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 then the number of electrons missing from each ion will be (e being the charge on an electron)
a)$\dfrac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{q}^{2}}}$
b)$\dfrac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{e}^{2}}}$
c)$\sqrt{\dfrac{4\pi {{\varepsilon }_{0}}F{{e}^{2}}}{{{d}^{2}}}}$
d)$\sqrt{\dfrac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{e}^{2}}}}$

Answer 1

Use coulomb's law formula. Every ion charge is an integral multiple of an electron. So, $q=ne$ . (where n is the no of electrons). By using both formulae we can get the solution.

Complete step by step answer:
According to Coulomb’s law, we know that:
$F=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}......(1)$
Also, we know that, every ion charge is an integral multiple of an electron, i.e.
$q=ne......(2)$
Now, substitute the value of charge from equation (1) n equation (1), we get:

& F=\dfrac{{{\left( ne \right)}^{2}}}{4\pi {{\varepsilon }_{0}}{{d}^{2}}} \\\ & =\dfrac{{{n}^{2}}{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}......(3) \end{aligned}$$ So, from equation (3) we get the number or electrons is: $n=\sqrt{\dfrac{4\pi {{\varepsilon }_{0}}F{{d}^{2}}}{{{e}^{2}}}}$ **Hence, option (d) is the correct answer.** **Note:** Always remember that charge of an electron is basic of all charges. Every charge is an integral multiple of charge of an electron. We can represent any charge in this world as an integral multiple of charge of an electron. This is the key point to get an answer.