Question

A parallel-plate capacitor in air has a plate separation of 1.43 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 before 363 545PC after 28.5666 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. pc (b) Determine the capacitance and potential difference after immersion. (c) Determine the change in energy of the capacitor Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nJ

Answer 1

Given data: - d = 1.43 cm lambda = 25 cm^2 v = 235 v Before Q = c v = 1.547 times 235 v Q = 363.545 pc After: - Q = 121.56 times 235 Q = 28.5666 nc Before: - Q = c v c = elementof_0 A/alpha = 8.85 times 10^-12 times 2 5 /100 times 100/1.43/100 = 8.85 times 10^-12 times 0.25/1.43 = 1.547 times 10^-12 F Q = 1.547 PF After c = 78.57 c = 78.57 times 8.85 times 10^-12 times 0.25/1.48 Therefore Q = 121.56 PF energy = 1/2 cv^2 change = 1/2 times (235)^2 (121.56 - 1.517) = 3313858.9625 times 10^-12 J = 3.3385 mu J