A parallel-plate capacitor in air has a plate separation of 1.74 cm and a plate
ID: 1524921 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.74 cm and a plate area of 25.0 cm2 . The plates are charged with a battery to a potential difference of 6.00 V. The capacitor is then immersed in a liquid with a dielectric constant of 8.00 while the capacitor still connected to the to the battery. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. (b) Determine the potential difference across the capacitor after immersion. (c) Determine the change in the stored energy in the capacitor.
Explanation / Answer
here,
plate sepration , d = 0.0174 m
area = 25 cm^2 = 0.0025 m^2
potential difference , V = 6 V
K = 8
a)
the charge on the plates before immersion , Q = C * V
Q = V * area * e0 /d
Q = 6 * 0.0025 * 8.85 * 10^-12 /0.0174
Q = 7.63 * 10^-12 C
the charge on the plates after immersion , Q' = C' * V
Q' = V * k * area * e0 /d
Q' = 6 * 8 * 0.0025 * 8.85 * 10^-12 /0.0174
Q' = 6.1 * 10^-11 C
b)
as the battery is connected after immersion
the potential difference remains same i.e 6 V
c)
the change in the stored energy in the capacitor , E = 0.5 * C' * V^2 - 0.5 * C * V^2
E = 0.5 * V^2 * ( k * area * e0 /d - area * e0 /d)
E = 0.5 * V^2 * area * e0/d * ( K - 1)
E = 0.5 * 6^2 * 0.0025 * 8.85 * 10^-12 /0.0174 * ( 8 - 1) J
E = 1.6 * 10^-10 J