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Q .1 dt 5 Oct 17: *Two point charges are separated by certain distance in air medium & experiences a force of F newton. If the charge strength of each charge is increased by 20% and distance between them reduced by 10 % ,what will be new force in terms of F ?*

Q .1 dt 5 Oct 17: *Two point charges are separated by certain distance in air medium & experiences a force of F newton. If the charge strength of each charge is increased by 20% and distance between them reduced by 10 % ,what will be new force in terms of F ?*

Grade:12th pass

1 Answers

Shailendra Kumar Sharma
188 Points
3 years ago
F=kq1q2/r2 
Now the charges are increased by 20% and distance reduced by 10% 
F*= k(1.2q1)(1.2q2)/(.9r)2 =1.6kq1q2/r2 
So F* =1.6F
 
 

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