For a uniformly charged insulating sphere of radius R and total charge Q, the electric field E at a distance r from the center (where r < R) is given by Gauss's Law: $$E = \frac{Q r}{4\pi\epsilon_{0} R^3}$$
The force F on a point charge -q placed at distance r is F = -qE. Substituting the expression for E: $$F = -\left(\frac{Qq}{4\pi\epsilon_{0} R^3}\right)r$$ This force is proportional to the displacement r and directed towards the center, which is the condition for Simple Harmonic Motion (SHM).
Comparing the force equation with the standard SHM equation F = -m\omega^2 r, we get: $$m\omega^2 = \frac{Qq}{4\pi\epsilon_{0} R^3}$$ $$\omega = \sqrt{\frac{Qq}{4\pi\epsilon_{0} m R^3}}$$
The maximum velocity vmax for an SHM is given by vmax = \omega A. Here, the amplitude A is the initial distance R. $$v_{max} = \sqrt{\frac{Qq}{4\pi\epsilon_{0} m R^3}} \times R$$ $$v_{max} = \sqrt{\frac{Qq}{4\pi\epsilon_{0} m R}}$$
Final Answer: (A)
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