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A parallel plate capacitor , each with plate area A and separation d , is charged to a potential difference V . The battery used to charge it remains connected . A dielectric slab of thickness of and dielectric constant K is now placed between the plates . What change , if any , will take placein : capacitance of the capacitor ? Justify your answer . |
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Answer» Solution :Given that plate area of either plate of parallel plate CAPACITOR = A distance between the plates = d and potential difference between the plates = V Hence initially CAPACITANCE ` C = (in_(0) A)/(d)` and CHARGE on plates Q = CV As the BATTERY remains connected THROUGHOUT the potential difference between the plates remains unchanged (V. = V)On placing a dielectric slab of the thickness .d. and dielectric constant .K. between the plates Electric field intensity between the plates `E. = (V.)/(d) = (V)/(d) = E` |
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