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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 potential difference between the plates? 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 : New potential difference between the plates `V = (Q.)/(C.) = (Q)/(KC) = (V)/(K)` . |
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