Determine the potential difference across the plates of the capacitor `C_1` of the network shown in the Fig. Assume `epsi_1 gt epsi_2`.

Explanation

As `epsi_1 gt epsi_2` hence the net effective voltage present in the network `epsi= epsi_1 - epsi_2`.
In the network `C_1 and C_2` are joined in series, hence equivalent capacitance of the network `C = (C_1C_2)/(C_1 + C_2)`.
`:.`Charge one each capacitor q `= C epsi =((C_1C_2)/(C_1 + C_2))(epsi_1 - epsi_2)`
`:.` Potential difference across the plate of the capacitor `C_1`, is
`V = q/(C_1) = (C_2/(C_1 + C_2))(epsi_1 - epsi_2)`