An air capacitor is first charged through a battery. The charging battery is then removed and a electric slab of dielectric constant `K=4` is inserted between the plates. Simultaneously, the distance between the plates is reduced to half, then find change, `C, E, V` and `U`.
A
C Increases , E decreases , V decreases, U decreases
B
C Increases , E decreases , V decreases, U increases
C
C decreases , E decreases , V decreases, U decreases
D
C Increases , E increases , V decreases, U decreases
Text Solution
Verified by Experts
The correct Answer is:
A
Change in capacitance `C=(Kepsilon_0A)/dpropK/d` `:.` Capacitance will become `8` times `(K=4, d^'=d/2)` Change in electric field `:. E=E_0//K` or `Eprop1/K` (if `q=constant`) or `Eprop1/K` or the electric field will become `1/4` times its initial value. Change in potential difference, `V=q/C` or `Vprop1/C` (if `q=constant`) Therefore, potential difference becomes `1/8` times of its initial value. Alternate method `V=Ed` Electric field has become `1/4` times its initial value and `d` is reduced to half. Hence `V` becomes `1/8` times. Change stored potential energy, `U=1/2q^2/C` or `Uprop1/C` (if `q=constant`) Capacitance has become `8` times. Therefore, the stored potential energy `U` will become `1/8` times.
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