• # question_answer 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 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). LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

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