Animated Solution for Physics - Electric Charges and Fields: If a charge $q$ is placed at the centre of the line joining two equal charges $Q$ such that the system is in equilibrium, then the value of $q$ is

A Tale of Three Charges

Imagine a straight line where two identical charges, both of magnitude $Q$, are firmly placed at points $A$ and $B$. Let the distance between them be $x$. Now, we introduce a third charge, $q$, and place it exactly at the midpoint $C$ of the line segment $AB$. Because it is at the midpoint, the distance from $A$ to $C$ is $\frac{x}{2}$, and the distance from $C$ to $B$ is also $\frac{x}{2}$.

The problem states a very crucial condition: the entire system is in equilibrium. This means that the net electrostatic force acting on every single charge in the system must be exactly zero.

Let's think about the middle charge $q$ first. Because it is placed exactly midway between two identical charges $Q$, the force exerted by the left charge $Q$ will be perfectly balanced by the force exerted by the right charge $Q$, regardless of the magnitude or sign of $q$. So, the middle charge is naturally in equilibrium due to symmetry.

To find the actual value of $q$, we must look at the outer charges. Let's analyze the forces acting on the charge $Q$ located at point $A$. For this charge to be in equilibrium, the net force on it must be zero:

Here, $F_{Qq}$ is the force exerted by the middle charge $q$, and $F_{QQ}$ is the force exerted by the other outer charge $Q$ at point $B$.

Now, we apply Coulomb's Law to express these forces mathematically.

The force between the charge $Q$ at $A$ and the charge $q$ at $C$ is:

The force between the two outer charges $Q$ at $A$ and $B$ is:

Substituting these into our equilibrium equation, we get:

Let's simplify this equation. We can immediately cancel out the common constant $\frac{1}{4\pi {\epsilon }_0}$. We can also divide the entire equation by one $Q$ (assuming $Q eq 0$) and multiply by $x^2$.

Notice that the denominator of the first term is $(\frac{x}{2}{)}^2=\frac{x^2}{4}$. When we bring the $4$ to the numerator, the equation simplifies beautifully to:

Solving for $q$, we find:

The negative sign is physically very significant. It tells us that the middle charge must be of the opposite sign to the outer charges. This makes perfect sense! The two outer charges $Q$ repel each other. To prevent them from flying apart, the middle charge must exert an attractive force, pulling them inwards. Thus, $q$ must be negative, and its magnitude must be exactly one-fourth of $Q$.