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        Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges so that the force between them equals the weight of a 50 kg person?
3 years ago

## Answers : (2)

Kevin Nash
332 Points
										Sol. charges ‘q’ each, AB = 1 m
Wt, of 50 kg person = 50 * g = 50 * 9.8 = 490 N
F base c = kq base 1q base 2/r^2 ∴ kq base 2/r^2 = 490 N
⇒ q^2 = 490 * r^2/9 * 10^9 = 490 * 1 * 1/9 * 10^9
⇒ q = √54.4 * 10^9 = 23.323 * 10^-5 coulomb = 2.3 * 10^-4 coulomb


3 years ago
Apoorva Arora
IIT Roorkee
181 Points
										a 50 kg person exerts a force mg i.e 500 newtonsSo,$F=500=kq^{2}/r^{2}$so$q^{2}=500/k=1/18\times 10^{6}$so $q=1/4.24\times 10^{3} coulumbs$Thanks and RegardsApoorva AroraIIT RoorkeeaskIITians Faculty

3 years ago
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