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        At what separation should two equal charge, 1.0 C each, be placed so that the force between them equals the weight of a 50 kg person?
5 years ago

							Sol. q = 1 C, Let the distance be x
F = 50 * 9.8 = 490
F = kq^2/x^2 ⇒ 460 = 9 * 10^9 * 1^2/x^2 or x^2 = 9 * 10^9/490 = 18.36 * 10^6
⇒ x = 4.29 * 10^3 m


5 years ago Apoorva Arora
IIT Roorkee
181 Points
							a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).So,so nearly 4.24 kilometresThanks and RegardsApoorva AroraIIT RoorkeeaskIITians Faculty

5 years ago Apoorva Arora
IIT Roorkee
181 Points


a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).

So,  so nearly 4.24 kilometres

Thanks and Regards

Apoorva Arora

IIT Roorkee


5 years ago
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