The plates of an air-filled parallel-plate capacitor with a plate area of 14.5 c

Question

The plates of an air-filled parallel-plate capacitor with a plate area of 14.5 cm^2 and a separation of 8.80 mm are charged to a 150-V potential difference. After the plates are disconnected from the source a porcelain dielectric with kappa = 6.5 is inserted between the plates of the capacitor. (a) What is the charge on the capacitor before and after the dielectric is inserted? Q_i = ____ C Q_f = _____ C (b) What is the capacitance of the capacitor after the dielectric is inserted? ______ F (c) What is the potential difference between the plates of the capacitor after the dielectric is inserted? ______ V (d) What is the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted? ______ J

Explanation / Answer

A = 14.5 cm^2, d = 8.8 mm, V = 150 V

Co = Aeo/d

Co = 14.5*10^-4*8.85*10^-12/0.0088

Co = 1.46*10^-12 F

(a) Qi = VC

Qi = 150*1.46*10^-12

Qi = 2.19*10^-10 C

(b) C = kCo

C = 6.5*1.46*10^-12

Qf = 150*6.5*1.46*10^-12

Qf = 1.42*10^-9 C

(c) Vf = 150 V

(d) U = 0.5CV^2

U = 0.5*6.5*1.46*10^-12*150^2

U = 1.1*10^-7 J

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