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

0

v

$\sqrt{2}$ v

$\sqrt{1.5}$ v

Solution

$\sqrt{1.5}$ v

Potential at infinity = V_∞ = 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 v₀ = speed of the particle at the centre of the sphere.

$\frac{1}{2} m v^{2} = - q [V_{\infty} - V_{S}]$ = qk $\frac{Q}{R}$ ....(i)

$\frac{1}{2} {mv}_{0}^{2} = - q [V_{\infty} - V_{c}] = q . \frac{3}{2} k \frac{Q}{R}$ ......(ii)

Dividing eqn. (ii) by eqn. (i),

$\frac{v_{0}^{2}}{v^{2}} = \frac{3}{2}$

= 1.5

v₀ = $\sqrt{1.5}$ v

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