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A parallel plate capacitor of capacitance C is charged to a potential V by a battery . Without disconnecting the battery , the distance between the plates is tripled and a dielectric medium of dielectric constant 10 is introduced between the plates of the capacitor . Explain giving reasons , how will the following be affected : energy density of the capacitor . |
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Answer» Solution :Initially capacitance of given parallel capacitor, `C = (in_(0) A)/(d)` and it has been charged to a potential DIFFERENCE V . HENCE charge on capacitor Q = CV and electric field between the plates of capacitor E `= (V)/(d)` and so the ENERGY density `u = 1/2 in_(0) E^2` As per question the new distance between the plate `d. = 3d` and a dielectric medium of dielectric constant K = 10 is introduced between the plates of capacitor . Moreover , as battery remains connected , the potential difference V remains unchanged . New electric field between the plates of capacitor , E. = `(V)/(KD.) = (V)/(10 xx 3d) = (E)/(30)` `therefore` New energy density u. `= (1)/(2) in_(0) E.^(2) = (1)/(2) in_(0) [(E)/(30)]^(2) = (1)/(90) xx (1)/(2) in_(0) , E^(2) = (u)/(90)` |
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