Q.

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 nonconducting sphere. 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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a

ρR2/4mε01/2

b

ρR2/24mε01/2

c

ρR2/6mε01/2

d

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

answer is A.

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Detailed Solution

If we throw the charged particle just right of the center of the tunnel, the particle will cross the tunnel. Hence, applying conservation of energy between starting point and center of tunnel we getΔK+ΔU=0or 0+12mv2+qVf−Vi=0or Vf=Vs23−r2R2=ρR26ε03−r2R2Here r=R2Vf=ρR26ε03−R24R2=11ρR224ε0; Vi=ρR23ε012mv2=111ρR224ε0−ρR23ε0=ρR2ε01124−13=ρR28ε0or V=ρR24mε01/2Hence velocity should be slightly greater than V.
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