The plates of an air-filled parallel-plate capacitor with a plate area of 17.5 c

Question

The plates of an air-filled parallel-plate capacitor with a plate area of 17.5 cm2 and a separation of 9.25 mm are charged to a 125-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?

Qi

Qf

(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

Qi

Qf

Explanation / Answer

Plate area = A = 17.5 cm2 = 17.5* 10-4 m2

d= 9.25mm = 9.25m2

V= 125V

K=6.5

S0lution

C= Q/V

Q = CV

C= AK0 /d

For air k=1

C= A0 /d

C =   ( 17.5* 10-4   * 8.85 * 10 -12 )/9.25 *10-3

C 1= 1.6743pF         -------------1

Capacitance when k=6.5                     

C= AK0 /d

C= =   ( 17.5* 10-4   *6.5 * 8.85 * 10 -12 )/9.25 *10-3

C2=10.883     pF        -------2

Charge without dielectric

Q= C1V

Q= 1.6743 *10-9 *125

Q=209.2875 *10-9 C

Q = 209.28 nC ----3

Charge with dielectric

Q=   C2V

Q = 10.883    *10-9 *125

Q= 1.3604nC -----4

b) See Eq 2

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