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 ?*
Saikat mallick
8 Years agoGrade 12th pass
1 Answer
Shailendra Kumar Sharma
8 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