A system has two charges $q_A = 2.5 \times 10^{-7}\,\text{C}$ and $q_B = -2.5 \times 10^{-7}\,\text{C}$ located at points A: $(0, 0, -15\,\text{cm})$ and B: $(0, 0, +15\,\text{cm})$, respectively. What are the total charge and electric dipole moment of the system?

> [!TLDR] > The total charge of the system is the algebraic sum of the individual charges, which is zero. The electric dipole moment is a vector quantity pointing from the negative charge to the positive charge, with a magnitude equal to the product of the charge magnitude and the separation distance; for this system, it is $\boxed{-7.5 \times 10^{-8}\hat{k}\,\text{C}\cdot\text{m}}$. When dealing with a system of charges, two fundamental properties are often of interest: the total charge and the electric dipole moment. The total charge tells us about the net amount of charge present, while the electric dipole moment provides insight into how the charges are distributed, particularly when there's a separation between positive and negative charges. ### Understanding the Concepts 1. **Total Charge:** This is the simplest property. Charge is a scalar quantity, meaning it only has magnitude. The total charge of a system is simply the algebraic sum of all individual charges within that system. If charges are positive and negative, they can cancel each other out. 2. **Electric Dipole Moment:** An electric dipole consists of two equal and opposite charges, $+q$ and $-q$, separated by a small distance. While the total charge of a dipole is zero, it still creates an electric field in its vicinity due to the separation of charges. The electric dipole moment, denoted by $\vec{p}$, is a vector quantity that quantifies this separation and the strength of the dipole. Its direction is conventionally defined from the negative charge to the positive charge. > [!FORMULA] > For a simple electric dipole consisting of charges $+q$ and $-q$ separated by a displacement vector $\vec{d}$ (from the negative charge to the positive charge), the electric dipole moment is given by: > $$\vec{p} = q\vec{d}$$ > For a general system of $N$ point charges $q_i$ located at position vectors $\vec{r}_i$, the electric dipole moment is defined as: > $$\vec{p} = \sum_{i=1}^{N} q_i

A system has two charges qA = 2.5 × 10⁻⁷C and qB = -2.5 × 10⁻⁷C located at points A: (0, 0, -15cm) and B: (0,

Exercises · 1.9

q_A = 2.5 \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.

--- …

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 total charge of the system is the algebraic sum of the individual charges, which is zero. The electric dipole moment is a vector quantity pointing from the negative charge to the positive charge, with a magnitude equal to the product of the charge magnitude and the separation distance; for this system, it is $\boxed{-7.5 \times 10^{-8}\hat{k}\,\text{C}\cdot\text{m}}$ .

When dealing with a system of charges, two fundamental properties are often of interest: the total charge and the electric dipole moment. The total charge tells us about the net amount of charge present, while the electric dipole moment provides insight into how the charges are distributed, particularly when there's a separation between positive and negative charges.

Understanding the Concepts

Total Charge: This is the simplest property. Charge is a scalar quantity, meaning it only has magnitude. The total charge of a system is simply the algebraic sum of all individual charges within that system. If charges are positive and negative, they can cancel each other out.
Electric Dipole Moment: An electric dipole consists of two equal and opposite charges, $+q$ and $-q$ , separated by a small distance. While the total charge of a dipole is zero, it still creates an electric field in its vicinity due to the separation of charges. The electric dipole moment, denoted by $\vec{p}$ , is a vector quantity that quantifies this separation and the strength of the dipole. Its direction is conventionally defined from the negative charge to the positive charge.
$+q$

Let's apply these concepts to the given system.

Step-by-Step Solution

First, let's list the given information and convert units to SI:

Charge $q_A = 2.5 \times 10^{-7}\,\text{C}$ at point A: $(0, 0, -15\,\text{cm}) = (0, 0, -0.15\,\text{m})$
Charge $q_B = -2.5 \times 10^{-7}\,\text{C}$ at point B: $(0, 0, +15\,\text{cm}) = (0, 0, +0.15\,\text{m})$

1. Calculate the Total Charge of the System

The total charge ( $Q_{total}$ ) is the algebraic sum of all charges in the system.

Q_{total} = q_A + q_B

Substitute the given values:

Q_{total} = (2.5 \times 10^{-7}\,\text{C}) + (-2.5 \times 10^{-7}\,\text{C})

Q_{total} = 0\,\text{C}

The system has a net charge of zero. This is characteristic of an electric dipole.

2. Calculate the Electric Dipole Moment of the System

This system consists of two equal and opposite charges, which perfectly fits the definition of an electric dipole. We can use the formula $\vec{p} = q\vec{d}$ .

Identify the magnitude of the charge ( $q$ ):

The magnitude of either charge is $q = |q_A| = |q_B| = 2.5 \times 10^{-7}\,\text{C}$ .
Determine the displacement vector ( $\vec{d}$ ):

The vector $\vec{d}$ points from the negative charge to the positive charge.

The negative charge is $q_B$ at $\vec{r}_B = (0, 0, 0.15\,\text{m})$ .

The positive charge is $q_A$ at $\vec{r}_A = (0, 0, -0.15\,\text{m})$ .

Therefore, $\vec{d} = \vec{r}_A - \vec{r}_B$ .

\vec{d} = (0\hat{i} + 0\hat{j} - 0.15\hat{k})\,\text{m} - (0\hat{i} + 0\hat{j} + 0.15\hat{k})\,\text{m}

\vec{d} = (0 - 0)\hat{i} + (0 - 0)\hat{j} + (-0.15 - 0.15)\hat{k}\,\text{m}

$$\vec{d} = -0.30\hat{k}\,\text{m}$$ …

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Here are the common mistakes students make on this exact problem, along with how to avoid each one.

1. Forgetting the Vector Nature of Dipole Moment

The Mistake:

Students often treat the dipole moment as a scalar and simply multiply the charge magnitude by the separation distance, writing:

p = q \times d = (2.5 \times 10^{-7}) \times (30 \times 10^{-2})

This gives a number but ignores direction.

How to Avoid:

Always remember: dipole moment is a vector from negative to positive charge.

\vec{p} = q \times \vec{d}

where $\vec{d}$ points from $-q$ to $+q$ . Here:

$q_B$ is negative (at $+15$ cm)
$q_A$ is positive (at $-15$ cm)

So $\vec{d}$ points from B to A, i.e., along the negative z-direction.

Correct result:

\vec{p} = (2.5 \times 10^{-7}) \times (0.30\,\text{m}) \times (-\hat{k}) = -7.5 \times 10^{-8}\,\text{C m}\,\hat{k}

2. Using Separation Distance in cm Instead of Metres

The Mistake:

Plugging $d = 30$ cm directly into the formula without converting to metres.

p = (2.5 \times 10^{-7}) \times 30 = 7.5 \times 10^{-6}

This is wrong by a factor of 100.

How to Avoid:

Always convert cm → m before calculation:

15\,\text{cm} = 0.15\,\text{m}

So separation $d = 0.15 + 0.15 = 0.30\,\text{m}$

Correct magnitude:

p = (2.5 \times 10^{-7}) \times 0.30 = 7.5 \times 10^{-8}\,\text{C m}

3. Confusing Total Charge with Net Charge

The Mistake:

Some students add magnitudes: $2.5 \times 10^{-7} + 2.5 \times 10^{-7} = 5.0 \times 10^{-7}\,\text{C}$ and call it "total charge."

How to Avoid:

"Total charge" means algebraic sum including signs:

q_{\text{total}} = q_A + q_B = (+2.5 \times 10^{-7}) + (-2.5 \times 10^{-7}) = 0

Key insight: This is an electric dipole — equal and opposite charges → net charge is zero.

4. Misidentifying Which Charge is Positive/Negative

The Mistake:

Assuming the charge at the positive coordinate ( $+15$ cm) is positive, and vice versa.

How to Avoid:

Read the given data carefully:

$q_A = +2.5 \times 10^{-7}\,\text{C}$ at $(0,0,-15\,\text{cm})$
$q_B = -2.5 \times 10^{-7}\,\text{C}$ at $(0,0,+15\,\text{cm})$ …

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Q.A system has two charges $q_A = 2.5 \times 10^{-7}\,\text{C}$ and $q_B = -2.5 \times 10^{-7}\,\text{C}$ located at points A: $(0, 0, -15\,\text{cm})$ and B: $(0, 0, +15\,\text{cm})$ , respectively. What are the total charge and electric dipole moment of the system?

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)

Understanding the Concepts

Step-by-Step Solution

1. Calculate the Total Charge of the System

2. Calculate the Electric Dipole Moment of the System

Method: Dipole Moment Calculation Using Superposition

Steps

1. Forgetting the Vector Nature of Dipole Moment

2. Using Separation Distance in cm Instead of Metres

3. Confusing Total Charge with Net Charge

4. Misidentifying Which Charge is Positive/Negative

Understanding the Concepts

Step-by-Step Solution

1. Calculate the Total Charge of the System

2. Calculate the Electric Dipole Moment of the System

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