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
(##BMS_V03_C03_E01_124_Q01.png" width="80%">

Options

A. `(pR^(2)//4mepsilon_(0)]^(1//2)`
B. `[pR^(2)//24mepsilon_(0)]^(1//2)`
C. `[pR^(2)//6mepsilon_(0)]^(1//2)`
D. zero because the initial and the final points are at same potential

Answer

A

Explanation

If we throw the charged particle just right off the centre of the tunnel the particle will cross the tunnel. Hence applying conservation of momentum between start point and centre of tunnel we get
`Delta K+Delta U=0`
Or `(0+1/2 mv^2)+q(V_f-V_i)=0 Or V_f-V_s/2(3-r^2/R^2)(3-r^2/R^2)=(pR^2)/(6e_0) (3-r^2/R^2)`
Hence `r=R/2 , V_f =(pR^2)/(6e_0) (3-R^2/(4R^2)) =(11pR^2)/(24e_0), V_i=(pR^2)/(3e_0)`
`1/2=mv^2=1 [(11pR^2)/(24e_0)- (pR^2)/(3e_0)] =(pR^2)/e_0[11/24-1/3]=(pR^2)/(8e_0) or V=((pR^2)/(4 me _0))^(1//2)`
Hence, velocity should be slightly greater than v.