(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.

Answer

a

Explanation

(a) Let at a particular instant charge on the plate of capacitor be q and its potential difference be `q/C`. If an additional charge dq is given to the capacitor plate, work done for it is given by
`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`