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A capacitor of capacitance C is charged by a battery of emf V & then disconnected . The work done by an external agent to slowly insert a di-electric of di -electric strength k and half the length of capaciitor plates is.........

MonishGayen , 7 Years ago
Grade 12
anser 1 Answers
Khimraj
then new capacitance will be C/2 + KC/2 = (1+K)C/2
and charge willremain constant
so work done will be change in energy
= Q2/(1+K)C – Q2/2C
= (Q2/2C)(1-K/1+K)
and Q = CV
SO
W = (1-K/1+K)(CV2/2)
Hope it clears.
Last Activity: 7 Years ago
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