Question

It is given that; two free positive charges are kept at a distance of $1$ apart. The charges are $4q$ and $q$ . Calculate the amount of $Q$ (charge) required to attain equilibrium for the entire system, also mention where the charge $Q$ should be kept from charge $q$ .
(a) $Q = \dfrac{4}{9}q,\;Q\;is\; - ve\;at\;\dfrac{1}{3}$
(b) $Q = \dfrac{4}{9}q,\;Q\;is\; + ve\;at\;\dfrac{1}{3}$
(c) $Q = q,\;Q\;is\; + ve\;at\;\dfrac{1}{3}$
(d) $Q = q,\;Q\;is\; - ve\;at\;\dfrac{1}{3}$

Answer 1

Hint : We must remember that between charges kept at a certain distance, there exists a non-contact force, that force can be called the Coulomb force. The formula for this Coulomb’s law is;
$F = \dfrac{{k \times Q1 \times Q2}}{{{d^2}}}$ ; here the notations stand for: $F$ - Coulomb’s force, $Q1,Q2$ - charges on objects, $k$ - proportionality constant, $d$ - is distance between charges
Also we should understand that for any charge $Q$ to make an equilibrium system, it means that the force present between each of the end charges and $Q$ will be equal to each other.

Complete Step By Step Answer:
Let us first note down what’s given to us;
The charges kept are: $q$ and $4q$
The distance between the given charges is: $1$ unit
Let us assume that the charge $Q$ is kept at a distance of $d$ from charge $q$
Then the charge $Q$ is kept at a distance of $(1 - d)$ from charge $4q$
According to Coulomb’s law, the force will be defined as;
$F = \dfrac{{k \times q1 \times q2}}{{{d^2}}}$
It is given that the systems are at equilibrium, so that means that the force present between charges $q$ and $Q$ must be equal to the force present between charges $4q$ and $Q$ .
So; $$ $\Rightarrow \dfrac{{k \times q \times Q}}{{{d^2}}} = \dfrac{{k \times 4q \times Q}}{{{{(1 - d)}^2}}}$
Solving the equation we get;
${(1 - d)^2} = 4{d^2}$ , now taking square roots we get,
$\Rightarrow 1 - d = 2d \\\ \Rightarrow 1 = 3d \\\ \Rightarrow d = \dfrac{1}{3} \\\$
$\therefore$ Distance between each charge and $Q$ will be $\dfrac{1}{3}$ .
Let us move on to find the value of $Q$ , by applying the equilibrium on $+ q$ condition;
$\Rightarrow \dfrac{{k \times q \times 4q}}{{{1^2}}} = \dfrac{{k \times q \times Q}}{{{{(\dfrac{1}{3})}^2}}} \\\ \Rightarrow 4q \times {(\dfrac{1}{3})^2} = Q \times {1^2} \\\ \Rightarrow Q = \dfrac{{4q \times {1^2}}}{{9 \times {1^2}}} \\\ \therefore Q = \dfrac{{4q}}{9} \\\$
Also, the charge $Q$ will take the negative sign in order to attain equilibrium, that is $Q = - \dfrac{{4q}}{9}$ and the distance between the charges is $d = \dfrac{1}{3}$ .
So the final answer is option (a) $Q = \dfrac{4}{9}q,\;Q\;is\; - ve\;at\;\dfrac{1}{3}$ .

Note :
Another name for Coulomb's force is Inverse square law. For instance, let us consider the disappearance of the coulomb’s force, what can happen then? If the coulomb force happens to disappear, then when the force holding the charges apart will be destroyed. This can cause the energy barrier to be removed. If this happens in reality, then elements will not be kept apart, they will fuse and it will lead to a violent release of energy.