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 Q/q should be...

Answer 1

Step 1: Given that:

Charge Q is divided into two parts.

First part = q

Second part = Q-q

Step 2: Determining the condition for the maximum force between the charges:

The force of attraction or repulsion between two charges $q_1$ and $q_2$ separated by a distance r is given by Coulomb's law, that is;

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Now, let both the charges; q and (Q-d) are separated by a distance d then, By Coulomb's law the force between them will be;

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q(Q-q)}{r^2}$

$F=\frac{1}{4\pi {\epsilon }_0}\frac{(Qq-q^2)}{r^2}$

Here, Q and d are constant, therefore;

For the maximum value of force, equating the differentiation of F with respect to q to zero, that is;

$\frac{dF}{dq}=0$

Now,

$\frac{d}{dq}\frac{1}{4\pi {\epsilon }_0}\frac{(Qq-q^2)}{r^2}=0$

$As,\frac{d}{dx}{x}^n=n{x}^{n-1}$

$Q-2q=0$

$Q=2q$

$\frac{Q}{q}=2$

Thus,

The ratio $\frac{Q}{q}$ = 2