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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3 V

$\frac{1}{3} \cdot V$

$\frac{3}{2} V$

$\frac{2}{3} V$

answer is B.

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Detailed Solution

Potential inside the sphere will be same as that on its surface i.e., $V = V_{surface} = \frac{q}{10}$ statvolt Also $V_{out} = \frac{q}{15} statvolt$

$∴ \frac{V_{o u t}}{V} = \frac{2}{3} \Rightarrow V_{o u t} = \frac{2}{3} V$

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