Four point charges $q_A = 2\,\mu\text{C}$, $q_B = -5\,\mu\text{C}$, $q_C = 2\,\mu\text{C}$, and $q_D = -5\,\mu\text{C}$ are located at the corners of a square ABCD of side $10\,\text{cm}$. What is the force on a charge of $1\,\mu\text{C}$ placed at the centre of the square?

> [!TLDR] > The problem involves calculating the net electrostatic force on a charge placed at the center of a square due to four charges at its corners. By applying Coulomb's Law and the Principle of Superposition, and recognizing the symmetry of the charge distribution, the forces from diagonally opposite charges cancel each other out, resulting in a **net force of zero** on the central charge. When multiple charges exert forces on a single charge, the net force is the vector sum of all individual forces. This is known as the **Principle of Superposition**. Each individual force is calculated using Coulomb's Law, which describes the magnitude and direction of the electrostatic force between two point charges. > [!FORMULA] > The magnitude of the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's Law: > $$F = k \frac{|q_1 q_2|}{r^2}$$ > where $k = 9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2$ is Coulomb's constant. The force is repulsive if the charges have the same sign and attractive if they have opposite signs. The key to solving this problem efficiently lies in understanding the vector nature of forces and recognizing the symmetry of the setup. Let's break down the solution step-by-step: 1. **Visualize the Setup and Determine Geometry** Imagine a square ABCD with side $s = 10\,\text{cm} = 0.1\,\text{m}$. Let the center of the square be point O. A charge $q_0 = 1\,\mu\text{C}$ is placed at O. The charges at the corners are: * $q_A = 2\,\mu\text{C}$ * $q_B = -5\,\mu\text{C}$ * $q_C = 2\,\mu\text{C}$ * $q_D = -5\,\mu\text{C}$ First, we need to find the distance from each corner to the center of the square. The diagonal of the square is $d = s\sqrt{2}$. The distance from a corner to the center, let's call it $r$, is half the diagonal: $r = \frac{d}{2} = \frac{s\sqrt{2}}{2}$ Substituting $s = 0.1\,\text{m}$: $r = \frac{0.1\sqrt{2}}{2} = 0.05\sqrt{2}\,\

Four point charges qA = 2muC, qB = -5muC, qC = 2muC, and qD = -5muC are located at the corners of a square ABC

Exercises · 1.6

q_A = 2\,\mu\text{C}

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★

27% · 18/67 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Coulomb Force Superposition

Coulomb Force Superposition – From Intuition to Precision

Imagine you're in a room with three friends. Each friend can push or pull you. If two friends push you from the same side, you feel a stronger push — the combined effect. If one pushes from the left and another from the right, you feel the net effect, which might be smaller or even zero if they push equally hard.

This is exactly how electric forces work. When multiple charged particles are present, each one exerts its own force on a given charge. The total force that charge feels is simply the vector sum of all the individual forces — as if each other charge were acting alone, completely ignoring the presence of the rest.

That's the core idea: forces add like arrows, not like numbers.

The Precise Statement

\vec{F}_{\text{net on } q_0} = \sum_{i=1}^{n} \vec{F}_{i \to 0} = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_0 q_i}{r_{i0}^2} \, \hat{r}_{i0}

Where:

$q_0$ is the charge you're calculating the force on
$q_i$ are all other charges (excluding $q_0$ itself)
$r_{i0}$ is the distance between $q_i$ and $q_0$
$\hat{r}_{i0}$ is a unit vector pointing from $q_i$ to $q_0$ (or away, depending on sign convention — be consistent)

The key point: Each pair of charges interacts independently. The presence of a third charge does not alter the force between the first two. This is what "superposition" means — the forces simply layer on top of each other.

Why This Matters (and a Common Trap)

Watch out

Never add the magnitudes of forces directly unless all forces are along the same line and in the same direction. Force is a vector — direction matters.

If two forces point in opposite directions, they partially cancel. If they're at right angles, the net force is found using the Pythagorean theorem, not simple addition.

Example: Three charges on a line:

$q_1 = +2\,\mu\text{C}$ at $x=0$
$q_2 = -1\,\mu\text{C}$ at $x=3\,\text{cm}$
$q_0 = +1\,\mu\text{C}$ at $x=1\,\text{cm}$

Step 1: Force from $q_1$ on $q_0$ — both positive, so repulsive. $q_0$ is pushed to the right.

Step 2: Force from $q_2$ on $q_0$ — opposite signs, so attractive. $q_0$ is pulled to the right (toward $q_2$ ).

Step 3: Both forces point right. Now you add magnitudes: $F_{\text{net}} = F_{1\to0} + F_{2\to0}$ .

If $q_2$ were also positive, the force from $q_2$ would push $q_0$ left, and you'd subtract.

The Deeper Reason

Coulomb's law is a linear law — the force is proportional to each charge individually. If you double $q_1$ , the force from $q_1$ doubles, but the force from $q_2$ stays the same. This linearity is what makes superposition possible. It's not a coincidence — it's a fundamental property of electromagnetic interactions at the classical level.

Important

Superposition works because electric forces obey a linear inverse-square law. If the force depended on products of three charges (like $q_0 q_1 q_2$ ), superposition would fail. It doesn't — and that's why we can break down any multi-charge problem into a series of two-charge calculations.

--- …

Why this formula?

Coulomb Force Superposition — Why the Formula Holds

The principle of superposition for Coulomb forces states that the net electrostatic force on a given charge due to a collection of other charges is the vector sum of the individual forces from each charge, as if the others were absent.

The Key Formula

If we have a charge $q_0$ at position $\vec{r}_0$ , and $N$ other point charges $q_1, q_2, \dots, q_N$ at positions $\vec{r}_1, \vec{r}_2, \dots, \vec{r}_N$ , the net force on $q_0$ is:

\vec{F}_{\text{net}} = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{N} \frac{q_0 q_i}{|\vec{r}_0 - \vec{r}_i|^2} \, \hat{r}_{0i}

where $\hat{r}_{0i}$ is the unit vector pointing from $q_i$ to $q_0$ .

Why This Works — The Physical Reasoning

1. Coulomb's Law is a Two-Body Interaction

Coulomb's law describes the force between exactly two point charges. It depends only on:

The product of their charges ( $q_0 q_i$ )
The inverse square of the distance between them
The direction along the line joining them

Crucially, the force between $q_0$ and $q_i$ does not depend on the presence of any other charges $q_j$ .

2. Forces Add as Vectors (Newton's Third Law + Linearity)

Electrostatic forces are real physical forces — they obey Newton's laws. If multiple forces act on the same charge, the net effect is the vector sum of each individual force. This is a fundamental property of forces in classical mechanics.

3. The Electric Field is Linear

A deeper reason: the electric field $\vec{E}$ obeys superposition. Since $\vec{F} = q_0 \vec{E}$ , and $\vec{E}$ from multiple sources adds linearly, the force automatically adds linearly.

The electric field at $\vec{r}_0$ due to $q_i$ is:

\vec{E}_i(\vec{r}_0) = \frac{1}{4\pi\varepsilon_0} \frac{q_i}{|\vec{r}_0 - \vec{r}_i|^2} \, \hat{r}_{0i}

Then:

\vec{F}_{\text{net}} = q_0 \sum_i \vec{E}_i = \sum_i \vec{F}_i

The Crucial Assumption (Why It's Not Trivial)

Superposition holds because Maxwell's equations are linear in the electric field. If the equations were nonlinear (e.g., if the field depended on $E^2$ ), then the force from two charges together would not be the sum of the individual forces.

In electrostatics, the electric field satisfies:

\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}, \quad \nabla \times \vec{E} = 0

Both equations are linear — if $\vec{E}_1$ and $\vec{E}_2$ are solutions, then $\vec{E}_1 + \vec{E}_2$ is also a solution. This linearity is the mathematical reason superposition works.

--- …

Start your 14-day free trial to unlock the full solution →

The problem involves calculating the net electrostatic force on a charge placed at the center of a square due to four charges at its corners. By applying Coulomb's Law and the Principle of Superposition, and recognizing the symmetry of the charge distribution, the forces from diagonally opposite charges cancel each other out, resulting in a net force of zero on the central charge.

When multiple charges exert forces on a single charge, the net force is the vector sum of all individual forces. This is known as the Principle of Superposition. Each individual force is calculated using Coulomb's Law, which describes the magnitude and direction of the electrostatic force between two point charges.

q_1

The key to solving this problem efficiently lies in understanding the vector nature of forces and recognizing the symmetry of the setup.

Let's break down the solution step-by-step:

Visualize the Setup and Determine Geometry

Imagine a square ABCD with side $s = 10\,\text{cm} = 0.1\,\text{m}$ . Let the center of the square be point O. A charge $q_0 = 1\,\mu\text{C}$ is placed at O. The charges at the corners are:
- $q_A = 2\,\mu\text{C}$
- $q_B = -5\,\mu\text{C}$
- $q_C = 2\,\mu\text{C}$
- $q_D = -5\,\mu\text{C}$
First, we need to find the distance from each corner to the center of the square. The diagonal of the square is $d = s\sqrt{2}$ . The distance from a corner to the center, let's call it $r$ , is half the diagonal:

$r = \frac{d}{2} = \frac{s\sqrt{2}}{2}$

Substituting $s = 0.1\,\text{m}$ :

$r = \frac{0.1\sqrt{2}}{2} = 0.05\sqrt{2}\,\text{m}$

It's often easier to work with $r^2$ :

$r^2 = (0.05\sqrt{2})^2 = (0.05)^2 \times 2 = 0.0025 \times 2 = 0.005\,\text{m}^2$ .

All four corner charges are equidistant from the center.
Calculate the Magnitudes of Individual Forces

We will calculate the magnitude of the force exerted by each corner charge on the central charge $q_0$ . Remember to use absolute values for charges in the magnitude calculation.
- Force from $q_A$ on $q_0$ ( $F_A$ ):
  
  $|F_A| = k \frac{|q_A q_0|}{r^2} = (9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2) \frac{|(2 \times 10^{-6}\,\text{C})(1 \times 10^{-6}\,\text{C})|}{0.005\,\text{m}^2}$
  
  $|F_A| = (9 \times 10^9) \frac{2 \times 10^{-12}}{0.005} = (9 \times 10^9) \frac{2 \times 10^{-12}}{5 \times 10^{-3}}$
  
  $|F_A| = \frac{18}{5} \times 10^{(9-12+3)} = 3.6 \times 10^0 = 3.6\,\text{N}$
- Force from $q_B$ on $q_0$ ( $F_B$ ):
  
  $|F_B| = k \frac{|q_B q_0|}{r^2} = (9 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2) \frac{|(-5 \times 10^{-6}\,\text{C})(1 \times 10^{-6}\,\text{C})|}{0.005\,\text{m}^2}$
  
  $|F_B| = (9 \times 10^9) \frac{5 \times 10^{-12}}{0.005} = (9 \times 10^9) \frac{5 \times 10^{-12}}{5 \times 10^{-3}}$
  
  $|F_B| = 9 \times 10^{(9-12+3)} = 9 \times 10^0 = 9\,\text{N}$
- Force from $q_C$ on $q_0$ ( $F_C$ ):
  
  Since $q_C = q_A = 2\,\mu\text{C}$ and the distance $r$ is the same, the magnitude of the force will be identical to $F_A$ :
  
  $|F_C| = |F_A| = 3.6\,\text{N}$
- Force from $q_D$ on $q_0$ ( $F_D$ ):
  
  Since $q_D = q_B = -5\,\mu\text{C}$ and the distance $r$ is the same, the magnitude of the force will be identical to $F_B$ :
  
  $|F_D| = |F_B| = 9\,\text{N}$
Determine the Directions of Individual Forces

The central charge $q_0 = 1\,\mu\text{C}$ is positive.

Let's assume the corners are labeled counter-clockwise starting from top-right: A (top-right), B (top-left), C (bottom-left), D (bottom-right). The center is O.
- Force $\vec{F}_A$ (from $q_A$ on $q_0$ ): $q_A$ is positive, $q_0$ is positive. The force is repulsive. This means $\vec{F}_A$ points away from $q_A$ , along the diagonal from A through O, towards corner C.
  
  (i.e., $\vec{F}_A$ points from O towards C).
- Force $\vec{F}_C$ (from $q_C$ on $q_0$ ): $q_C$ is positive, $q_0$ is positive. The force is repulsive. This means $\vec{F}_C$ points away from $q_C$ , along the diagonal from C through O, towards corner A.
  
  (i.e., $\vec{F}_C$ points from O towards A).
- Force $\vec{F}_B$ (from $q_B$ on $q_0$ ): $q_B$ is negative, $q_0$ is positive. The force is attractive. This means $\vec{F}_B$ points towards $q_B$ , along the diagonal from O towards corner B.
  
  (i.e., $\vec{F}_B$ points from O towards B). …

Start your 14-day free trial to unlock the full solution →

Here are the most common mistakes students make when solving this classic Coulomb force superposition problem, along with how to avoid each.

1. Forgetting the Vector Nature of Force

The Mistake:

Students often compute the magnitude of the force from each $q$ on $Q$ correctly, but then simply add them as scalars (e.g., $F_{\text{net}} = F_1 + F_2 + F_3$ ).

Why it’s wrong:

Coulomb force is a vector. Forces from different charges point in different directions. Adding magnitudes directly ignores direction and gives an incorrect (usually larger) result.

How to Avoid:

Always draw a clear diagram showing the direction of each force vector. Use vector addition (component method or symmetry) — never scalar addition.

2. Not Using Symmetry to Simplify

The Mistake:

Students calculate all three force vectors explicitly, resolve into components, and sum — a long, error-prone process.

Why it’s wrong:

It wastes time and increases the chance of algebraic mistakes. The problem has perfect symmetry.

How to Avoid:

Recognize that the three charges are identical and placed at vertices of an equilateral triangle. The centroid is equidistant from all vertices. By symmetry, the three force vectors are equal in magnitude and spaced $120^\circ$ apart. Their vector sum is zero.

Key result: The net force on $Q$ at the centroid is $\mathbf{F}_{\text{net}} = 0$ .

3. Incorrect Distance Calculation

The Mistake:

Using $l$ (side length) as the distance between a vertex charge and the centroid.

Why it’s wrong:

The distance from a vertex to the centroid of an equilateral triangle is not $l$ . It is $\frac{l}{\sqrt{3}}$ .

How to Avoid:

Memorize or derive:

Centroid divides the median in ratio $2:1$ .
Median length $= \frac{\sqrt{3}}{2} l$ .
Distance from vertex to centroid $= \frac{2}{3} \times \text{median} = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} l = \frac{l}{\sqrt{3}}$ .

Use $r = \frac{l}{\sqrt{3}}$ in Coulomb’s law.

4. Sign Confusion in Force Direction

The Mistake:

If $Q$ and $q$ have the same sign, students sometimes draw forces as attractive.

Why it’s wrong:

Like charges repel. All three forces on $Q$ are repulsive and point radially outward from each vertex.

How to Avoid:

Always check: same sign → repulsion (force away from the other charge). Opposite sign → attraction (force toward the other charge). Draw arrows accordingly.

--- …

Start your 14-day free trial to unlock the full solution →

←Previous question1.5 — When a glass rod is rubbed with a silk cloth, charges appear on both. …→Next question1.7 — (a) An electrostatic field line is a continuous curve. That is, a fiel…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days