Two identical metal balls of radius r are at | vedclass

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Question

$\mathrm{V}_{1}=\left(\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{a}}\right) \mathrm{K}$ and $\mathrm{V}_{2}=\left(\frac{\mathrm{q

Vedclass · Accepted Answer

${q_1} = \frac{1}{k}\left( {\frac{{r{V_2} - a{V_1}}}{{{r^2} - {a^2}}}} \right)ra,{q_2} = \frac{1}{k}\left( {\frac{{r{V_1} - a{V_2}}}{{{r^2} - {a^2}}}} \right)ra$

Vedclass · Answer

$\mathrm{V}_{1}=\left(\frac{\mathrm{q}_{1}}{\mathrm{r}}+\frac{\mathrm{q}_{2}}{\mathrm{a}}ight) \mathrm{K}$ and $\mathrm{V}_{2}=\left(\frac{\mathrm{q}_{2}}{\mathrm{r}}+\frac{\mathrm{q}}{\mathrm{a}}ight) \mathrm{K}$

Multiplying ${1^{st}}$ equation by a and ${2^{nd}}$ equation by $\mathrm{r}$ and then subtracting we get;

$\frac{\mathrm{rV}_{2}-\mathrm{aV}_{1}}{\mathrm{K}}=\mathrm{q}_{1}\left(\frac{\mathrm{r}}{\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{r}}ight)=\mathrm{q}_{1} \frac{\mathrm{r}^{2}-\mathrm{a}^{2}}{\mathrm{ar}}$

$	herefore {{m{q}}_1} = \frac{{\left( {{m{r}}{{m{V}}_2} - {m{a}}{{m{V}}_1}} ight){m{ra}}}}{{\left( {{{m{r}}^2} - {{m{a}}^2}} ight){m{K}}}}$

Similarly, we find, $q_{2}=\frac{\left(r V_{1}-a V_{2}ight) r a}{\left(r^{2}-a^{2}ight) K}$

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

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