Two silver spheres, each of 10 cm radius, are lying on a table with their center

Question

Two silver spheres, each of 10 cm radius, are lying on a table with their centers a half-meter apart. Each sphere contains an equal number of protons and electrons. If the magnitude of each electron’s charge were very slightly different than the magnitude of each proton’s charge, then the two spheres would repel each other with a substantial force, since there are so many charges in each ball. But the silver balls don’t noticeably repel, which implies that each of their electrons and protons have equal and opposite charges. But how equal? Suppose you could easily detect a force of 10-2 N (which is a bit less than the weight of a penny) on each ball. In the absence of a force that large, you can infer that the charge of an electron and a proton must be equal and opposite to within approximately one part in 10p . Compute the exponent p (rounded off to the nearest integer).

Explanation / Answer

using a coulomb's law, To compute the exponent p :

F = ke qe qp / r2

where, F = force = 10-2 N

ke = proportionality constant = 9 x 109 Nm2/C2

qe = charge on eectron = 1.6 x 10-19 C = qp

r = separation distance = 0.5 m

then, we get

(10-2 N) = (9 x 109 Nm2/C2) 10p (1.6 x 10-19 C)2 / (0.5 m)2

(0.25 x 10-2 Nm2) = (23.04 x 10-29 Nm2) 10p

10 p = (0.25 x 10-2 Nm2) / (23.04 x 10-29 Nm2)

p = 0.001085 x 1027

p = 1.085 x 1030

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