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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 : charge on 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 charge on the capacitor `Q. = C. V = `(10)/(3) C V = (10)/(3) Q` |
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