Question

A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel through its centre. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed $v$, and passes through the tunnel. Its speed at the centre of the sphere will be

• A. $0$
• B. $v$
• C. $\sqrt{2}v$
• D. $\sqrt{1.5}v$

Answer 1

The correct option is D $\sqrt{1.5}v$

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.
$\frac{1}{2}m{v}^2=-q[V_{\infty }-V_s]=qk\frac{Q}{R}....\left(i\right)$
$\frac{1}{2}m{v}_0^2=-q[V_{\infty }-V_c]=q.\frac{3}{2}k\frac{Q}{R}....\left(ii\right)$
Dividing eqn. (ii) by eqn. (i),
$\frac{v_0^2}{v^2}=\frac{3}{2}=1.5$ or $v_0=\sqrt{1.5}v$.