A) \[-\frac{(r+R)}{r}q\]
B) \[\frac{(r-R)}{r}q\]
C) \[-\frac{R}{r}q\]
D) \[\frac{r}{R}q\]
Correct Answer: C
Solution :
: Potential at surface\[=\frac{q}{r}\] \[\therefore \] \[{{V}_{1}}=q/r\] Potential at an inside point of a conducting sphere is equal to potential at its surface. \[\therefore \]Due to bigger sphere, \[{{V}_{2}}=\frac{Q}{R}\] \[V={{V}_{1}}+{{V}_{2}}\] Or \[0=\frac{q}{r}+\frac{Q}{R}\] Or \[Q=-\frac{qR}{r}=-\frac{R}{r}q\]You need to login to perform this action.
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