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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?

Shane Macguire , 10 Years ago
Grade upto college level
anser 3 Answers
Deepak Patra

Last Activity: 10 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

Apoorva Arora

Last Activity: 10 Years ago

a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).
So,
F=500= 9\times 10^{9}\times 1\times 1/r^{2}
r^{2}=18\times 10^{6}
so r\approx 4.24\times 10^{3}metres
nearly 4.24 kilometres
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty

Apoorva Arora

Last Activity: 10 Years ago

a 50 kg person exerts the force, F=mg=500 newtons assuming g=10 m/s(square).
So,
Description: F=500= 9\times 10^{9}\times 1\times 1/r^{2}
Description: r^{2}=18\times 10^{6}
so Description: r\approx 4.24\times 10^{3}metres
nearly 4.24 kilometres
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty

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