A charge of +2.0 x 10^-8 C is placed on the positive plate and a charge of -1.0 x 10^-8 C on the negative plate of a parallel plate capacitor of capacitance 1.2 x 10^-3 uF. Calculate the potential difference developed between the plates.Please add explanation!

Ans- 12.5 volt

Explanation-  charge appearing on right side of plate-2 = charge appearing of left side of plate-1= (2 x 10^-8-1 x10^-8)/2

                                                                                                                                        = 10^-8/2 Coulomb

                   now the charge appearing on right side of plate-1 will be-  2 x 10^-8-10^-8/2 = 1.5x10^-8coulomb

                   charge appearing on right side of plate-1 = charge of +ve plate of capacitor ( charge of left side of plate is not

                   taken as plate of charge be is does not give its electric field as well as potential inside the capacitor)

                               using  Q = charge of +ve plate = CV

                                         1.5x10^-8coulomb = 1.2x 10^-3 micro F x V= 1.2x 10^-9 Farad x Vg

                 solving this- We get V=12.5 volt