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Two identical metal balls of radius $r$ are at a distance $a (a >> r)$ from each other and are charged, one with potential $V_1$ and other with potential $V_2$. The charges $q_1$ and $q_2$ on these balls in $CGS$ esu are
${q_1} = \frac{{r{V_1} + a{V_1}}}{{{r^2} + {a^2}}},{q_2} = \frac{{r{V_1} + a{V_2}}}{{{r^2} + {a^2}}}$
${q_1} = \frac{1}{k}\left( {\frac{{r{V_2} - a{V_1}}}{{{r^2} - {a^2}}}} \right)ra,{q_2} = \frac{1}{k}\left( {\frac{{r{V_1} - a{V_2}}}{{{r^2} - {a^2}}}} \right)ra$
${q_1} = \frac{{a{V_2}}}{{k\left( {{r^2} - {a^2}} \right)}},{q_2} = \frac{{r{V_1}}}{{k\left( {{r^2} - {a^2}} \right)}}$
${q_1} = \frac{{r{V_1}}}{{k\left( {{r^2} - {a^2}} \right)}},{q_2} = \frac{{r{V_2}}}{{\left( {{r^2} - {a^2}} \right)k}}$
Solution
$\mathrm{V}_{1}=\left(\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{a}}\right) \mathrm{K}$ and $\mathrm{V}_{2}=\left(\frac{\mathrm{q}_{2}}{\mathrm{r}}+\frac{\mathrm{q}}{\mathrm{a}}\right) \mathrm{K}$
Multiplying ${1^{st}}$ equation by a and ${2^{nd}}$ equation by $\mathrm{r}$ and then subtracting we get;
$\frac{\mathrm{rV}_{2}-\mathrm{aV}_{1}}{\mathrm{K}}=\mathrm{q}_{1}\left(\frac{\mathrm{r}}{\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{r}}\right)=\mathrm{q}_{1} \frac{\mathrm{r}^{2}-\mathrm{a}^{2}}{\mathrm{ar}}$
$\therefore {{\rm{q}}_1} = \frac{{\left( {{\rm{r}}{{\rm{V}}_2} – {\rm{a}}{{\rm{V}}_1}} \right){\rm{ra}}}}{{\left( {{{\rm{r}}^2} – {{\rm{a}}^2}} \right){\rm{K}}}}$
Similarly, we find, $q_{2}=\frac{\left(r V_{1}-a V_{2}\right) r a}{\left(r^{2}-a^{2}\right) K}$
