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(a) Derive an expression for the energy stored in a parallel plate capacitor of capacitance C when charged up to voltage V. How is this energy stored in the capacitor ? (b) A capacitor of capacitance 1 μF is charged by connecting a battery of negligible internal resistance and emf 10 V across it. Calculate the amount of charge supplied by the battery in charging the capacitor fully. |
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Answer» `dW=(q/C).dq` Therefore, whole process of charging from Oto Q requires a work `W=int_(0)^(Q)(qdq)/C=1/C[(q^(2))/2]_(0)^(Q)=(Q^(2))/(2C)` This work done is stored as the electrostatic potential energy of the charged capacitor. Hence, potential energy of charged capacitor `u=(Q^(2))/(2C)` But Q = CV, where V be the potential difference between the plates of capacitor, hence `u=(Q^(2))/(2C)=1/2QV=1/2CV^(2)` This energy is stored within the DIELECTRIC of the capacitor in the form of electrostatic potential energy. (b) Here capacitance C =`1muF=1xx10^(-6)F` emf of battery E = 10 V and INTERNAL resistance of battery r= 0. Since r = 0, hence potential difference between the plates of capacitor, when it is FULLY charged is V = `epsi` = 10 V. `therefore` Charge supplied by the battery to the capacitor Q `=CV=(1xx10^(-6)F)xx10V` `=10xx10^(-6)C=10muC` 
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