Question
Two charged conducting spheres $S_{1}$ and $S_{2}$ of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $S_{1}$ and $S_{2}$ spheres are $E_{S1}$ and $E_{S2}$ respectively. The value of $\frac{E_{S1}}{E_{S2}}$ is $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Common potential $\Rightarrow \ Q \propto \ R$; surface field $E \propto \ Q/R^{2} \propto \ 1/R$.]
$$V_{1} = V_{2} $$
$$\frac{{kQ}_{1}}{8} = \frac{{kQ}_{2}}{18}\ \left\{ v_{\text{surface}\text{~}} = \frac{kQ}{R} \right\} $$
$$\frac{Q_{1}}{Q_{2}} = \frac{4}{9} $$
$$\frac{E_{S_{1}}}{E_{S_{2}}} = \frac{{kQ}_{1}}{r_{1}^{2}}\frac{r_{2}^{2}}{{kQ}_{2}}\left\{ E_{\text{surface}\text{~}} = \frac{kQ}{R^{2}} \right\} $$
$$= \frac{Q_{1}}{Q_{2}}\left( \frac{r_{2}}{r_{1}} \right)^{2} = \frac{4}{9} \times \left( \frac{9}{4} \right)^{2} = \frac{9}{4} $$
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