InterviewSolution
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InterviewSolution
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A parallel palte capacitor is filled by a dielectric whose relative permittively varies with the applied voltage (U) as `epsilon = alpha U` where alpha `= 2V^(-1)`. A similar capacitor with no dielectric is charged to `U_(0) = 78V`. It is then is connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. |
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Answer» On connecting the two given capacitors,let the final voltage be U. If capacity of capacitor of capacitor without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU` The other capacitor with dielectric has capacity `in C` . Therefore, charge on it is `Q_(2) = in CU`. As `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)` The initial charege on the capacitor (without the dielectric is C, then the charge on this capacitor is `Q_(1) = CU` The other capacitor with dielectric has capacity `in C`. Therefore, charge on it is `Q_(2) = in CU`. AS `in = alpha U`, therefore, `Q_(2) = alpha C U^(2)` The initial charge on the capacitor (without dielectric) that was charged si `Q_(0) = CU_(0)` From the conservation of charge, `Q_(0) = Q_(1) + Q_(2)` `CU_(0) = CU + alpha CU^(2) or alpha U^(2) + U - U_(0) = 0 :. U = (-1 +- sqrt(1+4 alpha U_(0)))/(2 alpha)` Using `alpha = 2 V^(-1) and U_(0) = 78 vol t`, we got `U = (-1 +- sqrt(1+4xx2xx78))/(2xx2) = (-1 +- sqrt(625))/(4)` As U is positive, therefore , `U = (sqrt(625) -1)/(4) = (24)/(4) = 6vol t` |
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