Question

A parallel plate capacitor is charged and then disconnected from a battery

Answer 1

hola there

A capacitor is initially charged by a battery and hence its 2 plates would have a +q and -q charge on them. Now, if a dielectric slab of thickness equal to d is inserted between the plates, the 

capacitance = KC , where C was the original capacitance of the capacitor and K is the dielectric constant. 

 

Also, since the charge would remain constant, and Q = CV

so, the V' = V/K

 

Now, the energy stored within the capacitor = Q^2/2C

Since Q is constant, and C' = KC, 

so, Energy (W') = W/K where W was the energy stored in the capacitor in the absence of dielectric. 

Answer 2

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