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Q.

If a charged spherical conductor of radius 10cm has potential ‘V’ at a point distant 5cm from its centre, then the potential at a point distant 15cm from the centre will be

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a

$\frac{1}{3} V$

b

3V

c

$\frac{2}{3} V$

d

$\frac{3}{2} V$

answer is B.

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Detailed Solution

$V = V_{s u r f a c e} = \frac{q}{10} s t a t v o l t$ , $V_{o u t} = \frac{q}{15} s t a t v o l t$

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

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