A parallel-plate capacitor in air has a plate separation of 1.59 cm and a plate
ID: 2075185 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.59 cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 275 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. If the current carried by a conductor is doubled, how are each of the following affected? (a) The charge carrier density (b) The current density (c) The electron drift velocity (d) The average time interval between collisionsExplanation / Answer
a) d = 1.59 cm = 0.0159 m , A = 25.0 cm2 = 0.0025 m2 , V = 275 volts
Capacitance C = e0 * A / d = 8.85 * 10-12 * 0.0025 / 0.0159 = 1.39 * 10-12 F
Charge = C V = 1.39 * 10-12 * 275 = 381 * 10-12 C = 382 pC
after immersion also the charge on the plates remain the same = 382 pC
b) Dielectric constant of distilled water = 80
Capacitance after immersion = capacitanceair * 80 = 1.39 * 10-12 * 80 = 111 * 10-12= 111 pF
after immersion = potential difference in air / 80 = 275/80 = 3.44 volts
c) Energy stored = Q2 / 2 C
Change in energy stored = (Q2 / 2 ) * (1/C2 - 1/C1)
= (382 * 10-12)2 / 2 * [1/(111 * 10-12) - 1/(1.39 * 10-12)]
= 5.18 * 10-8 J