Determine the potential difference `V_(A)-V_(B)` between point A and B of the circuit shown in figure . Under what condition is it equal to zero ?

Explanation

Mark the charge distribution, as shown in figure . In the circuit , 12561 , `SigmaDeltaV = 0` yields

`(q_(1))/(C_(4))+ (q_(1))/(C_(3))-epsilon=0` or `q_(1)=(epsilonC_(3)C_(4))/(C_(3)+C_(4))` and in the circuit 134561
`(q_(2))/(C_(2))+ (q_(2))/(C_(1))-epsilon=0` or `q_(2) = ( epsilonC_(1)C_(1)C_(2))/(C_(1)+C_(2))`
Now , `V_(A) - V_(B) = (q_(2))/(C_(1))-(q_(1))/(C_(3)) = epsilon[(C_(2))/(C_(1)+C_(2))-(c_(4))/(C_(3)+C_(4))]= epsilon[(C_(2)C_(3)-C_(1)C_(4))/((C_(1)C_(2))(C_(3)+C_(4)))]`
In becomes zero, when
`(C_(2)C_(3) - C_(1) C_(4)) =0` or `(C_(1))/(C_(2))= (C_(3))/(C_(4))`