A large solid sphere with uniformly distributed positive charge has a smooth narrow tunnel along its direction. A small particle with negative charge, initially at rest far from the sphere, approaches it along the line of the tunnel, reaches its surface with a speed `v`, and passes through the tunnel. Its speed at the centre of the sphere will be

Options

A. 0
B. `sqrt(1.5v)`
C. `sqrt(2)`
D. `v`

Answer

b

Explanation

Let `m` and `-q` be the mass and the charge of the particle respectively.
Let `v_(0)=`speed of the particle at the centre of the sphere.
Using conservation of mechanical energy
`Delta K+Delta U=0`
`((1)/(2)mv_(s)^(2)-0)+[-q(V_(s)-V_(oo))=0`
`(1)/(2)mv_(s)^(2)=q[V_(s)-0]=qV_(S)`
Similarly `(1)/(2)mv_(0)^(2)=q[V_(c )-0]=qV_(c )`
We know `V_(c )=(3)/(2)k(Q)/(R)V_(s)=k(Q)/(R)`
Hence `(1)/(2)mv_(s)^(2)=q.k(Q)/(R )`....(i)
and `(1)/(2)mv_(0)^(2)=q(3)/(2)(Q)/(R)`....(ii) Dividing Eq. (ii) by (i),
`(v_(0)^(2))/(v^(2)) = (3)/(2) =1.5` or `v_(0) = sqrt(1.5)v`