A charge q is placed at the centre of the line joining two equal charges Q. The system of the three

Q: Why do we only check the equilibrium of $Q$ and not $q$?

Because $q$ is placed exactly at the midpoint between two identical charges $Q$. The forces exerted by the two $Q$ charges on $q$ are equal in magnitude and opposite in direction, canceling each other out. Thus, $q$ is always in equilibrium regardless of its value.

Q: Why must $q$ be negative?

The two $Q$ charges have the same sign, so they repel each other. If $q$ were also of the same sign, it would repel the $Q$ charges, pushing them further apart. To keep the system stable, $q$ must pull the $Q$ charges inward, which requires it to have the opposite sign (i.e., negative).

Q: What if $q$ was not placed at the center?

If $q$ were not at the center, the forces from the two $Q$ charges on $q$ would not balance out, and $q$ itself would not be in equilibrium. The entire system would accelerate and fall apart.

Q: Does the magnitude of $Q$ matter for the ratio?

No, the magnitude of $Q$ determines the absolute value of $q$, but the ratio $q/Q$ is always $-1/4$ for this specific symmetric setup. The $Q$ term cancels out during the calculation.

Q: Can this system be in stable equilibrium?

Actually, Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. While the net force is zero (equilibrium), it is an unstable equilibrium. A slight nudge will cause the system to collapse.

Question

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to

Options:

(A) $-\frac{Q}{2}$

(B) $-\frac{Q}{4}$

(C) $+\frac{Q}{4}$

(D) $+\frac{Q}{2}$

Sigma Percentile · Accepted Answer

### The Art of Equilibrium

Imagine a delicate dance of invisible forces. We have a straight line, and at its ends, we place two identical charges, $Q$. They are separated by a distance $r$.

Right in the middle, at a distance of $r/2$ from each, we drop a third charge, $q$. The problem states a beautiful condition: the entire system is in equilibrium.

This means the net electrostatic force acting on every single charge must be exactly zero. No one is moving. It is a state of perfect electrostatic harmony.

### Analyzing the Middle Charge

Let's first look at the middle charge, $q$. Because it sits exactly halfway between two identical charges $Q$, the forces from the left and the right will always be equal in magnitude and opposite in direction.

They cancel out perfectly. So, $q$ is naturally in equilibrium, no matter what its value or sign is. The real challenge lies at the ends.

### The Struggle at the Ends

For the whole system to be stable, the charges on the ends must also be in equilibrium. Let's focus on the charge $Q$ at the right end.

The charge $Q$ at the left end is pushing it away with a repulsive force, $F_{QQ}$. If left alone, it would fly off to infinity!

To stop it from escaping, the middle charge $q$ must pull it back. This means $q$ must attract $Q$. Since opposite charges attract, the sign of $q$ must be opposite to the sign of $Q$. If $Q$ is positive, $q$ must be negative. We must keep this crucial minus sign in mind.

### The Master Equation

For perfect balance, the magnitude of the repulsive force from the far charge must exactly equal the magnitude of the attractive force from the middle charge.

$$|F_{QQ}| = |F_{Qq}|$$

Using Coulomb's law, we can write the expressions for these forces. The force between the two $Q$ charges separated by distance $r$ is equated to the force between $Q$ and $q$ separated by distance $r/2$.

$$\frac{1}{4\pi\varepsilon_0} \frac{Q \cdot Q}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q \cdot |q|}{(r/2)^2}$$

### The Final Calculation

Now, the magic of algebra takes over. We can cancel out the common terms. The constant $\frac{1}{4\pi\varepsilon_0}$ disappears. One $Q$ cancels out from both sides.

$$\frac{Q}{r^2} = \frac{|q|}{r^2 / 4}$$

$$\frac{Q}{r^2} = \frac{4|q|}{r^2}$$

The $r^2$ in the denominator also cancels out. We are left with a very simple relation.

$$|q| = \frac{Q}{4}$$

Solving this, we get the magnitude of $q$ as $Q/4$. But remember our earlier deduction? The charge $q$ must be negative to provide the necessary attractive force.

$$q = -\frac{Q}{4}$$

And there we have it! The perfect charge to hold the system together is $-\frac{Q}{4}$.

Animated Solution for Physics - Electrostatics: A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of the three charges will be in equilibrium if $q$ is equal to

Select Answer:

Visualized Solution

System Setup

Equilibrium Condition

Equilibrium of $q$

Equilibrium of $Q$

Forces on $Q$

Sign of $q$

Equating Force Magnitudes

Applying Coulomb's Law

Solving for $∣ q ∣$

Final Value of $q$

The Way Forward

The Sigma Insight: Coulomb's Law

The Art of Equilibrium

Analyzing the Middle Charge

The Struggle at the Ends

The Master Equation

The Final Calculation

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