What is `V_(A)-V_(B)` in the arrangement shown in.What is the condition such that `V_(A)-V_(B)=0`

Explanation

Let the charge be as shown in. We know that capacitors in series have the same charge. Considering the 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))]`
from the loop containting `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_(4)=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))]]`
For `V_(A)-V_(B)=0, C_(1)C_(4)=C_(2)C_(3)=0`. If `(C_(1)//C_(2))=(C_(3)//C_(4))`, the potential difference across `A` and `B` will be zero . This statement is very important is capacitor circuit solving. This type of the situation is callede balacned Wheatstone bridge.