My answer is V+(Q/C) but the correct answer is V+(Q/2C)

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now?

02-May-2016 12:32 AM

Answers (1)


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

S1 = Q+q1/A.   (q1 be the charge initially present on itdue to charging by v volt given by q1=CV)

Now, similarly,

S2= Q/A  

Now, Electric field b/w the plates would be  E = s1/2e0 + s2/2e0

Also, intial capacitance of the capacitor can be defined as C = Q1/V

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

VI= 2Q + q/ 2Ae0 X d                                       (suppose the distance b/w plates to be d)

= Q/C + V

14-09-2016 23:45

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