What is the force between two small charged spheres having charges of $2 \times 10^{-7}\,\text{C}$ and $3 \times 10^{-7}\,\text{C}$ placed $30\,\text{cm}$ apart in air?

> [!TLDR] > Two point charges interact via Coulomb's law: the force is proportional to the product of charges and inversely proportional to the square of their separation. Here $F = 6 \times 10^{-3}\,\text{N}$ (repulsive). ## Why Coulomb's Law Works Here When two charged objects are small compared to the distance between them, we can treat them as point charges. The electrostatic force between them follows Coulomb's law, one of the fundamental inverse-square laws in physics. The force is attractive if the charges have opposite signs and repulsive if they have the same sign. The magnitude of this force depends on three things: how much charge each sphere carries, how far apart they are, and the medium between them. In air (or vacuum), the proportionality constant is Coulomb's constant $k = 9 \times 10^9\,\text{N·m}^2/\text{C}^2$. > [!FORMULA] > $$F = k \frac{|q_1 q_2|}{r^2}$$ ## Step-by-Step Calculation **1. Identify the given quantities** We have: - First charge: $q_1 = 2 \times 10^{-7}\,\text{C}$ - Second charge: $q_2 = 3 \times 10^{-7}\,\text{C}$ - Separation: $r = 30\,\text{cm} = 0.30\,\text{m}$ - Medium: air, so $k = 9 \times 10^9\,\text{N·m}^2/\text{C}^2$ > [!WARNING] > Always convert centimeters to meters before substituting into Coulomb's law. The SI unit for distance in this formula is the meter, and mixing units is a common source of error. **2. Substitute into Coulomb's law** $$F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.30)^2}$$ **3. Simplify the numerator** The product of the charges: $$q_1 q_2 = 2 \times 10^{-7} \times 3 \times 10^{-7} = 6 \times 10^{-14}\,\text{C}^2$$ **4. Simplify the denominator** $$r^2 = (0.30)^2 = 0.09\,\text{m}^2$$ **5. Complete the calculation** $$F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09}$$ $$F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{9 \times 10^{-2}}$$ $$F = 9 \times 10^9 \times \frac{6}{9} \times 10^{-14} \times 10^{2}$$ $$F = 6 \times 10^{9 - 14 + 2}$$ $$F =

What is the force between two small charged spheres having charges of 2 × 10⁻⁷C and 3 × 10⁻⁷C placed 30cm apar

Exercises · 1.1

2 \times 10^{-7}\,\text{C}

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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.

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.

Two point charges interact via Coulomb's law: the force is proportional to the product of charges and inversely proportional to the square of their separation. Here $F = 6 \times 10^{-3}\,\text{N}$ (repulsive).

Why Coulomb's Law Works Here

When two charged objects are small compared to the distance between them, we can treat them as point charges. The electrostatic force between them follows Coulomb's law, one of the fundamental inverse-square laws in physics. The force is attractive if the charges have opposite signs and repulsive if they have the same sign.

The magnitude of this force depends on three things: how much charge each sphere carries, how far apart they are, and the medium between them. In air (or vacuum), the proportionality constant is Coulomb's constant $k = 9 \times 10^9\,\text{N·m}^2/\text{C}^2$ .

F = k \frac{|q_1 q_2|}{r^2}

Step-by-Step Calculation

1. Identify the given quantities

We have:

First charge: $q_1 = 2 \times 10^{-7}\,\text{C}$
Second charge: $q_2 = 3 \times 10^{-7}\,\text{C}$
Separation: $r = 30\,\text{cm} = 0.30\,\text{m}$
Medium: air, so $k = 9 \times 10^9\,\text{N·m}^2/\text{C}^2$

Watch out

Always convert centimeters to meters before substituting into Coulomb's law. The SI unit for distance in this formula is the meter, and mixing units is a common source of error.

2. Substitute into Coulomb's law

F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.30)^2}

3. Simplify the numerator

The product of the charges:

q_1 q_2 = 2 \times 10^{-7} \times 3 \times 10^{-7} = 6 \times 10^{-14}\,\text{C}^2

4. Simplify the denominator

r^2 = (0.30)^2 = 0.09\,\text{m}^2

5. Complete the calculation

F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09}

F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{9 \times 10^{-2}}

F = 9 \times 10^9 \times \frac{6}{9} \times 10^{-14} \times 10^{2}

F = 6 \times 10^{9 - 14 + 2}

F = 6 \times 10^{-3}\,\text{N}

6. Interpret the result

Both charges are positive, so the force is repulsive. The magnitude is $6 \times 10^{-3}\,\text{N}$ or $6\,\text{mN}$ .

Tip

When both charges have the same sign (both positive or both negative), the force pushes them apart. When they have opposite signs, the force pulls them together.

✓Final answer

\boxed{6 \times 10^{-3}\,\text{N}}

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.

5. Assuming the Net Force is Non-Zero Without Checking

The Mistake:

After computing magnitudes, students assume the forces don’t cancel and proceed to find a non-zero resultant.

Why it’s wrong:

Symmetry guarantees cancellation. The three equal-magnitude vectors at $120^\circ$ to each other always sum to zero.

How to Avoid:

Before doing heavy algebra, pause and check for symmetry. If the configuration is symmetric and all charges are identical, the net force at the center is zero.

Quick Summary Checklist

Mistake	How to Avoid
Scalar addition of forces	Always use vector addition
Ignoring symmetry	Use symmetry to simplify first
Wrong distance ( $l$ instead of $l/\sqrt{3}$ )	Derive or memorize centroid distance
Wrong force direction (attraction instead of repulsion)	Same sign → repulsion
Assuming net force is non-zero	Check symmetry — here it’s zero

Final takeaway: For this exact problem, the answer is zero — but only if you handle vectors, distances, and directions correctly.

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Q.What is the force between two small charged spheres having charges of $2 \times 10^{-7}\,\text{C}$ and $3 \times 10^{-7}\,\text{C}$ placed $30\,\text{cm}$ apart in air?

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

How to Use It in Exams

One Last Check

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)

Exam-Relevant Takeaway

Quick Example (To Cement the "Why")

Why Coulomb's Law Works Here

Step-by-Step Calculation

Concept: Coulomb Force Superposition

Steps

Final Answer

1. Forgetting the Vector Nature of Force

2. Not Using Symmetry to Simplify

3. Incorrect Distance Calculation

4. Sign Confusion in Force Direction

5. Assuming the Net Force is Non-Zero Without Checking

Quick Summary Checklist

Q.What is the force between two small charged spheres having charges of 2×10−7 C2 \times 10^{-7}\,\text{C}2×10−7C and 3×10−7 C3 \times 10^{-7}\,\text{C}3×10−7C placed 30 cm30\,\text{cm}30cm apart in air?

Why Coulomb's Law Works Here

Step-by-Step Calculation

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Q.What is the force between two small charged spheres having charges of $2 \times 10^{-7}\,\text{C}$ and $3 \times 10^{-7}\,\text{C}$ placed $30\,\text{cm}$ apart in air?

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