Physics - Electrostatics Question with Solution | TestHub

Potential at infinity $=V_{\infty }=0$ Potential at the surface of the sphere, $V_s=k\frac{Q}{R}$ Potential at the centre of the sphere, $V_c=\frac{3}{2}k\frac{Q}{R}$ Let $m$ and $-q$ be the mass and the charge of the particle respectively. Let $v_0=$ speed of the particle at the centre of the sphere. $\begin{array}{rl}& \frac{1}{2}m{v}^2=-q\left[V_{\infty }-V_s\right]=qk\frac{Q}{R}\\ & \frac{1}{2}m{v}_0^2=-q\left[V_{\infty }-V_c\right]=q\cdot \frac{3}{2}k\frac{Q}{R}\end{array}$ Dividing eqn. (ii) by eqn. (i), $\frac{v_0^2}{v^2}=\frac{3}{2}=1.5$ or $v_0=\sqrt{1.5}v$