Consider
.
The potential difference `V_(A)-V_(B)` is.

Options

A. `E[(C_(1)C_(4)-C_(2)C_(3))/((C_(1)+C_(2))(C_(3)+C_(4)))]`
B. `E[(C_(1)C_(4)+C_(2)C_(3))/((C_(1)+C_(3))(C_(2)+C_(4)))]`
C. `E[(C_(1)C_(3)+C_(2)C_(4))/((C_(1)+C_(2))(C_(3)+C_(4)))]`
D. `E[(C_(1)C_(3)+C_(2)C_(4))/((C_(1)+C_(3))(C_(2)+C_(4)))]`

Answer

A

Explanation

Let charge be as shown. (Capacitors in series have the same the charge.)
Taking loop containing `C_(1),C_(2),` and `E`, we get
`q/C_(1)+q/C_(2)-E=0` or `q=E[(C_(1)C_(2))/(C_(1)+C_(2))]`
Similarly, from loop containing `C_(3), C_(4)`, and `E`, we get
`(q')/C_(3)+(q')/C_(4)-E=0` or `q'=E[(C_(3)C_(4))/(C_(3)+C_(4))]`
Now, `V_(A)-V_(B)=q/C_(2)-(q')/C`
`=E[C_(1)/(C_(1)+C_(2))-C_(3)/(C_(3)+C_(4))]`
`=E[(C_(1)C_(4)-C_(3)C_(2))/((C_(1)+C_(2))(C_(3)+C_(4)))]`
Now `V_(A)-V_(B)=ErArrC_(1)C_(4)-C_(2)C_(3)=0`
or `C_(1)/C_(2)=C_(3)/V_(4)`.