A) V
B) \[V+\frac{Q}{C}\]
C) \[V+\frac{Q}{2C}\]
D) \[V-\frac{Q}{C},\]if \[V<CV\]
Correct Answer: C
Solution :
In the figure given in the next page, let X and Y b positive and negative plates. After charging from the cell, the inner faces of X and Y have charges \[\pm \,CV,\]as shown in (a). The outer surfaces have no charge. When charge Q is given to X, let the inner faces of \[X\]and Y have charges \[\pm q\](see Q. No.45). Then, by the principle of charge conservation, the outer faces have charges (Q + CV-q) for X and (q - CV) for V, as shown in (b). Now, the outer faces must have equal charges. \[\therefore \] \[Q+CV-q=q-CV\] or \[2q=2CV+Q\]or \[q=CV+\frac{Q}{2}.\] Potential difference \[=\frac{q}{C}=V+\frac{Q}{2C}.\] Hence, the correction option is (c).You need to login to perform this action.
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