A unit positive point charge of mass m is projected with a velocity v inside the tunnel as shown. The tunnel has been made inside a uniformly charged non-conducting sphere with charged density $ρ$ . The minimum velocity with which the point charge should be projected such it can it reach the opposite end of the tunnel is equal to.

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zero because the initial and the final points are at same potential

${[{ρR}^{2} / 4 {mε}_{0}]}^{1 / 2}$

${[{ρR}^{2} / 6 {mε}_{0}]}^{1 / 2}$

${[{ρR}^{2} / 24 {mε}_{0}]}^{1 / 2}$

answer is A.

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Detailed Solution

Let v be the initial velocity of unit charge so that its velocity at the mid point of the tunnel is zero (just reaches the tunnel) from work energy theorem

$W = ΔkE; q \times (V_{C} - V_{s}) = \frac{1}{2} {mv}^{2}$

Here $V_{s} = \frac{{ρR}^{2}}{3 ε_{0}} and$

V_C = potential at the centre of the tunnel

$\begin{array}{l} = \frac{ρ}{6 ε_{0}} [3 R^{2} - {(\frac{R}{2})}^{2}] = \frac{11 {ρR}^{2}}{24 ε_{0}} \\ ∴ q \times (\frac{11 {ρR}^{2}}{24 ε_{0}} - \frac{{ρR}^{2}}{3 ε_{0}}) = \frac{1}{2} {mv}^{2} (here q = 1 C) \\ v = {[\frac{{ρR}^{2}}{4 {mε}_{0}}]}^{1 / 2} \end{array}$