A 4.80 mm-diameter copper ball is charged to + 70.0 nC.?

Question

A 4.80 mm-diameter copper ball is charged to + 70.0 nC.?

What fraction of its electrons have been removed?

Answer 1

Radius r = dimater/3 = 4.80/2 mm = 2.40 mm = 0.24 cm

Volume V = 4/3 * pi * r^3 = 4/3 * 3.14 * 0.24^3 = 0.058 cm^3

Atomic weight of copper = 63.5 g/mol

Density = 8.96 g/cm^3

Therefore, volume of copper per mole = 63.5/8.96 = 7.087 cm^3/mol

In 7.087 cm^3, there is one mole

Therefore, in 0.058 cm^3(volume of the ball), there is 0.058/7.087 mole = 0.0082

Number of atoms in one mole = avogadro number = 6.022 * 10^23

Therefore, number of atoms in  0.0082 mole = 0.0082 * 6.022 * 10^23

= 0.049 * 10^23

There are 29 electrons in one atom of copper.

Therefore, number of electrons in the copper ball

= 0.049 * 10^23 * 29

= 1.421 * 10^23

If e is the magnitude of the charge on one electron, then e = 1.6 * 10^-19 C

70.0 nC = 70 * 10^-9 C = 70*10^-9/(1.6 * 10^-19)e

= 43.75 * 10^10 e

Therefore, 43.75 * 10^10 electrons have been removed.

Original number of electrons = 1.421 * 10^23

Fraction of electrons removed

= 43.75 * 10^10/(1.421 * 10^23)

= 30.788 * 10^-13

= 3.08 * 10^-12

Ans: 3.08 * 10^-12