Question
Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+ Q$. A point charge $- q$ $(q \ll Q)$ with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is:
( $\varepsilon_{0}$ is the permittivity of vacuum, neglect gravitational forces).
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Solution
Sol. [Key Concept: Field and potential of a uniformly charged sphere] 
$$\begin{matrix} & \ \Rightarrow \frac{1}{2}mv^{2} = q\left\lbrack \frac{KQ}{R} - \frac{KQ}{3R} \right\rbrack \\ & \ \Rightarrow \frac{1}{2}mv^{2} = \frac{kQq}{R}\left\lbrack \frac{2}{3} \right\rbrack \\ & \ \Rightarrow v^{2} = \frac{kQq \times 4}{3Rm} = \frac{1 \times Qq \times 4}{4\pi\varepsilon_{0} \times 3Rm} = \frac{qQ}{3\pi\varepsilon_{0}Rm} \\ & v = \sqrt{\frac{Qq}{3\pi\varepsilon_{0}mR}} \end{matrix}$$
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