# 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?

Grade:upto college level

## 3 Answers

Deepak Patra
askIITians Faculty 471 Points
8 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 IIT Roorkee
askIITians Faculty 181 Points
8 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 IIT Roorkee
askIITians Faculty 181 Points
8 years ago
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
askIITians Faculty

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