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two identical charges having the charge 2*10^(-4) C,and having the mass 10g are kept at separation by 10cm.what would be the velocity of particles when they the separation is maximum.
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
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