Question

Two free positive charges $4q$ and $q$ are kept at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?

Answer 1

Hint:Let us first talk about charge. When matter is put in an electromagnetic field, it acquires an electric charge, which causes it to undergo a force. Positive and negative charges are the two forms of electric charge commonly carried by protons and electrons respectively.

Complete step by step answer:
The electrostatic force $F$ between two point charges ${q_1}$ and ${q_2}$ is proportional to the product of their magnitudes and inversely proportional to the square of their distance. Like charges repel one another, while opposite charges draw one another.
$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, $F = $electric force, $k = $Coulomb constant, ${q_{1,}}{q_2} = $Charges and $r = $Distance of separation.

Now, let us solve the problem: Let "$r$" be the distance between the charge $q$ and the equilibrium state of $Q$.

Total Force acting on Charge $q$ and $4q$:
$F = \dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{(l - r)}^2}}}$
For \[Q\] to be in equilibrium, $F$ should be equated to zero.
$\dfrac{{kqQ}}{{{r^2}}} + \dfrac{{k4qQ}}{{{{(l - r)}^2}}} = 0$
$\Rightarrow {(l - r)^2} = 4{r^2}$
$\Rightarrow l - r = 2r$...................[By taking square root both the sides]
$\therefore r = \dfrac{l}{3}$
Let's compare the equilibrium of 4q charge to find out the value of charge:
$k4qQ\dfrac{1}{{{{(l - r)}^2}}} + 4kqq\dfrac{1}{{{l^2}}} = 0$
$\therefore Q = - \dfrac{{4q}}{9}$

Hence, the charge $Q$ needed to achieve equilibrium for the entire system is $-\dfrac{{4q}}{9}$ and it should be placed at a distance of $\dfrac{l}{3}$ from charge $q$.

Note:To determine the direction of the force, you must apply the rule of attraction, which states that opposite charges attract and equal charges repel. Two opposite charges (positive and negative) repel each other, while two positives (or negatives) attract each other.