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(B) The potential of the first ball is $V_1 = K \left( \frac{q_1}{r} + \frac{q_2}{a} \right)$ and the potential of the second ball is $V_2 = K \left(

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$q_1 = \frac{1}{k} \left( \frac{rV_1 - aV_2}{r^2 - a^2} \right) ra, q_2 = \frac{1}{k} \left( \frac{rV_2 - aV_1}{r^2 - a^2} \right) ra$

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(B) The potential of the first ball is $V_1 = K \left( \frac{q_1}{r} + \frac{q_2}{a} \right)$ and the potential of the second ball is $V_2 = K \left( \frac{q_2}{r} + \frac{q_1}{a} \right)$,where $K$ is the electrostatic constant.We have the system of equations:$V_1 = K \left( \frac{q_1}{r} + \frac{q_2}{a} \right) \implies \frac{V_1}{K} = \frac{q_1}{r} + \frac{q_2}{a} \quad (1)$$V_2 = K \left( \frac{q_2}{r} + \frac{q_1}{a} \right) \implies \frac{V_2}{K} = \frac{q_2}{r} + \frac{q_1}{a} \quad (2)$Multiply equation $(1)$ by $a$ and equation $(2)$ by $r$:$a \frac{V_1}{K} = \frac{aq_1}{r} + q_2$$r \frac{V_2}{K} = \frac{rq_2}{r} + \frac{rq_1}{a} = q_2 + \frac{rq_1}{a}$Subtracting the two equations:$\frac{rV_2 - aV_1}{K} = q_1 \left( \frac{r}{a} - \frac{a}{r} \right) = q_1 \left( \frac{r^2 - a^2}{ar} \right)$Solving for $q_1$:$q_1 = \frac{1}{K} \left( \frac{rV_2 - aV_1}{r^2 - a^2} \right) ra$Similarly,solving for $q_2$:$q_2 = \frac{1}{K} \left( \frac{rV_1 - aV_2}{r^2 - a^2} \right) ra$

Two identical metal balls of radius $r$ are at a distance $a$ $(a \gg r)$ 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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