(a) What total (excess) charge qmust the disk in Fig. 22-15 have for the electric field on the surface of the disk at its center to have magnitude $\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{N/C}}$, the Evalue at which air breaks down electrically, producing sparks? Take the disk radius as.$\mathbf{2}\mathbf{.5}\mathbf{\text{cm}}$ (b) Suppose each surface atom has an effective cross-sectional area of$\mathbf{0}\mathbf{.015}\mathbf{}{\mathbf{\text{nm}}}^{\mathbf{\text{2}}}$. How many atoms are needed to make up the disk surface? (c) The charge calculated in (a) results from some of the surface atoms having one excess electron. What fraction of these atoms must be so charged?

a) Electric field on the surface of the disk at its center to have magnitude,$E=3.0x{10}^6\text{N/C}$

b) Disk radius,$R=\text{2}.5\text{cm}$

c) Cross-sectional area,$A=0.015{\text{nm}}^{\text{2}}$

Using the concept of Gauss’s law and Coulomb’s law of electrostatics, we can get the value of the charge required. Now, by dividing the total area by the area of the atom, we can get the required number of atoms. Similarly, the fraction value can be determined by dividing the calculated value of charge by the product of the number of atoms with the charge of an electron.

Formulae:

The electric field on a particle due to a circular disk, $E=\frac{\left|q\right|}{4\pi {\epsilon }_0{R}^2}$ (i)

The total charge of a body due to number of atoms, $q=ne$ (ii)

Using equation (i), the value of the total charge on at the disk is given as:

$\begin{array}{c}\left|q\right|=2\pi {ϵ}_o{R}^{2}E\\ =\frac{{(2.5\times {10}^{-2}\text{m})}^2(3.0\times {10}^6\text{N/C})}{2(8.99\times {10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}})}\\ =1.0\times {10}^{-7}\text{C}\end{array}$

Hence, the value of the charge is.$1.0\times {10}^{-7}\text{C}$

Setting up a simple proportionality (with the areas), the number of atoms is estimated to be as follows:

$\begin{array}{c}n=\frac{\pi {(2.5\times {10}^{-2}\text{m})}^2}{0.015\times {10}^{-18}{\text{m}}^2}\\ =1.3\times {10}^{17}\end{array}$

Hence, the value of the number of the atoms to make up the surface is.$1.3\times {10}^{17}$

The fraction of the total charge to the charges on the present atoms can be given using equation (ii) as:

$\begin{array}{c}\frac{q}{Ne}=\frac{1.0\times {10}^{-7}\text{C}}{(1.3\times {10}^{17})(1.6\times {10}^{-19}\text{C})}\\ \approx 4.8\times {10}^{-6}\end{array}$

Hence, the value of the fraction is.$4.8\times {10}^{-6}$