Two charges q and -3q are placed fixed on x-axis separated by distance d. Where should a third charge 2q be pl

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Chhattisgarh CgbseShort· 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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The key idea is the Principle of Superposition of Coulomb Forces.

For the third charge $2q$ to experience no net force, the electrostatic forces exerted on it by $q$ and $-3q$ must be equal in magnitude and opposite in direction.

Identify possible regions: Since $q$ and $-3q$ have opposite signs, the forces on $2q$ will be in opposite directions only if $2q$ is placed outside the region between $q$ and $-3q$ .
Determine specific region: For the forces to cancel, $2q$ must be closer to the charge with the smaller magnitude. Here, $|q| < |-3q|$ , so $2q$ must be placed to the left of $q$ .
Set up coordinates and forces: Let $q$ be at $x=0$ and $-3q$ be at $x=d$ . Let the third charge $2q$ be placed at position $x$ (where $x<0$ ). The force due to $q$ on $2q$ is $F_1 = k \frac{q(2q)}{(-x)^2}$ (repulsive, to the left). The force due to $-3q$ on $2q$ is $F_2 = k \frac{|-3q|(2q)}{(d-x)^2}$ (attractive, to the right).
Equate magnitudes: For zero net force, $|F_1| = |F_2|$ :

k \frac{2q^2}{x^2} = k \frac{6q^2}{(d-x)^2}

\frac{1}{x^2} = \frac{3}{(d-x)^2}

$$(d-x)^2 = 3x^2$$ …

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For the net force on a third charge to be zero, the forces from the other two charges must be equal in magnitude and opposite in direction. This occurs at a point outside the segment connecting the two charges, closer to the charge with the smaller magnitude. The third charge $2q$ should be placed at a distance of $\frac{d(1+\sqrt{3})}{2}$ to the left of the charge $q$ .

When multiple charges exert forces on a single charge, the net force on that charge is the vector sum of all individual forces. This is known as the Principle of Superposition. For a third charge to experience no net force, the individual forces acting on it must perfectly cancel each other out. This means their magnitudes must be equal, and their directions must be opposite.

Let's set up the problem:

We have two fixed charges:

$q_1 = q$
$q_2 = -3q$ They are separated by a distance $d$ . Let's place $q_1$ at the origin ( $x=0$ ) and $q_2$ at $x=d$ on the x-axis. We want to find the position $x$ where a third charge $q_3 = 2q$ experiences zero net force.

Q_A

We need to consider three possible regions along the x-axis where the third charge $q_3$ could be placed:

Region I: To the left of $q_1$ ( $x < 0$ )
Region II: Between $q_1$ and $q_2$ ( $0 < x < d$ )
Region III: To the right of $q_2$ ( $x > d$ )

Let's analyze each region:

1. Analyze the possible regions for zero net force

We need the forces from $q_1$ and $q_2$ on $q_3$ to be equal in magnitude and opposite in direction.

Region II ( $0 < x < d$ ):
- $q_1$ (positive) will repel $q_3$ (positive) to the right.
- $q_2$ (negative) will attract $q_3$ (positive) to the right.
- In this region, both forces act in the same direction (towards the positive x-axis). Therefore, they cannot cancel each other out, and the net force will never be zero. No solution exists in this region.
Region I ( $x < 0$ ) and Region III ( $x > d$ ):
- In these regions, the forces from $q_1$ and $q_2$ on $q_3$ will be in opposite directions. For example, if $q_3$ is to the left of $q_1$ , $q_1$ repels $q_3$ to the left, while $q_2$ attracts $q_3$ to the right. If $q_3$ is to the right of $q_2$ , $q_1$ repels $q_3$ to the right, while $q_2$ attracts $q_3$ to the left. Thus, cancellation is possible in these regions.
Tip
For the forces to cancel, the third charge must be closer to the charge with the smaller magnitude. Here, $|q_1| = q$ and $|q_2| = |-3q| = 3q$ . Since $|q_1| < |q_2|$ , the null point must be closer to $q_1$ .
- In Region I ( $x < 0$ ), $q_3$ is closer to $q_1$ than to $q_2$ . This is consistent with the tip.
- In Region III ( $x > d$ ), $q_3$ is closer to $q_2$ than to $q_1$ . This contradicts the tip, meaning no solution exists in Region III.
Therefore, the only possible region for the null point is Region I ( $x < 0$ ).

2. Set up the force equations for Region I

Let the position of the third charge $q_3 = 2q$ be $x$ , where $x < 0$ .

Force from $q_1$ on $q_3$ ( $F_1$ ):
- Charges $q_1=q$ and $q_3=2q$ are both positive, so the force is repulsive.
- The distance between $q_1$ (at $x=0$ ) and $q_3$ (at $x$ ) is $r_1 = |x| = -x$ (since $x$ is negative).
- The force $F_1$ acts in the negative x-direction (to the left).
- Magnitude: $F_1 = k \frac{|q_1 q_3|}{r_1^2} = k \frac{q(2q)}{(-x)^2} = k \frac{2q^2}{x^2}$
Force from $q_2$ on $q_3$ ( $F_2$ ):
- Charges $q_2=-3q$ and $q_3=2q$ have opposite signs, so the force is attractive.
- The distance between $q_2$ (at $x=d$ ) and $q_3$ (at $x$ ) is $r_2 = |d-x| = d-x$ (since $x < 0$ , $d-x$ is positive).
- The force $F_2$ acts in the positive x-direction (to the right).
- Magnitude: $F_2 = k \frac{|q_2 q_3|}{r_2^2} = k \frac{|-3q|(2q)}{(d-x)^2} = k \frac{6q^2}{(d-x)^2}$

3. Equate the magnitudes of the forces …

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Concept: Superposition of Electrostatic Forces

The net force on a charge due to multiple other charges is the vector sum of the individual Coulomb forces. For equilibrium (zero net force), the forces from $q$ and $-3q$ on $2q$ must cancel each other.

Method: Force Balance with Sign Analysis

Step 1: Understand the force directions

$q$ is positive, $-3q$ is negative.
$2q$ is positive.
Like charges repel, unlike charges attract.

So:

Force from $q$ on $2q$ → repulsive (pushes $2q$ away from $q$ )
Force from $-3q$ on $2q$ → attractive (pulls $2q$ toward $-3q$ )

Step 2: Identify possible regions

Place $q$ at $x=0$ and $-3q$ at $x=d$ .

Region	Force from $q$	Force from $-3q$	Can they cancel?
Left of $q$ ( $x<0$ )	Left (repel)	Left (attract)	No (both same direction)
Between $q$ and $-3q$ ( $0<x<d$ )	Right (repel)	Right (attract)	No (both same direction)
Right of $-3q$ ( $x>d$ )	Right (repel)	Left (attract)	Yes (opposite directions)

Only the region $x > d$ works.

Step 3: Apply Coulomb’s law for equilibrium

Let the third charge be at $x = r$ (with $r > d$ ).

Distance from $q$ (at $x=0$ ) to $2q$ : $r$

Distance from $-3q$ (at $x=d$ ) to $2q$ : $r-d$

Force magnitudes must be equal:

\frac{k \cdot q \cdot 2q}{r^2} = \frac{k \cdot 3q \cdot 2q}{(r-d)^2}

Cancel $k$ , $2q^2$ (both sides):

\frac{1}{r^2} = \frac{3}{(r-d)^2}

Step 4: Solve for $r$

Take square roots (positive, since distances are positive):

\frac{1}{r} = \frac{\sqrt{3}}{r-d}

Cross-multiply:

r-d = \sqrt{3} \, r

-d = (\sqrt{3} - 1) r

r = \frac{-d}{\sqrt{3} - 1}

Since $r$ must be positive, and $\sqrt{3} - 1 > 0$ , the negative sign tells us we took the wrong sign in the square root. Let’s correct:

From $\frac{1}{r^2} = \frac{3}{(r-d)^2}$ , taking square root gives: …

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Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Sign of the Force

The error: Students treat all forces as repulsive. They forget that $-3q$ is negative, so the force between $2q$ and $-3q$ is attractive, not repulsive.

How to avoid:

Always write the sign of each charge explicitly.
Use the rule:
- Like charges → repulsion (force away from each other)
- Unlike charges → attraction (force toward each other)
Draw arrows on a diagram showing direction of each force on $2q$ .

Mistake 2: Placing the Third Charge Only Between the Two Fixed Charges

The error: Students assume the third charge must lie between $q$ and $-3q$ . They don't check the left side or right side regions.

How to avoid:

The net force is zero when the magnitudes of the two forces are equal and their directions are opposite.
On the $x$ $x$ -axis, there are three possible regions:
1. Left of $q$ ( $x < 0$ )
2. Between $q$ and $-3q$ ( $0 < x < d$ )
3. Right of $-3q$ ( $x > d$ )
Test each region systematically. For this problem, the correct location is left of $q$ .

Mistake 3: Using Wrong Distance in Coulomb’s Law

The error: Students take $d$ as the distance from $2q$ to both charges, or use $d$ incorrectly when the third charge is not between them.

How to avoid:

Let the position of $2q$ be at coordinate $x$ (with $q$ at $x=0$ , $-3q$ at $x=d$ ).
Distance to $q$ : $|x|$
Distance to $-3q$ : $|x-d|$
Write Coulomb’s law with these distances:

F_{\text{from }q} = \frac{k \cdot (2q)(q)}{x^2}

F_{\text{from }-3q} = \frac{k \cdot (2q)(-3q)}{(x-d)^2}

Mistake 4: Ignoring the Sign When Equating Forces

The error: Students set magnitudes equal but forget that the direction must also be opposite. This leads to wrong algebraic signs.

How to avoid:

Write the net force equation with proper signs:

\frac{k \cdot 2q \cdot q}{x^2} + \frac{k \cdot 2q \cdot (-3q)}{(x-d)^2} = 0

Simplify:

\frac{1}{x^2} - \frac{3}{(x-d)^2} = 0

Solve:

\frac{1}{x^2} = \frac{3}{(x-d)^2}

|x-d| = \sqrt{3}\,|x|

Mistake 5: Forgetting to Check Which Solution Is Valid

The error: Students solve the equation and get two mathematical solutions, but don't check whether each lies in a physically possible region.

How to avoid:

Solve $|x-d| = \sqrt{3}\,|x|$ $∣ x - d ∣ = 3 ∣ x ∣$ :
- Case 1: $x-d = \sqrt{3}\,x$ → $x = \frac{d}{1-\sqrt{3}}$ (negative, left of $q$ — valid)
- Case 2: $x-d = -\sqrt{3}\,x$ → $x = \frac{d}{1+\sqrt{3}}$ (positive, between $q$ and $-3q$ — invalid because forces would be in the same direction there)
Always test the direction of forces for each candidate position.

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Q.Two charges qqq and −3q-3q−3q are placed fixed on xxx-axis separated by distance ddd. Where should a third charge 2q2q2q be placed such that it will not experience any force?

1. Analyze the possible regions for zero net force

2. Set up the force equations for Region I

3. Equate the magnitudes of the forces …

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Q.Two charges $q$ and $-3q$ are placed fixed on $x$ -axis separated by distance $d$ . Where should a third charge $2q$ be placed such that it will not experience any force?

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