A) R
B) \[\frac{R}{2}\]
C) \[\frac{R}{3}\]
D) 2R
Correct Answer: C
Solution :
[c] :\[V=\frac{kQ}{2{{R}^{3}}}(3{{R}^{2}}-{{r}^{2}})\](for r < R) For r = 0, potential at the centre \[={{V}_{c}}=k\frac{3Q}{2R}.\] We require a point where\[V=\frac{{{V}_{c}}}{2}=k\frac{3Q}{4R}.\] This point cannot lie inside the sphere where\[V\ge k\frac{Q}{R}.\] Let the point lie outside the sphere, at a distance r from the centre. Then, \[V=k\frac{Q}{r}=\frac{3Q}{4R}\]or\[r=\frac{4}{3}R\] Distance from the surface\[=r-R=\frac{4}{3}R-R=\frac{R}{3}\]
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