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

If we throw the charged particle just right of the center of the tunnel, the particle will cross the tunnel. Hence, applying conservation of momentum between start point and center of tunnel we get ΔK+ΔU=0or (0+12mv2)+q(Vf−Vi)=0or Vf=Vs2(3−r2R2)=ρR26ε0(3−r2R2)Hence r=R2 Vf=ρR26ε0(3−R24R2)=11ρR224ε0;  Vi=ρR23ε012mv2=1[11ρR224ε0−ρR23ε0]=ρR2ε0[1124−13]=ρR28ε0or V=(ρR24mε0)1/2Hence velocity should be slightly greater than V.