A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V.
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 d and dielectric constant k is now placed between the plates. What change, if any, will take place in:
(i) charge on plates?
(ii) electric field intensity between the plates?
(iii) capacitance of the capacitor? Justify your answer in each case.
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(i) The charge Q =CV, V = same, C = increases; there, charge on plates increases.
(ii) A electric field E = V/d, and V = constant, d = constant; therefore, electric field strength remains the same.
(iii) The capacitance of capacitor increases as K >1.
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