Question

We know that the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by $\mathbf{0}\mathbf{.00010}\mathit{\%}\mathbf{.}$with what force would two copper coins, placed$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$apart, repel each other? Assume that each coin contains$\mathbf{3}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{22}}$copperatoms. Whatdo you conclude?

Short answer

The two coins will repel each other by a force equal to$1.7\times {10}^8\text{N}$

Step-by-step solution

Given information

The magnitude of the charge of an electron and proton differs from each other by 0.00010% distance between two copper coins$\mathrm{d}=1.0\text{m}$ and each coin contains $3\times {10}^{22}$copper atoms.

Understanding the concept

Charges are quantized. It means the charge on any object is equal to the integer multiple of an elementary charge. The charge will always be present in an integer number and never as a fraction.

Coulomb’s law states that the force of attraction or repulsion between the two charges is directly proportional to the product of the charges and inversely proportional to the square of a distance between two charges.

Calculate the force with which the two copper coins, placed 1.0 m apart, would repel each other 

If the relative difference between the proton and electron charges (in absolute

value) were,

$\begin{array}{c}\mathrm{n}=\frac{{\mathrm{q}}_{\mathrm{p}}-\left|{\mathrm{q}}_{\mathrm{e}}\right|}{\mathrm{e}}\\ =0.0000010\end{array}$

.

${\mathrm{q}}_{\mathrm{p}}-\left|{\mathrm{q}}_{\mathrm{e}}\right|=1.6\times {10}^{-25}\text{C}$

Amplified by a factor $29\times 3\times {10}^{22}$as indicated in the problem, this amounts to a deviation from perfect neutrality of,

$\begin{array}{c}\mathrm{\Delta q}=(29\times 3\times {10}^{22})(1.6\times {10}^{-25}\text{C})\\ =0.14\text{C}\end{array}$

in a copper penny.

Two such pennies, at r = 1.0 m, would therefore experience a very large force.

$\begin{array}{c}\mathrm{F}=\mathrm{k}\frac{{\left|\mathrm{\Delta q}\right|}^2}{{\mathrm{r}}^2}\\ =1.7\times {10}^8\text{N}\end{array}$

Therefore, the two coins will repel each other by a force equal to $1.7\times {10}^8\text{N}$.