Two non-conducting solid spheres of radii R a | vedclass

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(C) Let the smaller sphere have radius $R$ and center $C_1$,and the larger sphere have radius $2R$ and center $C_2$. The distance between $C_1$ and $C

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$(A, D)$

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(C) Let the smaller sphere have radius $R$ and center $C_1$,and the larger sphere have radius $2R$ and center $C_2$. The distance between $C_1$ and $C_2$ is $3R$.Case $1$: Point $P$ is at distance $2R$ from $C_1$ towards $C_2$. $P$ is inside the larger sphere at a distance $R$ from $C_2$.Electric field due to sphere $1$ at $P$: $E_1 = \frac{k Q_1}{(2R)^2} = \frac{k (\rho_1 \cdot \frac{4}{3} \pi R^3)}{4R^2} = \frac{k \rho_1 \pi R}{3}$.Electric field due to sphere $2$ at $P$: $E_2 = \frac{k Q_2 r}{R_{2}^3} = \frac{k (\rho_2 \cdot \frac{4}{3} \pi (2R)^3) \cdot R}{(2R)^3} = \frac{4}{3} k \rho_2 \pi R$.For $E_{net} = 0$,$E_1 = E_2 \implies \frac{\rho_1}{3} = \frac{4}{3} \rho_2 \implies \frac{\rho_1}{\rho_2} = 4$.Case $2$: Point $Q$ is at distance $2R$ from $C_1$ on the side opposite to $C_2$. $Q$ is outside both spheres.Distance of $Q$ from $C_1$ is $2R$,and from $C_2$ is $2R + 3R = 5R$.$E_1 = \frac{k Q_1}{(2R)^2} = \frac{k (\rho_1 \cdot \frac{4}{3} \pi R^3)}{4R^2} = \frac{k \rho_1 \pi R}{3}$.$E_2 = \frac{k Q_2}{(5R)^2} = \frac{k (\rho_2 \cdot \frac{4}{3} \pi (2R)^3)}{25R^2} = \frac{32}{75} k \rho_2 \pi R$.For $E_{net} = 0$,$E_1 + E_2 = 0 \implies \frac{\rho_1}{3} + \frac{32}{75} \rho_2 = 0 \implies \frac{\rho_1}{\rho_2} = -\frac{32}{25}$.Thus,the possible ratios are $4$ and $-\frac{32}{25}$.

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