Two conducting spheres of radii 3 cm and 1 cm are separated by distance of 10 cm in free space. If t

Question

Two conducting spheres of radii 3 cm and 1 cm are separated by distance of 10 cm in free space. If the spheres are charged to same potential of 10 V each, the force of repulsion between them is\

Professor PC Thomas Classes · Answer

Let $Q_{1}$ and $Q_{2}$ be the charges on the two conducting spheres of radii $\mathrm{R}_{1}(=3 \mathrm{~cm})$ and $\mathrm{R}_{2}(=1 \mathrm{~cm})$ respectively\ $	herefore \mathrm{V}_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}_{1}}{\mathrm{R}_{1}}$\ $10=\frac{9 	imes 10^{9} 	imes \mathrm{Q}_{1}}{9 	imes 10^{-2}}$ $$\begin{equation*} \mathrm{Q}_{1}=\frac{10 	imes 3 	imes 10^{-2}}{9 	imes 10^{9}} 	ag{i} \end{equation*}$$ and $\mathrm{V}_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}^{2}}{\mathrm{R}_{2}}$\ $10=\frac{9 	imes 10^{9} 	imes \mathrm{Q}_{2}}{1 	imes 10^{-2}}$\ $\mathrm{Q}_{1}=\frac{10 	imes 1 	imes 10^{-2}}{9 	imes 10^{9}}$\ According to Coulomb's law\ Force for repulsion between them is\ $\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}}$\ $=\frac{9 	imes 10^{9} 	imes 10 	imes 3 	imes 10^{-2} 	imes 10 	imes 1 	imes 10^{-2}}{9 	imes 10^{9} 	imes 9 	imes 10^{9} 	imes\left(10 	imes 10^{-2}ight)^{2}}$\ ( using (i) & ( ii))\