Question

a charged particle Q is fixed and another charged particle of mass m and charge q of same sign is
released from distance r. The impulse of force exerted by external agent on fixed charge by the time distance between Q and q becomes 2r is
options are given

Answer 1

We apply the conservation of energy principle here.  The electric field is conservative.

   The electric potential energy is converted into mechanical kinetic energy of the charged particle q.  The   change in PE when distance changes from r to 2 r :

         1/(4πε) * Q q [ 1/ r - 1/ (2r)  ]  = change in KE = 1/2 m v²

            v² = Qq / (4πε m)  * 1/ r
            v  = K / √r          let us say          K = √[Qq/(4πε m) ]

         v = velocity when the charge q is at a distance 2 r.
 
Change in linear momentum of the charge q = K m /√r

we know that the change in linear momentum = impulse = Force * Δt

So the impulse exerted by the external agent/force, on the Fixed charge Q, to keep it stationary, is equal to  k m /√r  = √ [Qqm / (4πε r) ]

So it is option B.
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We can find the time duration to travel, we can find the velocity function and displacement of position of charge q at any time t.
      
   Let x = distance of q  from charge Q.  Let  v be the velocity of q at  distance x.

Change in PE when q moves from  x  to x+Δx      =  change in KE
       1/(4πε) * Q q [  1/(x+Δx) - 1/x ] = 1/2 m [ (v+Δv)² - v² ]
       - (Qq / (2πε m)) * Δx / [x(x+Δx)]  =  2 v Δv + (Δv)²
       - 2 K²  / [ x (x+Δx) ]  =  2 v  Δv / Δx  + Δv/Δx * Δv

 Taking limits as Δx  => 0 and Δv => 0 , we get
   =>  - 2 K² / x²  = 2 v dv / dx

 Integrating wrt x,  from x = r:
        =>  v² - 0  = 2 K² [ 1 / x  - 1 / r ]
                v² (x) = [ Qq/(2πε m) ] * [1 / x - 1/ r ]

When x = 2 r,
           v² (2r) =  Qq/(4πε m r)
         v (2 r) =  √ [ Qq / (4πε m r) ]

Answer 2

Explanation:

here is the answer for you dear