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Step-by-step solution:
Step 1: State Coulomb’s Law
Coulomb’s law for the force of repulsion (or attraction) between two point charges states:
$F = \dfrac{1}{4\pi \varepsilon_0}\,\dfrac{q_1\,q_2}{d^2}.$
Here, $q_1$ and $q_2$ are the charges on the two ions, $d$ is the distance between them, $\varepsilon_0$ is the permittivity of free space, and $F$ is the electrostatic force.
Step 2: Apply the law to our case of two identical positive ions
Given that both ions carry the same charge $q$ and are separated by a distance $d$, the repulsive force is:
$F = \dfrac{1}{4\pi\,\varepsilon_0}\,\dfrac{q \times q}{d^2} = \dfrac{q^2}{4\pi\,\varepsilon_0\,d^2}.$
Step 3: Solve for the charge $q$
Rearrange to find $q$:
$q^2 = 4\pi\,\varepsilon_0\,F\,d^2.$
Taking the square root:
$q = \sqrt{4\pi\,\varepsilon_0\,F\,d^2}.$
Step 4: Express $q$ in terms of number of missing electrons
If $n$ is the number of electrons missing from each ion, then $q = n\,e$, where $e$ is the magnitude of charge on an electron. Thus:
$n = \dfrac{q}{e}.$
Step 5: Substitute the value of $q$ from Step 3
Substituting $q = \sqrt{4\pi\,\varepsilon_0\,F\,d^2}$ into $n = \dfrac{q}{e}$:
$n = \dfrac{\sqrt{4\pi\,\varepsilon_0\,F\,d^2}}{e}
= \sqrt{\dfrac{4\pi\,\varepsilon_0\,F\,d^2}{e^2}}.$
Hence, the number of electrons missing from each ion is
$\sqrt{\dfrac{4\pi\,\varepsilon_0\,F\,d^2}{e^2}}.$
