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A spherical body of radius 3 cm and having a charge of 48 nC is joined by a metal wire to an uncharged spherical body of radius 5 cm. The charges, in equilibrium on the two bodies respectively are;

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20 nC and 28 nC

28 nC and 20 nC

30 nC and 18 nC

18nC and 30 nC

answer is D.

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Detailed Solution

when they are joined they will have same potential
$\frac{q_{1}}{4 π ε_{0} r_{1}} = \frac{q_{2}}{4 π ε_{0} r_{2}}$ $\frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}} = \frac{3}{5}$

so total charge will be distributed in 3:5 ratio

$q_{1} = \frac{48}{8} 3 = 18 a n d q_{2} = \frac{48}{8} 5 = 30$

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