A) \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{3d}\]
B) \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{2d}\]
C) \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{4d}\]
D) \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{d}\]
E) \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{4d}\]
Correct Answer: D
Solution :
Work done in separating the plates = charge \[\times \] mean potential difference \[=Q\times {{V}_{0}}=C{{V}_{0}}\times {{V}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\times V_{0}^{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
