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`        two identical balls of charges Q1 and Q2 initally have equal velocity of d same magnitude and direction.after a uniform electric field is applied for sometym ,the direction of d velocity of d first ball changes by 60. and d magnitude is reduced by half.the direction of d velocity of d second ball changes there by 90. .in what proportion will the velocity of second ball changes???? the curect ans is v/<3>......where<\$>...means under-root \$...!! give me a curect solution wid a curect explaination!! THanKu... n plz do open my profile.....see the posted ques....n reply 'em.....i dint get reply f d ques i'hv sent`
7 years ago

110 Points
```										the solution is fairly simplethe point to be noticed here is the fact that their accelerations will be in the ratio to their charges since F=maand F=QEsince the field if uniformQE=maso a=QE/ma=kQnow,write the separate equation of of bothe the charges in x directionq1=>Vxi=v, Vxf=V/2cos60=V/4;so, Vxf=Vxi-a1xTV/4=V-a1T.................................1similarly for the charge q20=V-a2T.......................................2using 1 and 2, we geta1/a2=3/4.......................................3now write the equation for y direction in the same wayequate a1/a2(in x)=a1/a2(in y)and get the answer.Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.DO APPROVE IF THE ANSWER HELPS YOU.Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.Regards,Askiitians ExpertsRahul- IIT Roorkee
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7 years ago
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