Question

What is the smallest electric force between two charges placed at a distance of $1\text{m}$ ?

• A. $2.75\times {10}^{-28}\text{N}$
• B. $2.30\times {10}^{-28}\text{N}$
• C. $2.5\times {10}^{-25}\text{N}$
• D. $1.5\times {10}^{-24}\text{N}$

Answer 1

The correct option is B $2.30\times {10}^{-28}\text{N}$

The minimum possible charge = Charge of an electron, $e=1.6\times {10}^{-19}\text{C}$
Separation between the charge, $r=1\text{m}$
From Coulomb's law, the electric force between charge particles is given by:
$F=\frac{k{q}_1{q}_2}{r^2}$
For $F=F_{\text{min}},q_1=q_2=e$
$\Rightarrow F_{\text{min}}=\frac{ke.e}{r^2}$
or, $F_{\text{min}}=\frac{k{e}^2}{(1{)}^2}=k{e}^2$
($\because k=9\times {10}^9$)
$\Rightarrow F_{\text{min}}=9\times {10}^9\times (1.6\times {10}^{-19}{)}^2$
$\therefore F_{\text{min}}=2.30\times {10}^{-28}\text{N}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: We know charge follows quantization principle.}\\ \text{Thus, by putting n=1, in q=ne we get}\\ \text{minimum charge of body 'e'.}\end{array}$