Question

Two positive ions, each carrying a charge $q$ are separated by a distance $d$. If $F$ is the force of repulsion between the ions, the number of electrons missing from each ion will be ($\varepsilon_0$ = permittivity of free space, $e$ = charge on an electron)

• A. $\frac{4\pi\varepsilon_0 d^2}{e^2}$
• B. $\frac{4\pi\varepsilon_0 Fd}{e^2}$
• C. $\sqrt{\frac{4\pi\varepsilon_0 Fd^2}{e}}$
• D. $\sqrt{\frac{4\pi\varepsilon_0 Fd^2}{e^2}}$

Hint:

Perform a quick dimensional check! Since $n$ is a pure dimensionless number, everything inside the square root must cancel out structurally. Knowing that $F \propto \frac{q^2}{d^2} \rightarrow q^2 \propto Fd^2$, the term inside the radical simplifies to $\frac{q^2}{e^2} = n^2$, confirming that the formula must be under a square root block.

Answer 1

The Correct Option is D

Step 1: Understand the physical picture.
Two identical positive ions sit a distance $d$ apart and repel with force $F$. Each ion is positive because it has lost some electrons. We must count how many electrons, $n$, are missing from each.
Step 2: Relate charge to missing electrons.
Charge is quantised, so losing $n$ electrons leaves a charge $q = n e$ on each ion.
Step 3: Write Coulomb's law.
For two equal charges $q$ separated by $d$, $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{d^2}$.
Step 4: Substitute $q = ne$.
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(ne)^2}{d^2} = \dfrac{n^2 e^2}{4\pi\varepsilon_0 d^2}$.
Step 5: Solve for $n^2$.
Rearranging, $n^2 = \dfrac{4\pi\varepsilon_0 F d^2}{e^2}$.
Step 6: Take the square root.
Hence $n = \sqrt{\dfrac{4\pi\varepsilon_0 F d^2}{e^2}}$, which is option (4). Notice $e^2$ stays inside the root, so the $e$ in the denominator is squared.
\[ \boxed{n = \sqrt{\dfrac{4\pi\varepsilon_0 F d^2}{e^2}}} \]