Question

How do I solve this question?

Answer 1

(A)

Explanation:

Assumption: Charge distribution is located at the middle of non-conducting and uncharged pipe (not clearly shown or marked in the figure).

We observe that at the initial point when the ball is thrown in the pipe, the charge distribution repels the ball and impedes its motion. As the ball reaches just below the charge distribution the force of repulsion becomes orthogonal to the direction of motion of ball. Thereafter the force of repulsion aids motion along the rod towards the other end. As such the ball must be given initial kinetic energy slightly more than just to overcome the work done against the electric field and reach below the charge distribution.

We also know that the work done against electric forces in moving a charge `q` from infinity to a point a distance `r` from the charge `Q` is given by

`W_(oo->r)=qDeltaV=q1/(4piepsilon_0)(Q/r-Q/oo)=(qQ)/(4piepsilon_0)`
where `1/(4piepsilon_0)` is Coulombs constant.

We also observe that each element of the charge distribution is at a distance`=sqrt((4R)^2+(3R)^2)` from the initial position.
Now Initial total energy `=PE_"initial"+KE_"initial"`

Inserting given values we get
Initial total energy `=(+q)1/(4piepsilon_0)(+3q)/sqrt((4R)^2+(3R)^2)+1/2m u^2` ....(1)

When ball reaches the charge distribution its velocity `=0` and the distance between the ball and each charge element of distribution `=4R`

`:.` Total energy `=(+q)1/(4piepsilon_0)(+3q)/(4R)` ....(2)

Using Law of Conservation of Energy we equate RHSs (1) and (2) and solve for `u`

`(+q)1/(4piepsilon_0)(+3q)/sqrt((4R)^2+(3R)^2)+1/2m u^2=(+q)1/(4piepsilon_0)(+3q)/(4R)`

`=>1/2m u^2=1/(4piepsilon_0)(3q^2)/(4R)-1/(4piepsilon_0)(3q^2)/sqrt((4R)^2+(3R)^2)`
`=>1/2m u^2=1/(4piepsilon_0)(3q^2)/(R)(1/4-1/5)`
`=> u^2=(3q^2)/(40piepsilon_0Rm)`
`=> u=sqrt((3q^2)/(40piepsilon_0Rm))`

The ball will come out of the pipe for

`u>sqrt((3q^2)/(40piepsilon_0Rm))`