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Derive an expression for the energy stored in a parallel plate capacitor. On charging a parallel plate capacitor to a potential V , the spacing between the plates is halved and a dielectric medium of in_(r) = 10 is introduced between the plates , without disconnecting the d.c. source . explain , using suitable expressions , how the (i) capacitance , "" (ii) electric field , and (iii) energy density of the capacitor change . |
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Answer» SOLUTION :LET initial capacitance of a capacitor be C ` = (in_(0) A)/(d)` (i) When spacing between the plates is halved (i.e., `d. = (d)/(2))` and a dielectric medium of `in_r = 10` is INTRODUCED between the plates , new capacitance of the capacitor will be `C. = (in_(0) in_r A)/(d.) = (in_0 xx 10 xx A)/((d/2)) = 20 "" (in_(0) A)/(d) = 20 C ` (ii) Initial ELECTRIC field `E = (V)/(d)` As battery remains connected hence V. = V but `d. = (d)/(2)` `therefore` New electric field `E. = (V.)/(d.) = (V)/((d/2)) = 2 "" (V)/(d) = 2 E ` (iii) Energy density of capacitor U = `1/2 in_(0) E^(2)` `therefore` New energy density of capacitor `u. = 1/2 in_(0) , E.^(2) = (1)/(2) in_(0) xx 10 xx (2E)^(2) = 40 u` |
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