At what separation should two equal charges, 1.0 C each be placed so taht the force between them equals the weight of a 50 kg person?

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Solution

q = 1.C
Let distane be x
$F = 50 \times 9.8 = 490 o r 490 = 9 \times 10^{9} \frac{1}{x^{2}} o r x = 4.3 \times 10^{3}$

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At what separation should two equal charges, 1.0 C each be placed so taht the force between them equals the weight of a 50 kg person?

q = 1.C Let distane be x F=50×9.8=490or490=9×1091x2orx=4.3×103

q = 1.C
Let distane be x
$F = 50 \times 9.8 = 490 o r 490 = 9 \times 10^{9} \frac{1}{x^{2}} o r x = 4.3 \times 10^{3}$