Determine the potential differnece `varphi_(A) - varphi_(B)` between points `A` adn `B` of the circuit shown in Fig. Under what condition is it equal to zero ?

Explanation

Make the charge distribution, as shown in the fig. In the circuit, 12561
`-Delta varphi = 0` yields
`(q_(1))/(C_(4)) + (q_(1))/(C_(3)) - xi = 0` or, `q_(1) = (xi C_(3) C_(4))/(C_(3) + C_(4))`
and in the circuit 13461,
`(q_(2))/(C_(2)) + (q_(2))/(C_(1)) - xi = 0` or, `q_(2) = (xi C_(1) C_(2))/(C_(1) + C_(2))`
Now `varphi_(A) - varphi_(B) = (q_(2))/(C_(1)) - (q_(1))/(C_(3))`
`= xi [(C_(2))/(C_(1) + C_(2)) - (C_(4))/(C_(3) + C_(4))] = xi [(C_(2) C_(3) - C_(1) C_(4))/((C_(1) + C_(2)) (C_(3) + C_(4)))]`
It becomes zero, when
`(C_(2) C_(3) - C_(1) C_(4)) = 0`, or `(C_(1))/(C_(2)) = (C_(3))/(C_(4))`