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An air parallel plate capacitor having a capacitance C is charged by connecting it to a battery . Now it is disconnected from the battery and a dielectric constant K is inserted between its plates so as to completely fill the space between the plates compare : (i) Initial and final capacitance . (ii) Initial and final charge. (iii) Initial and final potential difference. (iv) Initial and final electric field between the plates . (v ) Initial and final energy stored in the capacitor |
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Answer» Solution :(i) Initial capacitance = C Final capacitance = KC ( after INSERTING dielectric ) `implies (C_(1))/(C_(i))= K` (ii) Charge on isolated capacitor remains same therefore `(Q_(t))/(Q_(i))` =1 (iii) As `Q= CVimplies V = (Q)/(C ) implies (V_(t))/(V_(t))= (C_(i))/(C_(t))= (1)/(K)` (iv) `E=(V)/(d) implies (E_(1))/(E_(i))=(V_(t))/(V_(i))=(1)/(K)` `U= (Q^(2))/(2C) implies (U_(r))/(U_(t))=(C_(i))/(C_(r))=(1)/(K)` |
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