Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre.

Q: Why don't we check the equilibrium of the center charge?

By symmetry, the forces from the four corners on the center charge cancel each other out perfectly, regardless of the value of $q$. It is always in equilibrium.

Q: Why did we resolve forces along the x-axis instead of the diagonal?

You can resolve along the diagonal too! The math works out exactly the same. Resolving along the edge (x-axis) is just one valid way to set up the equilibrium equation.

Q: How do we know the sign of $q$?

The corner charges are all $-Q$, so they repel each other. To keep them from flying apart, the center charge must pull them inwards. Therefore, $q$ must be positive.

Q: What is the distance from the center to a corner?

The diagonal of a square with side $a$ is $\sqrt{2}a$. The center is exactly halfway along the diagonal, so the distance is $\frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.

Q: Why is the force $F_{CA}$ multiplied by $\cos 45^\circ$?

$F_{CA}$ acts along the diagonal, which makes a $45^\circ$ angle with the edges (x and y axes). To find its contribution along the x-axis, we take the cosine component.

Question

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

Options:

(A) $-\frac{Q}{4}(1+2\sqrt{2})$

(B) $\frac{Q}{4}(1+2\sqrt{2})$

(C) $-\frac{Q}{2}(1+2\sqrt{2})$

(D) $\frac{Q}{2}(1+2\sqrt{2})$

Sigma Percentile · Accepted Answer

### The Core Concept: System Equilibrium

When a problem states that an entire system of charges is in equilibrium, it means the net electrostatic force acting on **every single charge** in the system must be exactly zero.

In our square setup, the central charge $q$ is surrounded symmetrically by four identical $-Q$ charges. Because of this perfect symmetry, the forces from the corners cancel each other out, meaning the central charge is naturally in equilibrium regardless of its magnitude.

Therefore, the real challenge is to ensure that the **corner charges** are also in equilibrium. Let's pick the charge at corner $C$ and analyze the forces acting on it.

### Visualizing the Forces

Imagine you are standing at corner $C$ with the $-Q$ charge. What forces do you feel?

1. **Repulsion from adjacent corners:** The $-Q$ charges at corners $B$ and $D$ push you away along the edges of the square with forces $\vec{F}_{CB}$ and $\vec{F}_{CD}$.
2. **Repulsion from the opposite corner:** The $-Q$ charge at corner $A$ pushes you away along the diagonal with force $\vec{F}_{CA}$.
3. **Attraction from the center:** To prevent you from flying away due to all this repulsion, the central charge $q$ must pull you inwards along the diagonal with an attractive force $\vec{F}_{CO}$. This immediately tells us that $q$ must be positive!

### The Master Equation: Resolving Forces

To mathematically enforce equilibrium at corner $C$, we can resolve all these force vectors along the $x$-axis (the edge $CD$). For the net force to be zero, the sum of the $x$-components must vanish:

$$F_x = F_{CD} + F_{CA} \cos 45^\circ + F_{CO} \cos 45^\circ = 0$$

Notice that $\vec{F}_{CD}$ lies entirely on the $x$-axis, while $\vec{F}_{CA}$ and $\vec{F}_{CO}$ lie on the diagonal, making a $45^\circ$ angle with the $x$-axis.

### Calculating the Magnitudes

Let's use Coulomb's Law to find the raw magnitudes of these forces. Let the side of the square be $a$.

The repulsive force from the adjacent corner $D$ is:
$$F_{CD} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{a^2}$$

The repulsive force from the opposite corner $A$ (distance is the diagonal $\sqrt{2}a$) is:
$$F_{CA} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{2a^2}$$

The attractive force from the center $O$ (distance is half the diagonal $\frac{a}{\sqrt{2}}$) is:
$$F_{CO} = \frac{1}{4\pi\varepsilon_0} \frac{(-Q)(q)}{(a/\sqrt{2})^2} = -\frac{1}{4\pi\varepsilon_0} \frac{2Qq}{a^2}$$
*(The negative sign naturally accounts for the attractive direction pointing opposite to the repulsive forces).*

### The Final Calculation

Now, we substitute these raw magnitudes back into our master equation:

$$\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{a^2} + \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{2a^2} \frac{1}{\sqrt{2}} - \frac{1}{4\pi\varepsilon_0} \frac{2Qq}{a^2} \frac{1}{\sqrt{2}} = 0$$

This looks dense, but we can easily simplify it by dividing the entire equation by the common factor $\frac{Q}{4\pi\varepsilon_0 a^2}$:

$$Q + \frac{Q}{2\sqrt{2}} - \sqrt{2}q = 0$$

Now, isolate $q$:

$$\sqrt{2}q = Q \left(1 + \frac{1}{2\sqrt{2}}\right)$$

Take the common denominator on the right side:

$$\sqrt{2}q = Q \left(\frac{2\sqrt{2} + 1}{2\sqrt{2}}\right)$$

Finally, divide by $\sqrt{2}$:

$$q = \frac{Q}{4} (1 + 2\sqrt{2})$$

And there we have it! The elegant balance of electrostatic forces yields our final answer.

Animated Solution for Physics - Electrostatics: Four charges equal to $- Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is

Select Answer:

Visualized Solution

System Equilibrium

Forces on Corner Charge

Resolving Forces

Magnitudes of Repulsive Forces

Force from Center Charge

Substituting into Equation

Final Answer

The Way Forward

The Sigma Insight: Coulomb's Law

The Core Concept

Visualizing the Forces

The Master Equation

Calculating the Magnitudes

The Final Calculation

Similar Questions

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

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 charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q / q$ equals

Three charges $+ Q, q, + Q$ are placed respectively at distance $0, d /2$ and $d$ from the origin on the X-axis. If the net force experienced by $+ Q$ placed at $x = 0$ is zero, then value of $q$ is

Five point charges, each of value $+ q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $- q$ coulomb placed at the centre of the hexagon is ………newton.

List-I

List-II

Three charges $- q_{1}, + q_{2}$ and $- q_{3}$ are placed as shown in the figure. The x-component of the force on $- q_{1}$ is proportional to

Two charges each equal to $q$ , are kept at $x = - a$ and $x = a$ on the x-axis. A particle of mass $m$ and charge $q_{0} = q /2$ is placed at the origin. If charge $q_{0}$ is given, a small displacement ( $y << a$ ) along the y-axis, the net force acting on the particle is proportional to