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. `[rhoR^2//4mepsilon_0]^1/2`
B. `[rhoR^2//24mepsilon_0]^1/2`
C. `[rhoR^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 of the `v` center of the tunnel, the particle will cross the tunnel. Hence applying conservation of `ME` between start point and center of tunnel,
`DeltaK+DeltaU=0`
or `(0+(1)/(2)mv^(2))+q(V_(f)-V_(i))=0`
or `V_(f)=(V_(s))/(2)(3-(r^(2))/(R^(2)))=(pR^(2))/(6epsilon_(0))(3-(r^(2))/(R^(2)))`
Hence `r=(R)/(2)`
`V_(f)=(pR^(2))/(6epsilon_(0))*(3-(R^(2))/(4R^(2)))=(11pR^(2))/(24epsilon_(0))`
`V_(i)=((pR^(2))/(3epsilon_90))`
`(1)/(2)mv^(2)=1[(11pR^(2))/(24epsilon_(0))-(pR^(2))/(3epsilon_90)]=(pR^(2))/(3epsilon_(0))[(11)/(24)-(1)/(3)]`
`=(pR^(2))/(8epsilon_(0))`
or `V=((pR^(2))/(4mepsilon_(0)))^(1//2)`
Hence, velocity should be slightly greater than `V`.