91影视

According to the Coulomb鈥檚 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,

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

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

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,${\mathrm{N}}_{\mathrm{A}}=6.022脳{10}^{23}{\mathrm{mol}}^{鈭�1}$

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

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

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

Consider the given data as below.

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

The Coulomb鈥檚 constant,$\mathrm{k}=9脳{10}^9\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{C}}^2$

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

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

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

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

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

From part (a) and (b),

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

Thus, the required fraction is,

$\begin{array}{r}\frac{{\mathrm{n}}_{\mathrm{e}}^{\mathrm{鈥�}}}{{\mathrm{n}}_e}=\frac{5.27脳{10}^{15}}{7.25脳{10}^{24}}\\ =7.27脳{10}^{鈭�10}\end{array}$

Hence, The fraction of electrons represented in each sphere is $7.27脳{10}^{鈭�10}$.