Answer
$ F = 1.74 \times 10^8 N$
Work Step by Step
We are assuming that the difference of the proton and electron charge is by $0.00010 \%$. And we assume the charge of proton is the one that differs by $0.00010 \%$.
Charge of electron,
$q_e = 1.6 \times 10^{-19}C$
Charge of proton,
$q_p = 1.6 \times 10^{-19} C \times 0.00010 \% $
$q_p = 1.6 \times 10^{-19} C \times 0.0000010 $
$q_p = 1.6 \times 10^{-25}C $
When two charges are alike, they will repel each other. In this case, we will assume that proton will repel other protons from the coins.
Taking into considerations
1. In a copper coin, we have $n = 3 \times 10^{22} $ atoms,
2. A neutral copper atom contains $29$ protons and $29$ electrons.
3. The distance between two copper coins is $r = 1.0 m$
We find the total charges of protons in the coins first,
$q_{total} = (3 \times 10^{22}) (1.6 \times 10^{-25}C)(29)$
$q_{total} = 0.1392 C$
Now, we can find the electric force between the two coins
$F = \frac{kq^2}{r^2} = \frac{(8.99 \times 10^{9} )(0.1392 C)^2}{(1.0 m)^2}$
$ F = 1.74 \times 10^8 N$
