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Electromagnetic Induction> Two equal charges are placed at a separat...

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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?

Radhika Batra , 10 Years ago

Grade 11

2 Answers

Kevin Nash

Last Activity: 10 Years ago

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

Apoorva Arora

Last Activity: 10 Years ago

a 50 kg person exerts a force mg i.e 500 newtons

So,

$F=500=kq^{2}/r^{2}$

so

$q^{2}=500/k=1/18\times 10^{6}$

so $q=1/4.24\times 10^{3} coulumbs$

Thanks and Regards

Apoorva Arora

IIT Roorkee

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