A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system the process of inserting the slab.
ε0AV22d(1−1K)
ε0AV2d(1K−1)
ε0AV22d(1K+1)
ε0AV2d(1K+1)
A
ε0AV22d(1−1K)
B
ε0AV22d(1K+1)
C
ε0AV2d(1K+1)
D
ε0AV2d(1K−1)
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Solution
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The correct option is Aε0AV22d(1−1K) after inserting slab
newcapacitanceC′=KC=KϵoAd
newpotentialdifferenceV′=VK
newelectricfieldE′=V′d=Vkd
newchargeQ′C′=ϵoAVd
work=final−initial
=12C′V′−12CV2
12(KC)=(1−1K)
[ω]=ϵoAV22d(1−1K)
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