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Two identical copper spheres are separated by 1 m in vacuum. How many electrons would have to be removed from one sphere and added to the other so that they now attract each other with a force of 0.9 N?

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$6.25 \times 10^{15}$

$62.5 \times 10^{15}$

$6.25 \times 10^{13}$

$0.65 \times 10^{13}$

answer is C.

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Detailed Solution

$F = \frac{9 \times 10^{9} (n e) (n e)}{r^{2}} = 0.9$

On solving

$n = 6.25 \times 10^{13}$

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