In a charged spherical conductor id radius $10 c m$ has potential $V$ at point distant $5 c m$ from its center, then the potential at a point $15 c m$ away from the center will be then

3 V

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\frac{3}{2} V

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\frac{2}{3} V

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\frac{1}{3} V

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Solution

The correct option is C $\frac{2}{3} V$
$V = \frac{K Q}{R}$ [Inside]

$V = \frac{K Q}{r}$ [Outside]

V at distance $10 c m$

$= \frac{K Q}{10} - - - - - (i)$

$= 15 c m$

$V^{'} = \frac{K Q}{15} - - - - (i i)$

$\frac{V}{V^{'}} = \frac{15}{10} = \frac{3}{2}$

$V^{'} = \frac{2}{3} V$

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In a charged spherical conductor id radius 10cm has potential V at point distant 5cm from its center, then the potential at a point 15cm away from the center will be then

The correct option is C 23VV=KQR [Inside] V=KQr [Outside] V at distance 10cm =KQ10−−−−−(i) =15cm V′=KQ15−−−−(ii) VV′=1510=32 V′=23V

In a charged spherical conductor id radius $10 c m$ has potential $V$ at point distant $5 c m$ from its center, then the potential at a point $15 c m$ away from the center will be then

The correct option is C $\frac{2}{3} V$
$V = \frac{K Q}{R}$ [Inside]

$V = \frac{K Q}{r}$ [Outside]

V at distance $10 c m$

$= \frac{K Q}{10} - - - - - (i)$

$= 15 c m$

$V^{'} = \frac{K Q}{15} - - - - (i i)$

$\frac{V}{V^{'}} = \frac{15}{10} = \frac{3}{2}$

$V^{'} = \frac{2}{3} V$