A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of th

Q: Why do we only calculate the force on one of the Q charges? What about the other one?

Because of the perfect symmetry of the square setup, the physical situation at one $Q$ corner is identical to the situation at the other $Q$ corner. If the net force is zero on one, it is guaranteed to be zero on the other.

Q: How do we know that the resultant of the two q forces acts exactly along the diagonal?

The two $q$ charges are at equal distances ($d$) from the $Q$ charge, so they exert forces of equal magnitude. When you add two vectors of equal magnitude that are at a 90-degree angle, their resultant perfectly bisects the angle, which aligns exactly with the 45-degree diagonal of the square.

Q: Why does the final answer have a negative sign?

The negative sign indicates that $Q$ and $q$ must have opposite charges. If they had the same sign, all forces would be repulsive, pushing the $Q$ charge away from the square. To achieve equilibrium, the $q$ charges must be attractive to pull the $Q$ charge back in, counteracting the repulsion from the other $Q$ charge.

Q: Why is the distance in the denominator for the Q-Q force $2d^2$ instead of $d^2$?

The distance between the two $Q$ charges is the diagonal of the square. By the Pythagorean theorem, the diagonal length is $\sqrt{d^2 + d^2} = \sqrt{2}d$. Coulomb's law requires the square of the distance, so $(\sqrt{2}d)^2 = 2d^2$.

Q: Can we make the net force zero on ALL four charges at the same time?

No, it is impossible. If you solve for the equilibrium of the $q$ charges, you get a different required ratio. You cannot satisfy both ratios simultaneously. This is a consequence of Earnshaw's theorem, which states that static charges cannot form a stable equilibrium using only electric forces.

Question

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

Options:

(A) $-2\sqrt{2}$

(B) $-1$

(C) $1$

(D) $-\frac{1}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine a microscopic tug-of-war happening at the corners of a perfect square. We have a geometric setup where four charges are placed at the vertices of a square of side length $d$. At two diagonally opposite corners, we place a charge $Q$. At the remaining two corners, we place a charge $q$.

Our mission is to find the exact ratio of $Q/q$ such that the net electrical force acting on the charge $Q$ is perfectly balanced, meaning it experiences zero net force. To solve this, we must rely on Coulomb's Law and the Principle of Superposition.

### The Principle of Superposition

Let's focus our attention on one specific charge $Q$, say the one located at the top-right corner of our square. According to the **Principle of Superposition**, the total force acting on this charge is simply the vector sum of the individual forces exerted by the other three charges in the system.

For this charge $Q$ to be in a state of equilibrium, the vector sum of these three forces must perfectly cancel out to zero:

$$\vec{F}_{	ext{net}} = \vec{F}_A + \vec{F}_C + \vec{F}_D = 0$$

### Resolving the Forces

Let's break down these forces one by one. First, consider the forces exerted by the two adjacent charges, $q$. Assuming for a moment that all charges are positive, the charge $q$ at the top-left corner will push our target $Q$ to the right. Similarly, the charge $q$ at the bottom-right corner will push it upwards.

Because the distance between adjacent corners is simply the side of the square $d$, both of these forces have the exact same magnitude:

$$F_A = F_C = \frac{1}{4\pi\epsilon_0} \frac{Qq}{d^2}$$

Since these two forces are perpendicular to each other (one along the x-axis, one along the y-axis), their resultant vector will point exactly along the diagonal of the square, bisecting the 90-degree angle. Using the Pythagorean theorem, the magnitude of this resultant force $\vec{F}_{AC}$ is $\sqrt{2}$ times the individual force:

$$F_{AC} = \sqrt{2} \frac{1}{4\pi\epsilon_0} \frac{Qq}{d^2}$$

Now, what about the third force? This comes from the other charge $Q$ located at the diagonally opposite corner. The distance between these two $Q$ charges is the diagonal of the square, which is $\sqrt{2}d$. The repulsive force from this charge also acts strictly along the diagonal line:

$$F_D = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(\sqrt{2}d)^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{2d^2}$$

### The Equilibrium Condition

We now have two force vectors acting along the exact same diagonal line: the resultant of the $q$ charges ($\vec{F}_{AC}$) and the direct force from the other $Q$ charge ($\vec{F}_D$). For the net force to be zero, these two vectors must sum to zero.

$$F_{AC} + F_D = 0$$

$$\sqrt{2} \frac{1}{4\pi\epsilon_0} \frac{Qq}{d^2} + \frac{1}{4\pi\epsilon_0} \frac{Q^2}{2d^2} = 0$$

**Here is the critical physical insight:** For this sum to be zero, the two forces must point in opposite directions. Since $Q$ repels $Q$ outwards along the diagonal, the resultant of the $q$ charges must pull $Q$ inwards. This mathematically proves that $q$ must have the **opposite sign** of $Q$!

### The Final Ratio

Let's execute the final algebra. We can divide the entire equation by the common non-zero terms $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2}$:

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

Rearranging the terms to isolate the ratio $Q/q$:

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

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

This elegant result tells us exactly how much stronger (and oppositely charged) the $Q$ charges must be compared to the $q$ charges to maintain equilibrium at the $Q$ corners.

### The Bigger Picture: Earnshaw's Theorem

While we successfully balanced the forces on the $Q$ charges, you might wonder: is the entire square system in equilibrium? If you repeat this exact calculation for the forces acting on one of the $q$ charges, you would find that it requires $q/Q = -2\sqrt{2}$.

These two conditions ($Q/q = -2\sqrt{2}$ and $q/Q = -2\sqrt{2}$) contradict each other! It is mathematically impossible to satisfy both simultaneously unless all charges are zero. This is a beautiful, classic demonstration of **Earnshaw's Theorem**, which states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.

Animated Solution for Physics - Electric Charges and Fields: 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

Select Answer:

Visualized Solution

System Setup

Principle of Superposition

Forces from Adjacent Charges

Resultant of Perpendicular Forces

Force from Diagonally Opposite Charge

Net Force is Zero

Solving for $Q / q$

System Equilibrium

The Sigma Insight: Coulomb's Law

Setting the Stage

The Principle of Superposition

Resolving the Forces

The Equilibrium Condition

The Final Ratio

The Bigger Picture

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