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Three identical uncharged metal spheres are at the vertices of an equilateral triangle. One at a time, a small sphere is connected by a conducting wire with a large metal sphere that is charged. The center of the large sphere is in the straight line perpendicular to the plane of equilateral triangle and passing through its centre (see figure). As a result, the first small sphere acquires charge $q_1$ and second charge $q_2 (q_2 < q_1)$ . The charge that the third sphere $q_3$ will acquire is (Assume $l >> R$ , $l >> r$ , $d >> R$ , $d >> r$ )
$\frac{{q_1^2}}{{{q_2}}}$
$\frac{{q_2^2}}{{{q_1}}}$
$\sqrt {{q_1}{q_2}} $
$\frac{{{q_1} + {q_2}}}{2}$
Solution
$\frac{\mathrm{q}_{1}}{\mathrm{r}}=\frac{\mathrm{Q}^{\prime}}{\mathrm{R}}$
$\mathrm{q}_{1}+\mathrm{Q}^{\prime}=\mathrm{Q}$
$Q^{\prime}\left(\frac{r}{R}+1\right)=Q$
$Q^{\prime}=\frac{Q R}{R+r}, q_{1}=\frac{Q r}{R+r}$
Similarly $Q^{\prime \prime}=\frac{Q R^{2}}{(R+r)^{2}}, \quad q_{2}=\frac{Q r R}{(R+r)^{2}}$
Similarly $\quad \mathrm{q}_{3}=\frac{\mathrm{QrR}^{2}}{(\mathrm{R}+\mathrm{r})^{3}}$
$\frac{\mathrm{q}_{2}^{2}}{\mathrm{q}_{1}}=\frac{\left[\frac{\mathrm{QrR}}{(\mathrm{R}+\mathrm{r})^{2}}\right]^{2}}{\left(\frac{\mathrm{Qr}}{\mathrm{R}+\mathrm{r}}\right)}=\frac{\mathrm{QR}^{2} \mathrm{r}}{(\mathrm{R}+\mathrm{r})^{3}}=\mathrm{q}_{3}$
