If a charged spherical conductor of radius $10\,cm$ has potential $V$ at a point distant $5\,cm$ from its centre, then the potential at a point distant $15\,cm$ from the centre will be
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(b) Potential inside the sphere will be same as that on its surface i.e. $V = {V_{surface}} = \frac{q}{{10}}stat\,volt$, ${V_{out}} = \frac{q}{{15}}stat\,volt$
$\frac{{{V_{out}}}}{V} = \frac{2}{3}$ $==>$ ${V_{out}} = \frac{2}{3}\,V$
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