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A parallel plate capacitor is charged by a battery . After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates . How will the capacitancee of the capacitor. ? Justify your answer |
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Answer» Solution :Let the capacitor has a CAPACITANCE C and charge on it , when charged by a battery , be Q . Hence potential difference between the PLATES of capacitor `V = (Q)/(C)`. When the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates , the charge Q remains conserved , Now : The new capacitance of the capacitor `C. = (K in_(0))/(Ad) = KC ` where K is the dielectric constant of GIVEN dielectric slab. |
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