In a circuit shown in fig find the potentail difference between the left and right plates of each capacitor.

Explanation

Let `xi_(2) gt xi_(1)`, then using `-Delta varphi = 0` in the closed circuit (Fig)
`(-q)/(C_(1)) + xi_(2) - (q)/(C_(2)) - xi_(1) = 0`
or, `q = ((xi_(2) - xi_(1)) C_(1) C_(2))/((C_(1) + C_(2))`
Hence the `P.D.` accross the left and right plates of capacitors,
`varphi_(1) = (q)/(C_(1)) = ((varphi_(2) - varphi_(1)) C_(2))/(C_(1) + C_(2))`
and similarly
`varphi_(2) = (-q)/(C_(2)) = ((varphi_(1) - varphi_(2)) C_(1))/(C_(1) + C_(2))`