An air-spaced parallel plate capacitor has an initial charge of .05 C after bein
ID: 1705403 • Letter: A
Question
An air-spaced parallel plate capacitor has an initial charge of .05 C after being connected to a 10 V battery.
a) what is the total energy stored between the plates of the capacitor?
b) if the battery is disconnected and the plate separation is tripled to .3 mm, what is the electric field before and after the plate separation change?
c) what is the final voltage across the plates and the final energy stored?
d) calculate the work done in pulling the plates apart. Does this fully account for the energy change in part B?
Explanation / Answer
Given
initial charge on the capacitor Q = 0.5C
voltage across the battery V = 10 V
Capacitance of the capacitor C = Q / V
= 0.5C / 10
= 0.05 F
a) energy stored betwen the plates of the capacitor
U = Q 2 / 2 C
= (0.5C ) 2 / 2*0.05F
= 5*10-6 J
b)Electric field before the plate seperation change is
E = V / d
= 10 / 0.1 mm
= 105 N /C
electric field after the plate seperation change
E = V / 3d
= 10 / 0.3mm
= 3.333*104 N /C
c) After the plate seperation the capacitance of the capacitor is
C' = C / 3
= 0.05F /3
= 0.016F
energy stored between the plates
U' = Q 2 / 2 C'
= ................... J
d) work done in pulling the plates apart
W = U' - U
= .......... J
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