Question

A charge $Q$ is divided into two parts of $q$ and $Q−q$. If the Coulomb repulsion between them when they are separated, is to be maximum, the ratio of $\dfrac{Q}{q}$​ should be:
(A) $2$
(B) $\dfrac{1}{2}$
(C) $4$
(D) $\dfrac{1}{4}$

Answer 1

Hint
Apply the coulombs law of repulsion i.e. $F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$ , $r$ is the distance of separation between the two charges. Substitute the value of charges we get the force of repulsion. And for maximum condition differentiate the required force with respect to $q$ then equate it to zero, we will get the required ratio.

Complete step by step solution
Let us consider the charges Q and Q-q are separated by the distance r then using Coulomb's law of repulsion, we get the required force of repulsion between these charges.
$F = \dfrac{{kQ\left( {Q - q} \right)}}{{{r^2}}}$ ……………… (1)
Now, it is given that the Coulomb's force of repulsion between these charges are maximum, for this condition we differentiate the equation (1) with respect to q because Q and r are constant. And equating it to zero, we get
$ \Rightarrow \dfrac{{dF}}{{dq}} = \dfrac{{k\left( {Q - 2q} \right)}}{{{r^2}}}$
For condition to be maximum, $\dfrac{{dF}}{{dq}} = 0$
$ \Rightarrow \dfrac{{k\left( {Q - 2q} \right)}}{{{r^2}}} = 0$
$ \Rightarrow Q - 2q = 0$
$ \Rightarrow \dfrac{Q}{q} = 2$
Hence, the ratio is equal to $2$.
Therefore, option (A) is correct.

Note
It should be noticed that, the coulomb’s force is directly proportional to the two charges and inversely proportional to the square of distance between them. After removing the proportionality sign, we added the constant $k$ which is known as proportionality constant. It must be remembered that for any quantity to be maximum, differentiate the equation and equate it to zero.