Question

A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8900kg/m^3.

Answer 1

`f = 2.6 * 10^(-13)`

Step-by-step explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

`N = (m)/(29) * 6.02 * 10^(23)`

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

`N_e = m(6.02 * 10^(29))`

now diameter of the ball is 7.0 mm

density of the ball =
`8900 kg/m^3`

now we have

`m = ((4)/(3)\pi r^3)(8900)`

`m = ((4)/(3)\pi((0.007)/(2))^3)(8900)`

`m = 1.6 gram`

now we have

`N_e = 9.63 * 10^(23)`

now the charge on the copper ball is 40 nC

so the number of electrons removed

`Q = ne`

`40 * 10^(-9) = n(1.6 * 10^(-19)`

`n = 2.5 * 10^(11)`

so the fraction of number of electrons removed is given as

`f = (n)/(N_e)`

`f = (2.5 * 10^(11))/(9.63 * 10^(23))`

`f = 2.6 * 10^(-13)`