Explanation

(a) [image:1_image375.emf]
Let $Q_{1}$ and $Q_{2}$ be the charge on the two conducting spheres of radii $R_{1}( = 3cm)$ and $R_{2}( = 1cm)$ respectively
$\therefore V_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{1}}{R_{1}} \Rightarrow 10 = \frac{9 \times 10^{9} \times Q_{1}}{3 \times 10^{- 2}}$
$\Rightarrow Q_{1} = \frac{10 \times 3 \times 10^{- 2}}{9 \times 10^{9}}$ …(i)
and $V_{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{2}}{R_{2}} \Rightarrow 10 = \frac{9 \times 10^{9} \times Q_{2}}{1 \times 10^{- 2}}$
$\Rightarrow Q_{2} = \frac{10 \times 1 \times 10^{- 2}}{9 \times 10^{9}}$ …(ii)
According to coulomb’s law
Force of repulsion between them is
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{1}Q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 10 \times 3 \times 10^{- 2} \times 10 \times 1 \times 10^{- 2}}{9 \times 10^{9} \times 9 \times 10^{9} \times \left( 10 \times 10^{- 2} \right)^{2}}$ [Using (i) & (ii)]
$= \left( \frac{1}{3} \right) \times 10^{- 9}N$