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(C) The potential at a distance $r$ from the center of a uniformly charged insulating sphere of radius $R$ and charge $Q$ is given by: For $r \ge R$,

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$\sqrt{\frac{Qq}{3\pi \epsilon_0 mR}}$

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(C) The potential at a distance $r$ from the center of a uniformly charged insulating sphere of radius $R$ and charge $Q$ is given by: For $r \ge R$, $V(r) = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}$. For $r At the surface $(r = R)$, $V(R) = \frac{1}{4\pi \epsilon_0} \frac{Q}{R}$. At the initial position $(r = 3R)$, $V(3R) = \frac{1}{4\pi \epsilon_0} \frac{Q}{3R}$. By the law of conservation of energy, the change in kinetic energy equals the change in potential energy: $\Delta K + \Delta U = 0 \implies \frac{1}{2}mv^2 - 0 = -q(V(R) - V(3R))$. $\frac{1}{2}mv^2 = q(V(3R) - V(R)) = q \left( \frac{Q}{12\pi \epsilon_0 R} - \frac{Q}{4\pi \epsilon_0 R} \right)$. $\frac{1}{2}mv^2 = \frac{Qq}{4\pi \epsilon_0 R} \left( \frac{1}{3} - 1 \right) = \frac{Qq}{4\pi \epsilon_0 R} \left( -\frac{2}{3} \right)$. Since the charge is negative, it moves towards the positive sphere, gaining kinetic energy. The magnitude of potential difference is $|V(3R) - V(R)| = \frac{2Qq}{12\pi \epsilon_0 R} = \frac{Qq}{6\pi \epsilon_0 R}$. $\frac{1}{2}mv^2 = \frac{Qq}{6\pi \epsilon_0 R} \implies v^2 = \frac{Qq}{3\pi \epsilon_0 mR} \implies v = \sqrt{\frac{Qq}{3\pi \epsilon_0 mR}}$.

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