 # Two positive charges A and B have charge +q and +2q, mass m and 2m. Both the charges are at rest when they are at distance x apart. Assume only force acting on either is the electrostatic force due to each other. Find: 1) Speed of charge A at instant separation b/w both charges is 2x. 2) Work done by electrostatic force on A while separation b/w charges change from x to 2x. 3) work done by electostatic force on A and B while separation b/w both charges changes from x to 2x.

11 years ago

here only internal forces are responsible for motion ...no external force acts on the system so

momentam of system will remain conserved...

initail momentam = 0 = final momentam

let final velocities of m,2m are v1 , v2 then

mv1 + 2mv2 = 0

v1 = -2v2          ....................1

now applying energy conservation

TE(total energy) = Kq.2q/x          (initial)

TE = kq.2q/2x + (1/2)mv12 + (1/2)2mv22           (final)

since energy is conserved so

(1/2)mv12 + (1/2)2mv22 + kq.2q/2x = kq.2q/x                 ....................2

from eq 1 substituting v1 in eq 2

2mv22 + mv22 = kqq/x

v2 = [kq2/3x]1/2

v1 = 2[kq2/3x]1/2                                                   (ans1)

total change in KE = (1/2)mv12 + (1/2)2mv22 = work done                      (this is called work energy theorem)

from eq 2 , change in KE = kq2/x = net workdone on A,B                                    (ans3)

work done on A is change in KE of A = (1/2)mv12 = 2mkq2/3x                              (ans2)

approve if u like my ans