Question

A particle of mass m and charge q is placed in uniform electric field It is given the initial velocity v opposite to the direction of electric field. After how much time will it return to its original position ?​

Answer 1

Answer:

2mv/qe

Explanation:

As initially acceleration (qe/m) is in opposite direction to initial velocity.

So particle will stop after travelling some distance in right direction. let it covers 'L' distance before stop and time taken by it is 'T1'

so by using equation of motion we get,

L = mv^2/2qe

and T1 = mv/qe

Now, particle start moving with acceleration in right direction towards its initial position.

Let suppose it covers T2 time to back go to its initial position.

So by using equation of motion we get T2 = mv/qe.

So total time = T1 + T2 = 2mv/qe

I hope it helps....