A) \[V\]
B) \[2V\]
C) \[4V\]
D) \[-2V\]
Correct Answer: D
Solution :
[d] When sphere has charge Q. The potential difference between'1'and '2'. \[{{V}_{1}}-{{V}_{2}}=K\frac{Q}{{{r}_{1}}}-K\frac{Q}{{{r}_{2}}}=KQ\left[ \frac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}} \right]=V\] When the sphere is given charge \[-3Q,\] the net charge in the sphere is \[(Q-3Q=-2Q)\] Hence new potential difference \[{{V}_{1}}-{{V}_{2}}=K\frac{(-2Q)}{{{r}_{1}}}-K\frac{(-2Q)}{{{r}_{2}}}\] \[=-2KQ\,\left[ \frac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}} \right]=-2V\]You need to login to perform this action.
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