Two positive charges $q_2$ and $q_3$ are fixed on the $y$-axis (one on the $+y$ side, one on the $-y$ side, symmetric about the origin O). A charge $q_1$ is fixed on the $x$-axis to the left of O (on the negative $x$-axis). Because of $q_2$ and $q_3$, the net electric force on $q_1$ points in the $+x$ direction (toward O). Now an additional positive charge $Q$ is placed on the positive $x$-axis at the point $(x,0)$, on the far side of O from $q_1$. After $Q$ is added, the force on $q_1$ (a) shall increase along the positive $x$-axis. (b) shall decrease along the positive $x$-axis. (c) shall point along the negative $x$-axis. (d) shall increase but the direction changes because of the interaction of $Q$ with $q_2$ and $q_3$.

Two positive charges q₂ and q₃ are fixed on the y-axis (one on the +y side, one on the -y side, symmetric abou

NCERT Exemplar · Q1

q_2

(a) shall increase along the positive

x

-axis.

(b) shall decrease along the positive

x

-axis.

x

-axis.

(d) shall increase but the direction changes because of the interaction of

Q

with

q_2

and

q_3

Gujarat GsebMCQ· 1mImportance★★★★★

54% · 36/67 Questions

✓ Free question

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.

How to Use It in Exams

Draw all charges and label distances.
For each other charge, sketch the direction of the force on your target charge (like charges repel, opposites attract).
Write the magnitude of each force using Coulomb's law.
Resolve into components if forces aren't along the same line.
Add components separately: $F_{\text{net},x} = \sum F_{i,x}$ , same for $y$ , $z$ .
Combine components to get the net force vector.

Tip

In symmetric arrangements (e.g., an equilateral triangle with equal charges), many components cancel. Always check for symmetry before diving into heavy algebra — it can save you minutes.

One Last Check

If you place a test charge $q_0$ at a point and there are 10 other charges around it, you calculate 10 separate Coulomb forces and add them as vectors. That's it. No extra physics, no hidden interactions. The universe, at this level, is beautifully simple: each pair talks only to each other, and you just listen to all the conversations at once.

"Coulomb's law superposition principle examples" and "electrostatics class 12 physics important questions" are frequently searched, both grounded in the Electrostatics chapter of the NCERT/CBSE Class 12 Physics curriculum. Multi-charge force problems using superposition are a near-guaranteed topic in JEE Main and NEET.

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.

Exam-Relevant Takeaway

Concept	Why It Holds
Superposition of forces	Coulomb force is a two-body interaction; forces add as vectors
Superposition of fields	Maxwell's equations are linear in $\vec{E}$
Net force formula	$\vec{F}_{\text{net}} = \sum \vec{F}_i$ — vector sum of individual Coulomb forces

Never forget: The unit vector $\hat{r}_{0i}$ points from the source charge to the test charge — this determines the correct direction of each term.

Quick Example (To Cement the "Why")

Suppose $q_0 = +1\,\mu\text{C}$ at the origin, $q_1 = +2\,\mu\text{C}$ at $(1,0)$ , $q_2 = -2\,\mu\text{C}$ at $(0,1)$ .

Force from $q_1$ : repulsive, along $+x$ direction
Force from $q_2$ : attractive, along $+y$ direction

The net force is not just the sum of magnitudes — it's the vector sum:

\vec{F}_{\text{net}} = F_1 \hat{x} + F_2 \hat{y}

This works because the two forces are independent — $q_1$ doesn't "know" about $q_2$ , and vice versa. The superposition principle is simply the statement that this independence holds.

The fact that $q_1$ is pushed in the $+x$ direction by the two positive charges tells us $q_1$ is negative (it is attracted toward them). Adding a positive charge $Q$ on the $+x$ side attracts the negative $q_1$ even more strongly toward $+x$ , so the net force simply increases along the positive $x$ -axis.

Concept

The direction of the Coulomb force reveals the sign of $q_1$ . Then superposition tells us how an extra charge changes the total force.

Why this reasoning

$q_2$ and $q_3$ are positive and lie symmetrically on the $y$ -axis. If $q_1$ (on the $-x$ axis) were positive, it would be repelled and pushed toward $-x$ (away from the charges). Instead it is pushed toward $+x$ , so it must be attracted — hence $q_1$ is negative.

Steps

From the given direction of the force, $q_1 < 0$ .
Place a positive charge $Q$ at $(x,0)$ , on the same side ( $+x$ ) that $q_1$ is already being pulled toward.
The force between $Q\,(>0)$ and $q_1\,(<0)$ is attractive, directed from $q_1$ toward $Q$ , i.e. along $+x$ .
By superposition this adds to the pre-existing $+x$ force, so the magnitude increases and the direction stays $+x$ .

Why the other options fail

(b) Wrong — the force grows, it does not decrease. (c) Wrong — nothing reverses the direction to $-x$ . (d) Wrong — $Q$ lies on the $x$ -axis, so its force on $q_1$ is purely along $x$ ; the direction does not change.

✓Final answer

q_1

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GUJCET 2026Set x1 markMCQ
$2R$
›Reveal solutionSolution

Fields of the $+\lambda$ and $-\lambda$ wires add at the midpoint: $E=\lambda/\pi\varepsilon_0 R$ .

Field of an infinite line at distance $r$ : $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$ . The midpoint is at $r=R$ from each wire.

At the midpoint the field of the positive wire points away from it, and the field of the negative wire points toward it — both in the same direction, so they add:
$E = 2\times\frac{\lambda}{2\pi\varepsilon_0 R} = \frac{\lambda}{\pi\varepsilon_0 R}.$
✓Final answer
$\dfrac{\lambda}{\pi\varepsilon_0 R}$

ANSWER: (A)
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge -0.8μC in air is 0.2 N. What is the distance between the two spheres?$
(a) 12 m
(b) 0.12 m
(c) 1.2 m
(d) 0.012 m
›Reveal solutionSolution

Coulomb's law relates the electrostatic force between two point charges to their separation: F = kq1q2/r².

Given q1 = 0.4 μC = 4 × 10⁻⁷ C, q2 = 0.8 μC = 8 × 10⁻⁷ C (magnitudes), F = 0.2 N, k = 9 × 10⁹ N m²/C².

r² = kq1q2/F = (9 × 10⁹)(4 × 10⁻⁷)(8 × 10⁻⁷)/0.2 = (9 × 10⁹)(3.2 × 10⁻¹³)/0.2 = 2.88 × 10⁻³/0.2 = 1.44 × 10⁻².

r = √(1.44 × 10⁻²) = 0.12 m.

✓Final answer
$(b) 0.12 m.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Two identical conducting spheres A and B having charges +q and -q are kept at 'd' distance apart experience coulombian force F between them. If 50% of charge is transferred from sphere B to A then the new coulombian force between them is ___.$
(a) F
(b) F/2
(c) F/4
(d) 2F/3
›Reveal solutionSolution

Coulomb's force is proportional to the product of the two charges; recompute the new charges after the transfer and rescale F accordingly.

Original force: F = k q (q)/d² (magnitude, using |+q| and |−q| = q each).

50% of sphere B's charge (−q) is transferred to A: transferred charge = −q/2.

New charge on B: −q − (−q/2) = −q/2.

New charge on A: q + (−q/2) = q/2.

New force F' = k |q/2| |q/2| / d² = k q²/(4d²) = F/4.

✓Final answer
$(c) F/4.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. As shown in figure charges +q each are placed at the four vertices of a square. Then the coulombian force acting on charge placed at vertex D is ___.$
(a) (√2 + 1/2) kq^2/a^2
(b) (√2 - 1/2) kq^2/a^2
(c) √2 kq^2/a^2
(d) kq^2/2a^2
›Reveal solutionSolution
The net force on a corner charge in a square of equal charges is the vector sum of two equal edge forces (perpendicular to each other) and one diagonal force.

Charge at D experiences:

Force from A (distance a, along DA): magnitude kq²/a²

Force from C (distance a, along DC): magnitude kq²/a², perpendicular to the A-force

Force from B (diagonal, distance a√2): magnitude kq²/(a√2)² = kq²/(2a²), directed along the diagonal DB

The two equal perpendicular edge forces combine (Pythagoras) to give a resultant of magnitude √2 × kq²/a², directed exactly along the diagonal — the same direction as the diagonal force from B.

Total force = √2 kq²/a² + kq²/(2a²) = (√2 + 1/2) kq²/a².

✓Final answer
$(a) (\sqrt2 + 1/2) kq²/a².$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The Coulombian repulsive force between two alpha particles kept at a distance of 3 cm in air is ___ N.$
(a) 1.024 x 10^-27
(b) 1.024 x 10^-25
(c) 1.024 x 10^-24
(d) 1.024 x 10^-23
›Reveal solutionSolution

Alpha charge = 2e = 3.2x10^-19 C; Coulomb's law with r = 0.03 m gives F = 1.024 x 10^-24 N.

Each alpha particle has charge q = 2e = 3.2 x 10^-19 C. Separation r = 3 cm = 0.03 m.

Coulomb force: F = k q^2 / r^2 = (9 x 10^9)(3.2 x 10^-19)^2/(0.03)^2.

(3.2 x 10^-19)^2 = 1.024 x 10^-37; (0.03)^2 = 9 x 10^-4.

F = (9 x 10^9)(1.024 x 10^-37)/(9 x 10^-4) = (1.024 x 10^-37)(10^13) = 1.024 x 10^-24 N.

✓Final answer
$(c) 1.024 x 10^-24 N.$
GUJCET 2021Set 151 markMCQ
$17.7 \times 10^{-22}$
›Reveal solutionSolution

Same-sign charged plates give $E=\sigma/\varepsilon_0$ in the outer region (fields add).

Concept: Each sheet produces $\dfrac{\sigma}{2\varepsilon_0}$ . In the region outside the second plate both fields point the same way and add:
$E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}=\frac{17.7\times10^{-22}}{8.85\times10^{-12}}\approx2\times10^{-10}\ \text{N C}^{-1}.$
✓Final answer
$2\times10^{-10}$

ANSWER: (B)
GUJCET 2020Set 071 markMCQ
$+10^{-8}$
›Reveal solutionSolution
At the centre of a dipole-like pair, both fields point the same way and add.

Concept — superposition of fields. At the midpoint, the field of $+q$ points away from it and the field of $-q$ points toward it — both in the same direction, so they add.

Steps.

Distance from each charge: $r=0.05$ m.

$E_{\text{one}}=\dfrac{kq}{r^2}=\dfrac{9\times10^{9}\times10^{-8}}{(0.05)^2}=\dfrac{90}{0.0025}=3.6\times10^{4}\ \text{NC}^{-1}$ .

Total: $E=2E_{\text{one}}=7.2\times10^{4}\ \text{NC}^{-1}$ .

✓Final answer
$7.2\times10^{4}\,NC^{-1}$

ANSWER: (C)
GUJCET 2019Set 131 markMCQ
$4Q$
›Reveal solutionSolution
Charges redistribute to Q each, and at half the separation the force is F/2.

Concept: When two conductors are joined by a wire the total charge shares equally. Coulomb force $F = \dfrac{k q_1 q_2}{r^2}$ .

Steps:

Initial magnitude: $F = \dfrac{k(4Q)(2Q)}{d^2} = \dfrac{8kQ^2}{d^2}$ .

After connection each sphere has $\dfrac{4Q+(-2Q)}{2} = Q$ .

New separation $d/2$ : $F' = \dfrac{k\,Q\cdot Q}{(d/2)^2} = \dfrac{4kQ^2}{d^2}$ .

Ratio: $\dfrac{F'}{F} = \dfrac{4}{8} = \dfrac{1}{2}$ , so $F' = \dfrac{F}{2}$ .

✓Final answer
$Option (B) — F/2$

ANSWER: (B)
GUJCET 2019Set 131 markMCQ
$1\,\mu\text{C}$
›Reveal solutionSolution
Missing one of six symmetric charges leaves a net field equal to a single charge's field, $10^{-6}K$ .

Concept: By symmetry, six equal charges at the vertices of a regular hexagon produce zero field at the centre (each field cancels its diametric opposite). Removing one charge is equivalent to superposing the full symmetric set (field 0) with a single negative-of-that charge at that vertex, leaving the field of one charge.

Steps:

For a regular hexagon, centre-to-vertex distance = side = 1 m.

Field of one charge: $E = \dfrac{kQ}{r^2} = \dfrac{K(10^{-6})}{1^2} = 10^{-6}K$ N/C.

✓Final answer
$10^{-6}K$

ANSWER: (A)
GUJCET 2015Set C1 markMCQ
$\dfrac{KQq}{r^2}$
›Reveal solutionSolution
[!TLDR] Integrating the point-charge force over the uniformly charged rod gives $F=\dfrac{KQq}{r(r+L)}$ .

Concept

A charge distributed along a line is handled by integration: split it into elements $dq$ , write the Coulomb force $dF = \dfrac{Kq\,dq}{x^2}$ from each element at distance $x$ , and integrate. Here all forces are collinear (rod on the axis), so they add as scalars.

Solution

Linear charge density $\lambda = \dfrac{Q}{L}$ , so $dq = \dfrac{Q}{L}dx$ . The near end of the rod is at distance $r$ , the far end at $r+L$ .

$F = \int_r^{r+L} \dfrac{Kq}{x^2}\cdot\dfrac{Q}{L}\,dx = \dfrac{KqQ}{L}\left[-\dfrac{1}{x}\right]_r^{r+L} = \dfrac{KqQ}{L}\left(\dfrac{1}{r}-\dfrac{1}{r+L}\right).$

$= \dfrac{KqQ}{L}\cdot\dfrac{(r+L)-r}{r(r+L)} = \dfrac{KqQ}{L}\cdot\dfrac{L}{r(r+L)} = \dfrac{KqQ}{r(r+L)}.$

[!ANSWER] (D) $\dfrac{KQq}{r(r+L)}$

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Concept

Why this reasoning

Steps

Why the other options fail

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