0

0

Calculate the electric feild. Suppose the area of p;late to be A. Then, the charge density on plate 1 would be

S_{1} = Q+q1/A. (q1 be the charge initially present on itdue to charging by v volt given by q1=CV)

Now, similarly,

S_{2}= Q/A

Now, Electric field b/w the plates would be E = s_{1}/2e_{0} + s_{2}/2e_{0}

Also, intial capacitance of the capacitor can be defined as C = Q_{1}/V

Hence by the Formula V= Edr calculate the the new Potential difference;

V^{I}= 2Q + q_{1 }/ 2Ae_{0} X d (suppose the distance b/w plates to be d)

= Q/C + V