The plates of an air-filled parallel-plate capacitor with a plate area of 15.0 c

Question

The plates of an air-filled parallel-plate capacitor with a plate area of 15.0 cm2 and a separation of 8.95 mm are charged to a 170-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

Given

capacitor with  

A = 15.0 cm^2 , separation of 8.95 mm are charged to a

dV = 170-V potential difference.

k = 6.5

we know that the capacitance of a parallel plate capacitor is C = epsilon not *A/d

C = 8.854*10^-12*(15*10^-4)/(8.95*10^-3)F

C = 1.4839106145251*10^-12 F

C = 1.483911*10^-12 F

charge on the capacitor before the dielectric inserted is  

Q = C*V

Q = 1.483911*10^-12*170 C = 2.5226487*10^-10 C

after the dielectric inserted  

the capacitance changes ( increases) by k times  

C' = k*C = 6.5*1.483911*10^-12 F = 9.6454215*10^-12 F

when the capacitor is disconnected from the battery there is no change in the charee  

We know Q = C*V

Capacitance increases so as the potential difference decreases by k times to have constant

a) so

Qi = 2.5226487*10^-10 C

Qf = 2.5226487*10^-10 C

b)the capacitance of the capacitor after the dielectric is inserted is c' = k*C =

c' = 6.5*1.483911*10^-12 F

C' = 9.6454215*10^-12 F

c) the potential difference between the plates of the capacitor after the dielectric is inserted is decreased to 1/k times

dV' = (1/k*V) = 170/6.5 V = 26.153846 V

d)

d) before dielectric inserted

the energy stored is U = 0.5*C*V^2

U = 0.5* 9.6454215*10^-12*170^2 J

U = 1.39376341*10^-7 J

after the dielectric inserted

the energy stored is U = 0.5*C*V^2 = 0.5*C'*dv^2 = 0.5* 9.6454215*10^-12*26.153846^2 J

U' = 3.29884826*10^-9 J

now the energy difference is

U-U' = 1.39376341*10^-7 - 3.29884826*10^-9 J

dU = 1.3607749274*10^-7 J

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