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A parallel - plate capacitor of plate ar...

A parallel - plate capacitor of plate area `A` and plate separation d is charged to a potential difference `V` and then the battery is disconnected . `A` slab of dielectric constant `K` is then inserted between the plate of the capacitor so as to fill the space between the plate .Find the work done on the system in the process of inserting the slab.

Text Solution

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The correct Answer is:
`W= (1)/(2)C_(0)V_(0)^(2) (1-(1)/(K))`
`C = (in_(0)A)/(d)`

`U_(i) = (1)/(2)CV_(0)^(2)`
`U_(f) = (q^(2))/(2KC) = ((CV_(0))^(2))/(2KC) = (V_(0)^(2))/(2KC)`
`W_("ext.agent") = U_(f) -U_(i) = (V_(0)^(2))/(2C) ((1)/(K)-1)`
`W_(("capacitance+slab")) = (V_(0)^(2))/(2C) (1-(1)/(k))`
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