Two point charges qA = 3muC and qB = -3muC are located 20cm apart in vacuum. (a) What is the electric field at

Exercises · 1.8

q_A = 3\,\mu\text{C}

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude

1.5 \times 10^{-9}\,\text{C}

is placed at this point, what is the force experienced by the test charge?

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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.

--- …

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 electric field at the midpoint is the vector sum of the fields from each charge. Since the charges are equal in magnitude but opposite in sign, their fields at the midpoint are equal in magnitude and point in the same direction (from positive to negative). The net field is $5.4 \times 10^{6}\,\text{N/C}$ directed from $q_A$ to $q_B$ . The force on a negative test charge is opposite to the field direction, so it is $8.1 \times 10^{-3}\,\text{N}$ directed from $q_B$ to $q_A$ .

Concept and Intuition

The electric field at a point due to multiple charges is the vector sum of the fields each charge would produce alone — this is the principle of superposition. For a point charge, the field magnitude is given by Coulomb's law:

E = \frac{k |q|}{r^2}

where $k = 9 \times 10^{9}\,\text{N·m}^2/\text{C}^2$ in vacuum. The direction of the field is away from a positive charge and toward a negative charge.

Here, the two charges are equal in magnitude but opposite in sign. At the midpoint O, each charge is at the same distance $r = 10\,\text{cm} = 0.1\,\text{m}$ . So the magnitudes of the individual fields are identical. The key is their directions: the field from $q_A$ (positive) points away from $q_A$ , i.e., toward $q_B$ ; the field from $q_B$ (negative) points toward $q_B$ as well. So both fields point in the same direction — from A to B. They add, not cancel.

Watch out

A common mistake is to think opposite charges produce opposite fields at the midpoint. That would be true if both charges had the same sign. Here, because one is positive and the other negative, both fields point the same way.

Step-by-step solution

1. Find the distance from each charge to the midpoint.

The charges are 20 cm apart. The midpoint O is 10 cm from each charge. Convert to SI units:

r = 10\,\text{cm} = 0.1\,\text{m}

2. Compute the magnitude of the electric field due to one charge at O.

Using $k = 9 \times 10^{9}$ and $|q| = 3\,\mu\text{C} = 3 \times 10^{-6}\,\text{C}$ :

E_{\text{single}} = \frac{k |q|}{r^2} = \frac{(9 \times 10^{9})(3 \times 10^{-6})}{(0.1)^2}

First, $(0.1)^2 = 0.01$ . Then:

E_{\text{single}} = \frac{27 \times 10^{3}}{0.01} = 27 \times 10^{5} = 2.7 \times 10^{6}\,\text{N/C}

3. Determine the direction of each field.

For $q_A = +3\,\mu\text{C}$ : field at O points away from $q_A$ , i.e., from A toward B.
For $q_B = -3\,\mu\text{C}$ : field at O points toward $q_B$ , i.e., again from A toward B.

So both fields point in the same direction — along AB from A to B.

4. Apply superposition: add the vectors.

Since they are in the same direction, the net field magnitude is the sum:

E_{\text{net}} = E_A + E_B = 2.7 \times 10^{6} + 2.7 \times 10^{6} = 5.4 \times 10^{6}\,\text{N/C}

Direction: from $q_A$ to $q_B$ (i.e., along AB toward the negative charge). …

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Common Mistakes & How to Avoid Them (Coulomb Force Superposition)

Mistake 1: Treating Electric Field as a Scalar (Ignoring Direction)

The error: Students compute the magnitude of $E$ from each charge and simply add them numerically, forgetting that electric field is a vector.

Example of wrong work:

$E_A = \frac{k q_A}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = 2.7\times10^6\ \text{N/C}$
$E_B = \frac{k q_B}{r^2} = \frac{9\times10^9 \times (-3\times10^{-6})}{(0.1)^2} = -2.7\times10^6\ \text{N/C}$
Then: $E_\text{net} = 2.7\times10^6 + (-2.7\times10^6) = 0$ ✗

Why this is wrong: The negative sign on $q_B$ is charge, not direction. The field direction depends on the sign of the charge and the position relative to it.

How to avoid:

Always draw a diagram showing the direction of each field vector at the point.
For a positive charge, field points away from it.
For a negative charge, field points toward it.
At the midpoint O:
- $q_A$ (positive) → field points away from A, i.e., toward B (rightward).
- $q_B$ (negative) → field points toward B, i.e., also rightward.
So both fields are in the same direction — they add, not cancel.

Correct approach:

E_A = \frac{k|q_A|}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = 2.7\times10^6\ \text{N/C (→)}

E_B = \frac{k|q_B|}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = 2.7\times10^6\ \text{N/C (→)}

E_\text{net} = 2.7\times10^6 + 2.7\times10^6 = \mathbf{5.4\times10^6\ N/C\ (→)}

Mistake 2: Using the Wrong Distance

The error: Using the full separation (20 cm) instead of the distance from each charge to the midpoint (10 cm).

Example: $r = 20\ \text{cm} = 0.2\ \text{m}$ instead of $r = 10\ \text{cm} = 0.1\ \text{m}$ .

Why this is wrong: The field at O depends on the distance from each charge to O, not the distance between the charges.

How to avoid:

Mark the point O clearly on your diagram.
Measure the distance from each charge to O.
Here, O is the midpoint → distance from A to O = distance from B to O = $10\ \text{cm} = 0.1\ \text{m}$ .
Always convert cm to m: $1\ \text{cm} = 10^{-2}\ \text{m}$ .

Mistake 3: Forgetting Unit Conversion

The error: Using $\mu\text{C}$ as $10^{-6}\ \text{C}$ incorrectly, or forgetting to convert cm to m.

Common slip: $20\ \text{cm} = 0.2\ \text{m}$ is correct, but then using $r = 0.2\ \text{m}$ instead of $0.1\ \text{m}$ (see Mistake 2).

How to avoid:

Write all quantities in SI units before plugging into formulas.
$1\ \mu\text{C} = 10^{-6}\ \text{C}$
$1\ \text{cm} = 10^{-2}\ \text{m}$
Use scientific notation consistently.

Mistake 4: Confusing Force on Test Charge with Field Direction

The error: For part (b), students compute force as $F = q_\text{test} \times E$ but get the direction wrong.

Why this is wrong: The force on a negative test charge is opposite to the field direction.

How to avoid:

Remember: $\vec{F} = q\vec{E}$
If $q$ is negative, $\vec{F}$ is opposite to $\vec{E}$ . …

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Q.Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\,\text{cm}$ apart in vacuum.

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)

Concept and Intuition

Step-by-step solution

Method: Superposition Principle for Electric Fields

Step-by-step solution

Common Mistakes & How to Avoid Them (Coulomb Force Superposition)

Mistake 1: Treating Electric Field as a Scalar (Ignoring Direction)

Mistake 2: Using the Wrong Distance

Mistake 3: Forgetting Unit Conversion

Mistake 4: Confusing Force on Test Charge with Field Direction

Q.Two point charges qA=3 μCq_A = 3\,\mu\text{C}qA​=3μC and qB=−3 μCq_B = -3\,\mu\text{C}qB​=−3μC are located 20 cm20\,\text{cm}20cm apart in vacuum.

Concept and Intuition

Step-by-step solution

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Q.Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\,\text{cm}$ apart in vacuum.

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