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A parallel plate capacitor whose capacit...

A parallel plate capacitor whose capacitance C is `13.5 pF ` is charged by a battery to a potential difference V=12.5V between its plates. The charging battery is now disconnected, and a porcelain slab (k=6.50) is slipped between the plates.
(a) What is the potential energy of the capacitor before the slab is inserted?
(b) What is the potential energy of the capacitor slab device after the slab is inserted?

Text Solution

Verified by Experts

a) Because we are given the initial potential V(=12.5V) we use Eq.25-22 to find the initial stored energy.
`U_1=1/2 CV^2=1/2 (13.5 times 10^-12 F)(12.5V)^2`
`=1.055 times 10^-9 J=1055 pJ approx 1100 pJ`
b) `U_f= q^2/(2kC)=U_1/k=(1055 pJ)/6.50`
`=162 pJ approx 160 pJ`
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