Need Help. Consider a small, conducting sphere of 0.0010 kg mass. Extra electron

Question

Need Help.
Consider a small, conducting sphere of 0.0010 kg mass. Extra electrons are placed on this sphere and on an identical sphere 3.0 m below it so the repulsion between these extra electrons provide a force equal to the weight of the top sphere. How many electrons must be added to each sphere?

a. 5.0 x 10^-21
b. 3.1 x 10^-6
c. 2.0 x 10^13
d. 3.9 x 10^13
e. none of these

What I've done is tried Coulomb's Law: F=k(q1q2/r^2), w=mg. I am not getting anything, and I want to make sure that there is no answer (e). Please help. Thank you

Explanation / Answer

Mass of the sphere, m = 0.0010 kg Distance between the spheres, r = 3 m No. of electrons added to each sphere = n Charge of each sphere, q = ne = 1.6 x 10^-19 n C Force of repulsion between spheres, F = k q1 q2 / r^2 = 9 x 10^9 * 1.6 x 10^-19 n * 1.6 x 10^-19 n / 9 = 2.56 x 10^-19 n Weight of the sphere, W = m g = 0.001 g = 0.001 * 9.8 = 9.8 x 10^-3 N But, F = F' 2.56 x 10^-19 n = 9.8 x 10^-3 So, n = 0.26 x 10^16 Ans: None of these (e)

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