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```        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. ```
8 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
```
8 years ago
```							But why in finding out work done, we have not considered the electrostatic potential energy due to the other charge?
```
8 years ago
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