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# hello Please Solve this questionTwo 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 ?Answer :600m/s 10 years ago

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