Question

A charge +Q is uniformly distributed in a spherical volume of radius R. A particle of charge + q and mass m projected with velocity $v_o$ from the surface of the spherical volume to its centre inside a smooth tunnel dug across the sphere, The minimum value of $v_o$ such that it just reaches the centre (assume thatthere is no resistance on the particle except electrostatic force) of the spherical volume is :

• A. $\sqrt{\frac{Qq}{2\pi {\epsilon }_{0}mR}}$
• B. $\sqrt{\frac{Qq}{\pi {\epsilon }_{0}mR}}$
• C. $\sqrt{\frac{2Qq}{\pi {\epsilon }_{0}mR}}$
• D. $\sqrt{\frac{Qq}{4\pi {\epsilon }_{0}mR}}$

Answer 1

The correct option is D $\sqrt{\frac{Qq}{4\pi {\epsilon }_{0}mR}}$

Electric field inside the sphere is given by: $\frac{kQr}{R^3}$

Hence the potential difference between the surface of sphere and the center of the sphere $={\int }_0^R\frac{kQr}{R^3}dr$ $=\frac{kQ}{2R}$

Hence, the change in kinetic energy of the charge $q$ that takes place when it starts with minimum speed $v_0$ on surface, and just stops at the center $=\frac{kQq}{2R}$

$⟹\frac{1}{2}m{v}_0^2=\frac{Qq}{4\pi {ϵ}_0\left(2R\right)}$

$⟹v_0=\sqrt{\frac{Qq}{4\pi {ϵ}_{0}mR}}$