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```        hello Please Solve this question
Two identical particles,each having a charge of 2.0 X 10-4 C and mass of 10 g,are kept at a seperation of 10cm and then released.What would bee the speeds of the particle When the seperation becomes large ?
7 years ago

510 Points
```

initial potential energy is (PE)i = kq2 /r                  (charges are of same magnitude q)
k = 9 . 109 units
(PE)i  =3600j                 (after substituting value of charge and K)
finally when distance is maximum then (PE)=kq2 /r
r=r(max)=infinity
so final potential energy is 0...
initially kinetic energy is zero ....
finally kinetic energy is (KE)f =  mv12 /2   +  mv2 2 /2
=m(v12 + v22 )/2 ................1
from conservation of energy ,total energy initial = total energy final
therefore ,     m(v12 +v22 )/2 =3600 ..................2
since no external; force is acting so momentam will be conserved..
initial momentam =0
final momentam=mv1 +mv2
mv1 +mv2 =0
v1=-v2 ..................1
now from eq 2 putting v1=-v2
v1=600m/s
v2=-600m/s

```
7 years ago
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