Consider two copper balls, each of one molar mass, and 10.0 m apart. Suppose 0.01% of the atoms in one ball acquire an excess positive charge of +e each, and 0.01% of the atoms in the other ball acquire an excess negative charge of -e each. What is the electric force between them?

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The electric force between two charged objects is given by Coulomb's law:$$F = k\frac{q_1q_2}{r^2}$$where k is Coulomb's constant, q1 and q2 are the charges of the objects, and r is the distance between them.In this problem, we have two copper balls, each of one molar mass, and 10.0 m apart. Suppose 0.01% of the atoms in one ball acquire an excess positive charge of +e each, and 0.01% of the atoms in the other ball acquire an excess negative charge of -e each.The number of atoms in one molar mass of copper is:$$N = 6.022 	imes 10^{23} 	ext{ atoms/mol}$$So, the number of atoms in each ball is:$$n = 6.022 	imes 10^{23} 	ext{ atoms/mol} 	imes 1 	ext{ mol} = 6.022 	imes 10^{23} 	ext{ atoms}$$The number of atoms in each ball that acquire an excess charge is:$$n_c = 0.01% 	imes 6.022 	imes 10^{23} 	ext{ atoms} = 6.022 	imes 10^{19} 	ext{ atoms}$$The charge on each ball is:$$q = n_c e = 6.022 	imes 10^{19} 	ext{ atoms} 	imes 1.602 	imes 10^{-19} 	ext{ C} = 9.635 	imes 10^{-1} 	ext{ C}$$The distance between the balls is:$$r = 10.0 	ext{ m}$$So, the electric force between the balls is:$$F = k\frac{q_1q_2}{r^2} = 9.0 	imes 10^9 \frac{(9.635 	imes 10^{-1} 	ext{ C})^2}{(10.0 	ext{ m})^2} = \boxed{8.3 	imes 10^9 	ext{ N}}$$

Consider Two Copper Balls, Each of One Molar Mass, and 10.0