Question

Two small aluminum spheres, each having mass 0.0250 kg , are separated by 80.0 cm . (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 g/mol, and its atomic number is 13.) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude $\mathbf{1}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{N}$(roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Short answer

1. The number of electrons on the sphere is $7.25\times {10}^{24}\text{electrons.}$
2. The number electrons to be removed from one sphere and added to the other is localid="1668319958837" $5.27\times {10}^{15}$$5.27\times {10}^{15}$electrons.
3. The fraction of electrons represented in each sphere is $7.27\times {10}^{-10}$.

Step-by-step solution

Step 1: Concept of Coulomb’s Law

According to the Coulomb’s law, the force between two charges spaced at a distance is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

Mathematically,

$\mathbf{F}\mathbf{=}\mathbf{k}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}^{\mathbf{2}}}$ ….. (1)

Here F, is the force, ${\mathrm{q}}_1$and ${\mathrm{q}}_2$are the charges, k is the Coulomb’s charge, and r is the distance between two charges.

(a) Determination of the number of electrons on the sphere.

Consider the given data as below.

Mass, m = 0.0250 kg

The atomic mass, M = 26.982 g/mol = 0.026982 kg/mol

Avogadro number,${\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}{\mathrm{mol}}^{-1}$

A single Aluminium atom has 13 electrons. So the required quantity is the total number of Al atom on the sphere at first using the sphere mass and mass of single Al atom.

Thus, the number of electrons is equal to the number of proton on the sphere surface.

localid="1668319967001" $\begin{array}{r}{\mathrm{n}}_{\mathrm{e}}={\mathrm{n}}_{\mathrm{p}}\\ =\left(13\right)\left({\mathrm{N}}_{\mathrm{A}}\right)\left(\frac{\mathrm{m}}{\mathrm{M}}\right)\\ =\left(13\right)\left(6.022\times {10}^{23}{\mathrm{mol}}^{-1}\right)\left(\frac{0.0250\mathrm{kg}}{0.026982\mathrm{kg}/\mathrm{mol}}\right)\\ =7.25\times {10}^{24}\text{electrons}\end{array}$

(b) Determination of the number electrons to be removed from one sphere and added to the other.

Coulomb force of attraction between the opposite charges is given as,

$\mathrm{F}=-\mathrm{k}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}$

Consider the given data as below.

The force, $\mathrm{F}=1.00\times {10}^4\mathrm{N}$

The Coulomb’s constant,$\mathrm{k}=9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$

The distance between charges, r = 80 cm = 0.8 cm

Rewrite equation (1) as below.

localid="1668319977179" $\begin{array}{r}\mathrm{F}=\frac{{\mathrm{ke}}^2}{{\mathrm{r}}^2}\\ \left|\mathrm{e}\right|=\sqrt{\frac{{\mathrm{Fr}}^2}{\mathrm{k}}}\end{array}$

Substitute known values in the above equation, and you have

$\begin{array}{r}\left|\mathrm{e}\right|=\sqrt{\frac{\left(1.00\times {10}^4\mathrm{N}\right)(0.800\mathrm{m}{)}^2}{9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2}}\\ =8.43\times {10}^{-4}\mathrm{C}\end{array}$

Thus, the number of electrons removed from one sphere and added to the other is

localid="1668319982850" $\begin{array}{r}{\mathrm{n}}_{\mathrm{e}}^{\mathrm{\prime }}=\frac{\mathrm{e}}{\mathrm{q}}\\ =\frac{8.43\times {10}^{-4}\mathrm{C}}{1.6\times {10}^{-19}\mathrm{C}}\\ =5.27\times {10}^{15}\text{electrons.}\end{array}$

Hence, the number electrons to be removed from one sphere and added to the other is$5.27\times {10}^{15}\mathrm{electrons}$localid="1668319985767" $5.27\times {10}^{15}\text{electrons}$

(c) Determination of the fraction of electrons represented in each sphere.

From part (a) and (b),

$\begin{array}{r}{\mathrm{n}}_{\mathrm{e}}=7.25\times {10}^{24}\text{electrons}\\ {\mathrm{n}}_{\mathrm{e}}^{\mathrm{\prime }}=5.27\times {10}^{15}\text{electrons}\end{array}$

Thus, the required fraction is,

$\begin{array}{r}\frac{{\mathrm{n}}_{\mathrm{e}}^{\mathrm{\prime }}}{{\mathrm{n}}_e}=\frac{5.27\times {10}^{15}}{7.25\times {10}^{24}}\\ =7.27\times {10}^{-10}\end{array}$

Hence, The fraction of electrons represented in each sphere is $7.27\times {10}^{-10}$.