Question

Electric field?

kindly exlain the situation in detail?

Answer 1

(C)

Explanation:

Ignoring Coulomb's force of repulsion between the charges.
Taking position of charge `q_1` at `t=2s` as `(2a,0)`

Let `vecE=Ehat x`. Hence all motion is along `x`-axis

For charge `q_1`

Force in the electric field`=q_1E`

Using Newton's Second law of Motion, Acceleration `a=(q_1E)/m`

Using kinematic expression

`s=s_0+ut+1/2at^2`

and inserting given values

`2a=a+0xx2+1/2((q_1E)/m)2^2`
`=>a=(2q_1E)/m` .....(1)

Similarly for charge `q_2`
`x_(tq_2)=-a+0xx2+1/2((q_2E)/m)2^2`

Simplifying

`x_(tq_2)=-a+((2q_2E)/m)`

Adding and subtracting `(2q_1E)/m` in the RHS

`x_(tq_2)=-a+((2q_2E)/m)+(2q_1E)/m-(2q_1E)/m`
`=>x_(tq_2)=-a+2((q_1+q_2)/m)E-(2q_1E)/m`

Inserting value of `a` from (1) we get

`x_(tq_2)=2((q_1+q_2)/m)E-2a`