A cubic unit cell of cesium chloride (CsCl) has cesium atoms situated at the eight corners of a cube of side $0.40\text{ nm}$, and a chlorine atom situated at the centre of the cube. Each Cs atom is deficient by one electron, so it carries a charge $+e$; the Cl atom carries one excess electron, so it carries a charge $-e$. (i) What is the net electric field at the Cl atom (the cube centre) due to the eight Cs atoms? (ii) Suppose the Cs atom at one corner A is removed. What is the net force on the Cl atom due to the seven remaining Cs atoms?

A cubic unit cell of cesium chloride (CsCl) has cesium atoms situated at the eight corners of a cube of side 0

NCERT Exemplar · Q22

0.40\text{ nm}

(i) What is the net electric field at the Cl atom (the cube centre) due to the eight Cs atoms?

(ii) Suppose the Cs atom at one corner A is removed. What is the net force on the Cl atom due to the seven remaining Cs atoms?

Uttarakhand UbseShort· 3mImportance★★★★★

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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.

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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.

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Because the eight Cs $^+$ charges sit at corners that pair up diametrically across the centre, their fields at the Cl atom cancel and the net field is zero. When one Cs $^+$ is removed, the remaining seven give a net force equal in magnitude and opposite in direction to the force the removed charge alone would have exerted — about $1.92\times10^{-9}\,$ N, pointing away from the missing corner.

(i) Net field at Cl

The eight corners occur in four pairs, each pair lying on a body diagonal on opposite sides of the centre and equidistant from it. The two Cs $^+$ of a pair produce equal and opposite fields at the centre, which cancel. Summing over the four pairs:

\vec{E}_{net} = 0.

(ii) Net force after removing the corner charge at A

Use superposition. With all eight present the force on Cl is zero:

\vec{F}_{8} = \vec{F}_A + \vec{F}_{other\ 7} = 0 \;\Rightarrow\; \vec{F}_{other\ 7} = -\vec{F}_A.

So the seven remaining charges give a force equal and opposite to the force $\vec{F}_A$ that the removed Cs $^+$ at A exerted on Cl. Since Cs $^+$ attracted the Cl $^-$ toward A, $-\vec{F}_A$ points away from A (along the body diagonal, toward the diagonally opposite corner). Equivalently, removing a $+e$ is like adding a $-e$ at A, which repels the Cl $^-$ .

Magnitude. The corner-to-centre distance is half the body diagonal: …

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Concept understanding — Coulomb Force Superposition

Coulomb Force Superposition – From Intuition to Precision

The Precise Statement

Why This Matters (and a Common Trap)

The Deeper Reason

Why this formula?

Coulomb Force Superposition — Why the Formula Holds

The Key Formula

Why This Works — The Physical Reasoning

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

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

3. The Electric Field is Linear

The Crucial Assumption (Why It's Not Trivial)

(i) Net field at Cl

(ii) Net force after removing the corner charge at A

Method: Using Symmetry to Simplify Superposition, Then Removing One Term

Steps

(i) Net field at Cl

(ii) Net force after removing the corner charge at A

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