJ

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