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A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R ( R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {R - r} \right)Q}}{{\left( {{R^2} + {r^2}} \right)}}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {R + r} \right)Q}}{{2\left( {{R^3} + {r^3}} \right)}}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {R + r} \right)Q}}{{\left( {{R^2} + {r^2}} \right)}}$
$\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {R - r} \right)Q}}{{2\left( {{R^2} + {r^2}} \right)}}$
Solution
Let $q_{1}$ and $q_{2}$ be charge on two spheres of radius $'r'$ and $'R'$ respectively As, $q_{1}+q_{2}=Q$
and $\sigma_{1}=\sigma_{2} \quad$ [Surface charge density are cqual $]$
$\therefore \frac{q_{1}}{r \pi r^{2}}=\frac{q_{2}}{4 \pi R^{2}}$
So, $q_{1}=\frac{Q r^{2}}{R^{2}+r^{2}}$ and $q_{2}=\frac{Q R^{2}}{R^{2}+r^{2}}$
Now, potential, $V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r}+\frac{q_{2}}{R}\right]$
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{Q r}{R^{2}+r^{2}}+\frac{Q R}{R^{2}+r^{2}}\right]$
$=\frac{Q(R+r)}{R^{2}+r^{2}} \frac{1}{4 \pi \varepsilon_{0}}$
