Potential difference between the points B and E of the circuit is -

Options

A. `((C_(2)-C_(1)))/(V)`
B. `((C_(4)-C_(3)))/(V)`
C. `{(C_(2)C_(3)-C_(1)C_(4))/(C_(1)+C_(2)+C_(3)+C_(4))}`
D. `{(C_(1)C_(4)-C_(2)C_(3))/((C_(1)+C_(2))(C_(3)+C_(4)))}`

Answer

D

Explanation

Charge on `C_(1)` and `C_(2)`
`Q_(1)=Q_(2)=((C_(1)C_(2))/(C_(1)+C_(2)))V`
and charge on `C_(3)` and `C_(4)`
`Q_(3)=Q_(4)=((C_(3)C_(4))/(C_(3)+C_(4)))V`
so `V_(B)-V_(E)=(Q_(2))/(C_(2))-(Q_(4))/(C_(4))`
`=(C_(1)V)/(C_(1)+C_(2))-(C_(3)V)/(C_(3)+C_(4))`
`={(C_(1)C_(4)-C_(2)C_(3))/((C_(1)+C_(2))(C_(3)+C_(4)))}V`