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peyyala vamsi krishna Grade: 12
        

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 ?


Answer :600m/s

6 years ago

Answers : (1)

vikas askiitian expert
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


6 years ago
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