A copper sphere with a diameter of 3.0 mm is charged to 50 nC. What fraction of its electrons have been removed? Use a copper density of 8900 kg/m^3.

• A. 9.039e-12
• B. 1.000e-9
• C. 1.000e-6
• D. 9.000e-9

Answer: (A)

Removing electrons changes the net charge by one elementary charge per electron, so the fraction of electrons removed is the net charge divided by the total number of electrons in the sphere.

First find how many electrons are in the copper sphere. The volume of a sphere with diameter 3.0 mm (radius 1.5e-3 m) is V = (4/3)π(1.5e-3)^3 ≈ 1.41e-8 m^3. With density ρ = 8900 kg/m^3, the mass is m ≈ ρV ≈ 8900×1.41e-8 ≈ 1.26e-4 kg. Copper’s molar mass is about 0.063546 kg/mol, so the number of moles is n ≈ m/M ≈ 1.26e-4 / 0.063546 ≈ 1.98e-3 mol. The number of atoms is N_atoms ≈ n×N_A ≈ 1.98e-3 × 6.022e23 ≈ 1.19e21 atoms. Each copper atom has 29 electrons, so the total number of electrons is N_e ≈ N_atoms×29 ≈ 1.19e21×29 ≈ 3.45e22 electrons.

The net charge corresponds to removing n_removed = Q/e electrons, where Q = 50 nC = 50×10^-9 C and e = 1.602×10^-19 C. So n_removed ≈ (50×10^-9) / (1.602×10^-19) ≈ 3.12×10^11 electrons.

The fraction removed is f = n_removed / N_e ≈ (3.12×10^11) / (3.45×10^22) ≈ 9.0×10^-12 (numerically about 9.039×10^-12).