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 that it can reach the opposite end of the tunnel is equal to

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${[{ρR}^{2} / 24 m ε_{0}]}^{\frac{1}{2}}$

${[{ρR}^{2} / 4 m ε_{0}]}^{\frac{1}{2}}$

${[{ρR}^{2} / 6 m ε_{0}]}^{\frac{1}{2}}$

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

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

$W = Δ k E; q \times (V_{c} - V_{s}) = \frac{1}{2} m v^{2}$ Here $V_{s} = \frac{{ρR}^{2}}{{3ε}_{0}} and$ $V_{c} =$ potential at the centre of the tunnel $= \frac{ρ}{6 ε_{0}} [3 R^{2} - {(\frac{R}{2})}^{2}] = \frac{11 ρ R^{2}}{24 ε_{0}}$ $∴ 1 \times (\frac{11 ρ R^{2}}{24 ε_{0}} - \frac{ρ R^{2}}{3 ε_{0}}) = \frac{1}{2} m v^{2}$ $v = {[\frac{ρ R^{2}}{4 m ε_{0}}]}^{1 / 2}$

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