A parallel-plate capacitor in air has a plate separation of 1.79 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.79 cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 260 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine the charge on the plates before and after immersion, Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. pC Determine the capacitance and potential difference after immersion. C_f = F delta V_f= V Determine the change in energy of the capacitor. nJ

Explanation / Answer

a)

here

C = e0 * A / d

C = (8.854 * 10^-12 * 25 * 10^-4) / (1.79 * 10^-2)

C = 1.236 * 10^-12 F

then

Q = C * V

Q = 1.236 * 10^-12 * 260

Q = 3.21 * 10^-10 C

the charge is same before and after immersion

b)

C1 = k * C

C1 = 80 * 1.23 * 10^-12

C1 = 9.84 * 10^-10 C

then

V1 = V / k

V1 = 260 / 80 = 3.25 V

c)

U = Uf - Ui

U = 0.5 * C * V^2 * (1/k - 1)

U = 0.5 * 1.23 * 10^-12 * 260^2 * (1/80 - 1)

U = 41 * 10^-9 J

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