In chemistry and physics courses we learn that atoms are electrically neutral because the number of electrons equals the number of protons (atomic number \(Z\)) in an atom. In addition, we know that the charges of the proton and electron are equal in magnitude. As a result, macroscopic objects do not experience considerable electrostatic interactions. Have you ever considered what would happen if the magnitudes of the charges of the electron and proton were slightly different? Suppose that the charge of the proton differs in \(1\%\) from the charge of the electron. That is, \[Q(\mbox{proton})=|Q(\mbox{electron})|(1+1/100).\] Therefore the net electric charge on an atom with atomic number \(Z\) is \(Q_{Net}=Q(\mbox{proton})-|Q(\mbox{electron})|=0.01Z|Q(\mbox{electron})|\).

In this case, what would be the magnitude of the force of interaction **in Newtons** between two small 1-gram copper balls separated by one meter? There are \(N=9.5 \times 10^{21} \) atoms in one gram of copper and its atomic number is \(Z=29\). Hint: the answer's big... real big.

**Details and assumptions**

- \(|Q(\text{electron})|=e=1.6 \times 10^{-19}~\mbox{C}\)
- \(k=9\times 10^{9}~\mbox{N m}^{2}/\mbox{C}^{2} \)
- Since the balls are small and spherical you may treat them as point charges and apply Coulomb's law.

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