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 ?*

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Shailendra Kumar Sharma · Answer

F=kq 1 q 2 /r 2 Now the charges are increased by 20% and distance reduced by 10% F*= k(1.2q 1 )(1.2q 2 )/(.9r) 2 =1.6kq 1 q 2 /r 2 So F* =1.6F

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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 ?

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