Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2}$, respectively. The electric fields $E_{1}$ (on $S_{1}$ ) and $E_{2}$ (on $S_{2}$ ) on their surfaces are such that $E_{1} / E_{2}=R_{1} / R_{2}$. Then the ratio $V_{1}\left(\right.$ on $\left.S_{1}\right) / V_{2}\left(\right.$ on $\left.S_{2}\right)$ of the electrostatic potentials on each sphere is:
Correct Option: , 2
(2) Electric field at a point outside the sphere is given by
$E=\frac{1 Q}{4 \pi \in_{0} r^{2}}$
But $\rho=\frac{Q}{\frac{4}{3} \pi R^{3}}$
$\therefore E=\frac{\rho R^{3}}{3 \in_{0} r^{2}}$
At surface $r=R$
$\therefore E=\frac{\rho R^{3}}{3 \epsilon_{0}}$
Let $\rho_{1}$ and $\rho_{2}$ are the charge densities of two sphere.
$E_{1}=\frac{\rho R_{1}}{3 \varepsilon_{0}}$ and $E_{2}=\frac{\rho_{2} R_{2}}{3 \varepsilon_{0}}$
$\because \frac{E_{1}}{E_{2}}=\frac{\rho_{1} R_{1}}{\rho_{2} R_{2}}=\frac{R_{1}}{R_{2}}$
This gives $\rho_{1}=\rho_{2}=\rho$
Potential at a point outside the sphere
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}$
$=\frac{\rho R^{3}}{3 \varepsilon_{0} r}\left(\because \rho=\frac{Q}{\frac{4}{3} \pi R^{3}}\right)$
At surface, $r=R$
$V=\frac{\rho R^{2}}{3 \varepsilon_{0}}$ so, $V_{1}=\frac{\rho R_{1}^{2}}{3 \varepsilon_{0}}$ and $V_{2}=\frac{\rho R_{2}^{2}}{3 \varepsilon_{0}}$
$\therefore \frac{V_{1}}{V_{2}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}$
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