Q. Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge −q (q ≪ 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: (ε₀ is the permittivity of vacuum, neglect gravitational forces).
A) $\sqrt{\frac{Qq}{4\pi\varepsilon_0 mR}}$
B) $\sqrt{\frac{3Qq}{4\pi\varepsilon_0 mR}}$
C) $\sqrt{\frac{2Qq}{3\pi\varepsilon_0 mR}}$
D) $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$
Explanation
[Solution from source] Loss of potential energy $=\frac{Qq}{4\pi\varepsilon_0}\left(\frac1R-\frac1{3R}\right)=\frac{Qq}{4\pi\varepsilon_0}\cdot\frac{2}{3R}=\frac12mv^2$, so $v^2=\frac{Qq}{3\pi\varepsilon_0mR}$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.