A parallel plate capacitor is connected to a 12V battery. The capacitor\'s plate

Question

A parallel plate capacitor is connected to a 12V battery. The capacitor's plate area is 1.77 * 10^-4 m^2. The separation between the plates is 2*10^-4 m. A capacitor remains connected to the battery while a dielectric with dielectric constant of k=10 is inserted between the plates of the capacitor. a) Please calculate the capacitance of the capacitor before the dielectric is inserted. b) Please calculate the charge of the capacitor before the dielectric is instant. c) Please calculate the capacitance of the capacitor that is filled with the dielectric. d) Please calculate the energy stored in the capacitor with dielectric.

Explanation / Answer

(a)

capacitance of capacitor C0 = e0*A/d

C0 = 8.85*10^-12*1.77*10^-4/(2*10^-4)

C0 = 7.83*10^-12 F

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part(b)

capacitance C0 = Q/v

Q = C0*v = 7.83*10^-12*12 = 9.396*10^-11 C

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part(c)

capacitance C = k*C0

K = dielectric constant = 10

C = 7.83*10^-11 F

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energy stored U = (1/2)*C*v^2 = (1/2)*7.83*10^-11*12^2 = 5.64*10^-9 J

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