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A parallel plate capacitor is charged by a battery . After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates . How will the energy stored in the capacitor be affected ?Justify your answer |
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Answer» Solution :Let the CAPACITOR has a CAPACITANCE C and charge on it , when charged by a battery , be Q . Hence potential difference between the plates of capacitor `V = (Q)/(C)`. When the battery is disconnected and a dielectric SLAB with its thickness equal to the plate separation is inserted between the plates , the charge Q remains conserved , Now : The energy stored in the capacitor `U. = (Q.^(2))/(2 C.) = (Q^(2))/(2C.) = (Q^(2))/(2 KC) = (U)/(K.)` where U = initial value of the energy stored . |
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