Aparallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
25.0

c
m
2
. 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.

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

(c) determine the change in energy of the capacitor.

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

Final energy

magnets not strong enough

explanation: