6. ÷ 315 points 1 Previous Answers KatzPSE11 27 P.047 My Notes Ask Your Teacher

Question

6. ÷ 315 points 1 Previous Answers KatzPSE11 27 P.047 My Notes Ask Your Teacher The plates of an air-filled parallel-plate capacitor with a plate area of 17.0 cm2 and a separation of 9.25 mm are charged to a 150-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with 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? Q2.44e-10 Qf = 2.44e-10 (b) What is the capacitance of the capacitor after the dielectric is inserted? 8.85e-12 | Your response differs from the correct answer by more than 10%, Double check your calculations, F (c) What is the potential difference between the plates of the capacitor after the dielectric is inserted? 23 (d) What is the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted? Need Help? 11 Read it Submit Answer Save Pro Practice Another Version

Explanation / Answer

Given area A = 17 cm^2 = 17 * 10^-4 m^2

distance between the plates d = 9.25 mm = 0.00925 m

initial potential difference V0 = 150 V

a) capacitance C0 = epsilonnot * A / d

C0 = 8.854 * 10^-12 * 17 * 10^-4 / 0.00925

C0 = 1.627 * 10^-12 F

the charge is Qi = C0 * V0

Qi = 1.627 * 10^-12 * 150

Qi = 2.44 * 10^-10 C

before and after the charge is same so

Qi = Qf = 2.44 * 10^-10 C

b)

capacitance after the dielectric is inserted

C = k * C0

C = 6.5 * 1.627 * 10^-12

C = 1.057 * 10^-11 F

c)

The potential difference after the dielectric is inserted is

V = V0 / k

V = 150 / 6.5

V = 23.07 V

d)

the energy stored in the capacitor is

without dielectric

U1 = 1/2 * C0 * V0^2

U1 = 1/2 * 1.627 * 10^-12 * (150)^2

U1 = 1.8303 * 10^-8 J

after the dielectric is inserted

U2 = 1/2 C * V^2

U2 = 1/2 * 1.057 * 10^-11 * (23.07)^2

U2 = 2.8128 * 10^-9 J

the difference is

U = U1 - U2

U = (1.8303 * 10^-8) - (2.8128 * 10^-9)

U = 1.549 * 10^-8 J

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