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A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U ) as epsilon = alpha U wherealpha=2V^(-1). A similar capacitor with no dielectricis charged to U_(0) = 78 V. lt is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. |
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Answer» Solution :Let the capacitance of capacitor without dielectric is C and hence, charge on capacitor `Q_(1)=CU` where U is the FINAL potential of a capacitor As the capacitor with dielectric having relative permittivity `epsilon` and its capacitance `in` C. Hence, charge on the capacitor is, `Q_(2)= in CU = aUxx CU= aCU^(2)` `[ because in = AU]` The initial charge on the capacitor is, `Q_(0)=CU_(0)` From the conservation of charge, `Q_(0)= Q_(1)+Q_(2)` `CU_(0)=CU+aCU^(2)` `:.aU^(2)+U-U_(0)=0` Now a = `2V^(-1)` and `U_(0)` = 78 V `:. 2U^(2) + U-78=0` is a binomial equation of U. `:. 2U^(2) +13U -12 U -78 =0` `:. U(2U+13) - 6 (2U+13) =0` `:.(2U+13)` (U-6) =0 `:. U= -(13)/(2) "or" U =6` `:. U = -(13)/(2)` is impossible `:. U = 6V` final voltage |
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