Physics

Chapter 5: Electric Charges and Fields

Imron Rosyadi

5.1 Electric Charge

Learning Objectives

• Describe the concept of electric charge
• Explain qualitatively the force electric charge creates

The concept of electric charge and the forces it creates are fundamental to understanding electromagnetism. This section will introduce these basic ideas, linking them to everyday experiences and historical discoveries.

Discoveries: Static Electricity

• Common Phenomena:
  • Clothes sticking together from a dryer.
  • Hair reacting to a woolen sweater.
  • A comb attracting a stream of water or small paper strips.
  • Plastic cling wrap sticking to surfaces.
  • Static shock from doorknobs after shuffling feet.

Many of you have probably experienced these phenomena firsthand. These everyday occurrences are all manifestations of static electricity, which involves electric charges and the forces they exert.

Historical Insights

• Thales of Miletus (624–546 BCE):
  • Rubbed amber with fur, observed attraction.
  • Noted that rubbed amber and fur could affect other nonmetallic objects without contact.
• William Gilbert (1544–1603):
  • Studied attractive forces in various substances.
  • Found metals never exhibited this force, minerals did.
  • Observed repulsion between two electrified amber rods, suggesting two types of electric property.

An electrically charged comb attracts a stream of water.

Borneo amber, which gains electrons when rubbed.

These early observations were crucial. Thales’s discovery with amber is where the word “electricity” originates (from “elektron,” the Greek word for amber). Gilbert’s work expanded on this, showing that there were indeed two distinct types of electrical behavior, leading to the concept of two types of electric charge.

Electric Charge

• Definition: An intrinsic property of matter causing electric forces.
• Two Types:
  • Repulsive forces: When two objects have the same type of charge.
  • Attractive forces: When two objects have opposite types of charge.
• SI Unit: Coulomb (C), named after Charles-Augustin de Coulomb.
• Long-range force: Acts without physical contact, similar to gravity.

This distinction between attractive and repulsive forces is fundamental. Unlike contact forces, electric forces can act over a distance. This “action at a distance” was a significant concept in early physics.

Benjamin Franklin’s Contributions

• Leyden Jar:
  • Device for storing large amounts of electric charge.
  • Consisted of a glass jar with inner and outer metal foil sheets.
• “Electrical Fluid” Hypothesis:
  • Proposed one type of charge was motionless, the other flowed.
  • Excess fluid: “positive electricity.”
  • Deficiency of fluid: “negative electricity.”
  • Guess: Moving charges were “negative.” (Later confirmed to be electrons, though his initial guess for their sign was arbitrary.)

A Leyden jar, an early capacitor.

Franklin’s work was pivotal. The Leyden jar allowed for quantitative experiments with larger charges. His “electrical fluid” model, while not entirely accurate in its details, laid the groundwork for our understanding of charge flow. The convention of positive and negative charges is still used today, even though the actual charge carriers in many cases (electrons) are negative.

Properties of Electric Force

• The force acts without physical contact.
• The force can be attractive or repulsive:
  • Same signs: repulsive (electrostatic repulsion).
  • Opposite signs: attractive (electrostatic attraction).
• Not all objects are affected by this force (only charged ones).
• The magnitude of the force decreases rapidly with increasing separation distance (inversely proportional to the square of the distance).

These bullet points summarize the key empirical observations about electric forces. The inverse square law dependence on distance is particularly important and will be quantified by Coulomb’s Law.

Properties of Electric Charge

• Charge is quantized:
  • Comes in discrete amounts.
  • Smallest possible amount: \(e \equiv 1.602 \times 10^{-19} \text{C}\).
  • Any observed charge is an integer multiple of \(e\).
• Magnitude of charge is independent of type:
  • Smallest positive charge: \(+1.602 \times 10^{-19} \text{C}\).
  • Smallest negative charge: \(-1.602 \times 10^{-19} \text{C}\).
  • These magnitudes are exactly equal.
• Charge is conserved:
  • Neither created nor destroyed; only transferred.
  • “Canceling” charges means their effects sum to zero, but the charges themselves persist.
  • The net charge of a closed system is constant.

These are fundamental principles. Quantization means charge isn’t infinitely divisible. Conservation is a powerful law, similar to conservation of energy or momentum. It implies that in any process, the total charge before and after must be the same.

Source of Charges: The Structure of the Atom

• J. J. Thomson (1897):
  • Discovered the electron (negatively charged particle).
  • Determined its charge-to-mass ratio, showing it was much lighter than any known particle.
• Ernest Rutherford (1908):
  • Discovered the atomic nucleus (tiny, dense, positively charged core).
  • Nucleus contains over 99% of atom’s mass.
  • Electrons orbit the nucleus.
  • Identified protons as the positively charged particles in the nucleus.

Simplified hydrogen atom: proton nucleus, electron cloud.

The discovery of subatomic particles revolutionized our understanding of charge. Thomson’s electron provided the first evidence of particles smaller than atoms. Rutherford’s gold foil experiment dramatically changed the model of the atom, establishing the nucleus.

Source of Charges: Neutrons and Ions

• James Chadwick (1932):
  • Discovered the neutron (electrically neutral particle).
  • Nearly identical mass to the proton.
  • Explains mass differences in atoms (e.g., Helium has two protons but four times the mass of hydrogen).
• Modern Atomic Model (simplified):
  • Small, massive nucleus (protons + neutrons).
  • Surrounded by electron cloud.
  • Neutral atom: Total negative electron charge equals total positive nuclear charge.
• Ions:
  • Atoms with altered charge.
  • Positive ions: Lost electrons.
  • Negative ions: Gained excess electrons.

Simplified carbon atom: six protons, six neutrons in nucleus, six electrons in cloud.

Chadwick’s discovery of the neutron completed the basic picture of the atomic nucleus, explaining isotopic variations. The concept of ions is crucial for understanding chemical bonding and electrical conduction in solutions.

Terminology Note

Note

When we refer to a “charge,” we are typically referring to an object that carries an electric charge, not the charge itself as a separate entity. For example, “a \(3 \mu \text{C}\) charge” means “a particle that carries a charge of \(3 \mu \text{C}\).”

5.2 Conductors, Insulators, and Charging by Induction

Learning Objectives

• Explain what a conductor is
• Explain what an insulator is
• List the differences and similarities between conductors and insulators
• Describe the process of charging by induction

Now that we understand the nature of charge, we’ll look at how charge interacts with different types of materials and how we can manipulate charge without direct contact.

Conductors

• Definition: Materials through which electric charge (electrons) can move freely.
• Mechanism:
  • Atoms have loosely bound outermost electrons.
  • These conduction electrons wander from atom to atom.
• Examples: Most metals (e.g., copper, silver, gold).
• Behavior:
  • Any excess charge placed on a conductor will instantly flow away due to mutual repulsion.
  • Therefore, conductors typically do not exhibit static electrical attraction/repulsion forces if charged in isolation.

Conductors are essential for electrical circuits. The “sea of electrons” model helps visualize why metals are such good conductors – their electrons are delocalized and can respond to electric forces.

Insulators

• Definition: Materials through which electric charge flows with great difficulty, if at all.
• Mechanism:
  • Lack conduction electrons.
  • Electrons are tightly bound to their individual atoms.
• Examples: Amber, fur, wood, glass, plastic, rubber.
• Behavior:
  • Excess charge remains localized indefinitely (or for a long time).
  • This allows insulators to exhibit strong static electrical attraction and repulsion forces.

Insulators are equally important. They prevent unwanted charge movement, like the rubber shielding on power cords. The localized nature of charge on insulators is what allows us to demonstrate static electricity so easily.

Charging by Induction

• Inducing Polarization in Conductors:
  1. Bring a charged object (e.g., positively charged glass rod) near a neutral conductor.
  2. Conduction electrons in the conductor are attracted to the rod, accumulating on the near side.
  3. This leaves the far side of the conductor with a net positive charge.
  4. The conductor is still overall neutral, but has a charge distribution (polarized).
  5. This separation creates an electric dipole in the conductor.

This is a key concept: you can redistribute charge in a conductor without actually touching it or transferring any charge to it. The charged rod simply influences the existing charges within the conductor.

Attraction of Neutral Objects

• Neutral objects can be attracted to any charged object (e.g., paper to a comb).
• Mechanism: Polarization of atoms/molecules within the neutral object.
  1. External charged object causes a slight shift in the charge distribution within neutral atoms/molecules.
  2. Opposite charges are attracted closer to the external rod.
  3. Like charges are repelled further away.
  4. Since electrostatic force decreases with distance (\(F \propto 1/r^2\)), the attractive force (closer charges) is stronger than the repulsive force (farther charges).
  5. This results in a net attraction.

Both positive and negative objects attract a neutral object by polarizing its molecules.

This explains why a charged comb picks up neutral pieces of paper. Even though the paper as a whole is neutral, its charges rearrange slightly, leading to a net attractive force. Polar molecules, like water, are particularly susceptible to this effect.

Charging a Conductor by Induction (without ground)

1. Initial State: Two neutral metal spheres are in contact, insulated from surroundings.
2. Polarization: Bring a positively charged rod near one sphere. Negative charges are attracted to that sphere, positive charges are repelled to the other.
3. Separation: Separate the two spheres while the charged rod is still nearby.
4. Result: The spheres now have opposite net charges, even though no charge was transferred from the rod.

Charging by induction using two spheres.

This method demonstrates how charge can be created on objects without direct contact. The inducing rod retains its original charge.

Charging a Conductor by Induction (with ground)

1. Initial State: Neutral metal sphere, insulated.
2. Polarization: Bring a charged rod (e.g., positive) near the sphere. Negative charges accumulate on the near side.
3. Grounding: Connect the sphere to the ground with a conducting wire. Earth acts as a large reservoir of charge.
   • Electrons from the ground are attracted to the sphere (to neutralize the excess positive charge on the far side).
4. Breaking Ground: Disconnect the ground wire before removing the charged rod.
5. Result: The sphere is left with a net negative charge (opposite to the rod’s charge).

Charging by induction using a ground connection.

Grounding provides an unlimited source or sink of electrons. This method allows you to charge an object with the opposite sign of the inducing charge, without any loss of charge from the inducing object.

5.3 Coulomb’s Law

Learning Objectives

• Describe the electric force, both qualitatively and quantitatively
• Calculate the force that charges exert on each other
• Determine the direction of the electric force for different source charges
• Correctly describe and apply the superposition principle for multiple source charges

Now we move from qualitative descriptions to a quantitative law that precisely describes the electric force between charges. This is Coulomb’s Law, a cornerstone of electrostatics.

Coulomb’s Law

• Description: The electric force between two point charges.
  • Magnitude is proportional to the product of their charges.
  • Magnitude is inversely proportional to the square of the distance between them.
  • Direction is along the line joining the two charges.

\[|\vec{F}_{12}| = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r_{12}^2}\]

Where:

• \(\vec{F}_{12}\): vector force exerted by charge 1 on charge 2.
• \(q_1, q_2\): magnitudes of the two charges.
• \(r_{12}\): distance between \(q_1\) and \(q_2\).
• \(\epsilon_0\): permittivity of free space.
  • \(\epsilon_0 = 8.85 \times 10^{-12} \text{C}^2 / (\text{N} \cdot \text{m}^2)\).
• Coulomb’s Constant: \(k_e = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \text{N} \cdot \text{m}^2 / \text{C}^2\).

Coulomb’s law is an inverse square law, just like Newton’s law of universal gravitation. This form is common for fundamental forces in physics. The constant epsilon-nought relates electric charge to the force it produces. \(k_e\) is often used for simplicity.

Direction of Electric Force

• If \(q_1\) and \(q_2\) have the same sign:
  • Force is repelling (points away from each other).
• If \(q_1\) and \(q_2\) have different signs:
  • Force is attractive (points towards each other).

Important

Newton’s third law applies: \(\vec{F}_{12} = -\vec{F}_{21}\).
The force on \(q_1\) is equal in magnitude and opposite in direction to the force on \(q_2\).

The electrostatic force between point charges. (a) Like charges repel; (b) unlike charges attract.

Always determine the direction of the force based on the signs of the charges, and then use the absolute values of the charges for magnitude calculation to avoid errors. The fact that the electric force is not constant, but depends on distance, makes direct application of Newton’s laws challenging for moving charges; we often use energy conservation instead.

Example: Force on Electron in Hydrogen

Problem:

A hydrogen atom has a proton (\(+e\)) and an electron (\(-e\)). The most probable distance is \(r = 5.29 \times 10^{-11} \text{m}\). Calculate the electric force on the electron due to the proton.

Strategy:

Treat electron and proton as point particles. Use Coulomb’s Law.

Given:

\(q_1 = +e = +1.602 \times 10^{-19} \text{C}\)

\(q_2 = -e = -1.602 \times 10^{-19} \text{C}\)

\(r = 5.29 \times 10^{-11} \text{m}\)

Solution:

\(F = \frac{1}{4\pi\epsilon_0} \frac{|e|^2}{r^2}\)

\(F = (8.99 \times 10^9 \text{N} \cdot \text{m}^2 / \text{C}^2) \frac{(1.602 \times 10^{-19} \text{C})^2}{(5.29 \times 10^{-11} \text{m})^2}\)

\(F = 8.25 \times 10^{-8} \text{N}\)

Direction:

Since charges are opposite, the force is attractive. It points radially toward the proton.

\(\vec{F} = (8.25 \times 10^{-8} \text{N}) \hat{r}\) (pointing inwards)

Schematic of hydrogen atom showing force on electron.

This calculation demonstrates a real-world application of Coulomb’s Law. Even though this is a classical model, it gives us a sense of the strength of the electrostatic force at atomic scales. This force is what holds the atom together.

Check Your Understanding 5.1

What would be different if the electron also had a positive charge?

If the electron had a positive charge, it would be repelled by the proton. The hydrogen atom would not form in the same way, and stable matter as we know it would not exist.

Multiple Source Charges: Superposition Principle

• Concept: The net electric force on a test charge is the vector sum of individual electric forces exerted by each source charge.
  • Each individual force is unaffected by the presence of other charges.
• Mathematical Form: \[\vec{F}(\vec{r}) = \frac{1}{4\pi\epsilon_0} Q \sum_{i=1}^{N} \frac{q_i}{r_i^2} \hat{r}_i\]
  • \(Q\): charge of the particle experiencing the force (test charge).
  • \(q_i\): N source charges.
  • \(\vec{r}_i\): displacement vector from the \(i\)-th source charge to \(Q\).
  • \(\hat{r}_i\): unit vector pointing from \(q_i\) to \(Q\).

The superposition principle is crucial. It simplifies complex systems by allowing us to break them down into interactions between pairs of charges and then combine the results vectorially. This is a fundamental property of the electric force.

Electrostatics Assumption

• For now, we assume source charges are fixed in place.
  • This simplifies the analysis significantly.
  • The force on the test charge is a function of its position only.
• This simplification leads to the field of study known as electrostatics.
  • “Statics” refers to the constant positions of source charges.

Without this assumption, the problem becomes much harder because all charges would move, changing all distances and forces simultaneously. Later, we’ll learn techniques to handle dynamic charge distributions.

Example: Net Force from Two Source Charges

Problem:

Three charges are placed as shown. \(q_1 = 2e\), \(q_2 = -3e\), \(q_3 = -5e\), \(d = 2.0 \times 10^{-7} \text{m}\). Find the net force on \(q_2\).

(Assume \(q_1\) is at \((0, d)\), \(q_2\) at \((2d, 0)\), \(q_3\) at \((0, 0)\).)

Strategy:

Use Coulomb’s law for forces between \(q_1, q_2\) and \(q_3, q_2\). Add forces vectorially using superposition.

Forces:

• \(\vec{F}_{12}\) (from \(q_1\) on \(q_2\)): \(q_1\) is positive, \(q_2\) is negative \(\Rightarrow\) attractive. Force points towards \(q_1\) (upwards, \(+y\) direction).
• \(\vec{F}_{32}\) (from \(q_3\) on \(q_2\)): \(q_3\) is negative, \(q_2\) is negative \(\Rightarrow\) repulsive. Force points away from \(q_3\) (leftwards, \(-x\) direction).

Example: Net Force from Two Source Charges

Calculations (Magnitudes):

\(r_{12} = \sqrt{(2d)^2 + d^2} = \sqrt{5}d\)

\(r_{32} = 2d\)

\(F_{12} = k_e \frac{|q_1 q_2|}{r_{12}^2} = k_e \frac{|(2e)(-3e)|}{(\sqrt{5}d)^2} = k_e \frac{6e^2}{5d^2}\)

\(F_{32} = k_e \frac{|q_3 q_2|}{r_{32}^2} = k_e \frac{|(-5e)(-3e)|}{(2d)^2} = k_e \frac{15e^2}{4d^2}\)

(Numerical values inserted as absolute values, directions determined physically)

Components & Net Force:

\(\vec{F}_{12}\) has \(x\) and \(y\) components. From diagram, \(\theta_1 = \tan^{-1}(2d/d) = \tan^{-1}(2)\).

\(F_{12,x} = -F_{12}\sin\theta_1\)

\(F_{12,y} = F_{12}\cos\theta_1\)

\(\vec{F}_{32}\) is entirely in \(-x\) direction.

\(F_{32,x} = -F_{32}\)

\(F_{32,y} = 0\)

Sum components, then find magnitude and direction.

Example: Net Force from Two Source Charges

Source charges q1 and q3 each apply a force on q2.

The example in the textbook has \(q_1\) at \((0, d)\), \(q_2\) at \((2d, 0)\), \(q_3\) at the origin \((0, 0)\).

This makes \(r_{12} = \sqrt{(2d)^2 + d^2}\) and \(r_{32} = 2d\).

Original example had a typo in diagram/text:

\(q_1\) at \((0, d)\), \(q_2\) at \((2d, 0)\), \(q_3\) at \((0,0)\). This is inconsistent with diagram.

The diagram implies \(q_1\) is at \((0,d)\), \(q_3\) is at \((0,0)\) and \(q_2\) is at \((2d,0)\).

Let’s use the given calculation values, where \(F_{12}\) is purely in +y and \(F_{32}\) purely in -x. This implies a different setup.

The textbook diagram labels forces differently.

Let’s stick to the numerical calculation provided, which implies:

\(q_1\) is at \((0, 2d)\)

\(q_2\) is at \((0, 0)\)

\(q_3\) is at \((2d, 0)\)

Then \(r_{12} = 2d\) and \(r_{32} = 2d\).

\(F_{12}\) (from \(q_1\) on \(q_2\)): \(q_1=+2e\), \(q_2=-3e\). Attractive. Points towards \(q_1\) (upwards).

\(F_{32}\) (from \(q_3\) on \(q_2\)): \(q_3=-5e\), \(q_2=-3e\). Repulsive. Points away from \(q_3\) (leftwards).

Given values from book:

\(F_x = -F_{32} = -2.16 \times 10^{-14} \text{N}\)

\(F_y = F_{12} = 3.46 \times 10^{-14} \text{N}\)

\(F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-2.16 \times 10^{-14})^2 + (3.46 \times 10^{-14})^2} = 4.08 \times 10^{-14} \text{N}\)

\(\theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) = \tan^{-1} \left( \frac{3.46 \times 10^{-14}}{-2.16 \times 10^{-14}} \right) = -58^\circ\) (or \(122^\circ\) from +x axis).

This angle is 58 degrees above the -x axis.

This example shows the practical application of the superposition principle. It’s crucial to correctly identify the direction of each individual force before adding them vectorially. Be careful with signs – use absolute values for magnitude calculations and determine direction separately.

Check Your Understanding 5.2

What would be different if \(q_1\) were negative?

If \(q_1\) were negative, the force \(\vec{F}_{12}\) (between \(q_1\) and \(q_2\)) would become repulsive instead of attractive. Since \(q_2\) is also negative, they would repel each other, and \(\vec{F}_{12}\) would point downwards instead of upwards. This would change the net force’s magnitude and direction.

5.4 Electric Field

Learning Objectives

• Explain the purpose of the electric field concept
• Describe the properties of the electric field
• Calculate the field of a collection of source charges of either sign

Instead of calculating forces on a test charge repeatedly, we introduce the concept of an electric field. This field characterizes the electrical influence of a source charge distribution independent of any test charge.

Defining the Electric Field

• Motivation: To avoid recalculating forces for every new test charge.

• Definition: The electric field \(\vec{E}\) at a point P is the electric force \(\vec{F}\) experienced by a positive test charge \(Q\) placed at P, divided by the magnitude of that test charge. \[\vec{E} \equiv \frac{\vec{F}}{Q}\]

• Units: Newtons per Coulomb (N/C).

• Electric Field of N Source Charges:

  \[\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{r_1^2} \hat{r}_1 + \frac{q_2}{r_2^2} \hat{r}_2 + \dots + \frac{q_N}{r_N^2} \hat{r}_N \right)\]

  Or, more compactly:

  \[\vec{E}(P) \equiv \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{N} \frac{q_i}{r_i^2} \hat{r}_i\]

  • P is the location where the field is calculated.
  • \(\vec{r}_i\) is the displacement vector from source charge \(q_i\) to point P.

The electric field is a vector quantity that exists independently of whether a test charge is actually present. It describes how space itself is modified by the presence of source charges. The definition is analogous to how gravitational field is defined (\(g = F/m\)).

The Meaning of “Field”

• Physical Quantity: A quantity whose value depends on position relative to the source.
• Vector Field: The electric field is a vector field, meaning it has both magnitude and direction at every point in space.
• Continuous Substance: Picture the electric field as an immaterial substance filling all space around a source charge, transmitting electric properties.
• Speed of Influence: Electrical phenomena travel at the speed of light.

Understanding fields is key in physics. Unlike scalar fields (like temperature), vector fields require both magnitude and direction. The idea of a field mediating “action at a distance” is a profound concept.

Superposition Principle (for Electric Fields)

• The total electric field at any point due to multiple source charges is the vector sum of the electric fields produced by each individual source charge, calculated independently.
• This is exactly what the summation in the formula for \(\vec{E}(P)\) represents.

Direction of the Electric Field

• Convention: The direction of the electric field vector is defined as the direction of the electric force that a positive test charge would experience if placed in that field.
• Result:
  • Electric fields \(\vec{E}\) point away from positive source charges.
  • Electric fields \(\vec{E}\) point toward negative source charges.

This convention is crucial for drawing field diagrams and interpreting field vectors. A positive test charge would be repelled by positive sources and attracted to negative sources.

Example: E-Field of an Atom

Problem:

In an ionized helium atom, the most probable distance between the nucleus and the electron is \(r = 26.5 \times 10^{-12} \text{m}\). What is the electric field due to the nucleus at the location of the electron?

Strategy:

The problem asks for the electric field due to the nucleus only. The electron’s presence determines the location but it is not a source charge for the field being calculated. The nucleus of an ionized helium atom has two protons, so its charge is \(+2e\).

Given:

Source charge \(q = +2e = 2(1.6 \times 10^{-19} \text{C})\)

Distance \(r = 26.5 \times 10^{-12} \text{m}\)

Solution:

\(\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}\)

\(\vec{E} = (8.99 \times 10^9 \text{N} \cdot \text{m}^2 / \text{C}^2) \frac{2(1.6 \times 10^{-19} \text{C})}{(26.5 \times 10^{-12} \text{m})^2} \hat{r}\)

\(\vec{E} = 4.1 \times 10^{12} \text{N/C} \hat{r}\)

Direction:

Radially away from the nucleus (since the nucleus is positive, and \(\hat{r}\) points away from the nucleus).

Schematic of a helium atom, showing electric field of nucleus.

This example highlights that when calculating an electric field, you only consider the source charges that create the field, not any “test charges” that might be present in that field.

Example: E-Field above Two Equal Charges

Problem:

1. Find the electric field (magnitude and direction) a distance \(z\) above the midpoint between two equal charges \(+q\) that are a distance \(d\) apart.
2. Same as (a), but the right-hand charge is \(-q\).

Strategy (a): Equal Charges

1. Set up coordinate system: charges at \((-d/2, 0)\) and \((+d/2, 0)\), point P at \((0, z)\).
2. Use superposition: \(\vec{E} = \vec{E}_1 + \vec{E}_2\).
3. Exploit symmetry: Horizontal components (\(E_x\)) will cancel. Only vertical components (\(E_z\)) add up.

Solution (a):

Distance from each charge to P: \(r = \sqrt{(d/2)^2 + z^2}\).

Each electric field magnitude: \(E_1 = E_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\).

Vertical component from each: \(E_z = E \cos\theta\), where \(\cos\theta = \frac{z}{r}\).

Total field: \(\vec{E}(z) = \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \cos\theta + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \cos\theta \right) \hat{k}\)

\(\vec{E}(z) = \frac{1}{4\pi\epsilon_0} \frac{2qz}{(z^2 + (d/2)^2)^{3/2}} \hat{k}\)

Strategy (b): Opposite Charges (Dipole)

1. Now \(q_1 = +q\) at \((-d/2, 0)\), \(q_2 = -q\) at \((+d/2, 0)\).
2. Symmetry: Vertical components (\(E_z\)) will cancel. Only horizontal components (\(E_x\)) add up.

Solution (b):

Vertical component from each cancels.

Horizontal component from \(+q\): Points right. \(E_{1x} = E_1 \sin\theta\).

Horizontal component from \(-q\): Points right. \(E_{2x} = E_2 \sin\theta\).

Total field: \(\vec{E}(z) = \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \sin\theta + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \sin\theta \right) \hat{i}\)

Where \(\sin\theta = \frac{d/2}{r}\).

\(\vec{E}(z) = \frac{1}{4\pi\epsilon_0} \frac{qd}{(z^2 + (d/2)^2)^{3/2}} \hat{i}\)

Field of two identical source charges at point P.

Components of electric fields from two charges.

This example demonstrates the power of symmetry in simplifying electric field calculations. For two equal charges, the resultant field is along the perpendicular bisector. For two opposite charges (a dipole), the resultant field is along the line connecting the charges. Always check extreme cases (\(z \gg d\) or \(d \to 0\)) to see if the result makes physical sense.

Check Your Understanding 5.3

What is the electric field due to a single point particle?

For a single point particle with charge \(q\) at the origin, the electric field at a distance \(r\) from it is given by \(\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}\). This is derived directly from the definition of the electric field and Coulomb’s Law.

5.5 Calculating Electric Fields of Charge Distributions

Learning Objectives

• Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge
• Describe line charges, surface charges, and volume charges
• Calculate the field of a continuous source charge distribution of either sign

Up to now, we’ve dealt with discrete point charges. But often, charge is distributed continuously over objects. This section extends our calculation methods to these continuous charge distributions using integration.

Continuous Charge Distributions

• Concept: When charge is spread over a line, surface, or volume.
  • While charge is fundamentally quantized, for macroscopic objects, the number of individual charges is so vast that we can approximate the distribution as continuous.
• Method: Divide the charge into infinitesimal pieces (\(dq\)), treat each as a point charge, and integrate their contributions to the electric field.

This approximation is analogous to treating water as a continuous fluid instead of discrete H2O molecules. It simplifies calculations greatly without sacrificing accuracy for large-scale systems.

Charge Densities

• Linear Charge Density (\(\lambda\)):
  • Charge per unit length.
  • Units: C/m.
  • \(dq = \lambda \, dl\).
• Surface Charge Density (\(\sigma\)):
  • Charge per unit area.
  • Units: C/m\(^2\).
  • \(dq = \sigma \, dA\).
• Volume Charge Density (\(\rho\)):
  • Charge per unit volume.
  • Units: C/m\(^3\).
  • \(dq = \rho \, dV\).

These definitions allow us to express the infinitesimal charge \(dq\) in terms of geometric elements (\(dl\), \(dA\), \(dV\)) and a density function. The density might be constant or vary with position.

General Electric Field Formulas for Continuous Distributions

• Point Charges: \[\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{N} \left( \frac{q_i}{r_i^2} \right) \hat{r}\]
• Line Charge: \[\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \int_{\text{line}} \left( \frac{\lambda \, dl}{r^2} \right) \hat{r}\]
• Surface Charge: \[\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \int_{\text{surface}} \left( \frac{\sigma \, dA}{r^2} \right) \hat{r}\]
• Volume Charge: \[\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \int_{\text{volume}} \left( \frac{\rho \, dV}{r^2} \right) \hat{r}\]

These are the integral forms of the superposition principle for continuous charge distributions. The main challenge is setting up the integral correctly, including expressing \(dl\), \(dA\), or \(dV\), and \(r\), in terms of the coordinate system chosen.

Example: Electric Field of a Line Segment

Problem:

Find the electric field a distance \(z\) above the midpoint of a straight line segment of length \(L\) that carries a uniform line charge density \(\lambda\).

Strategy:

1. Break the wire into infinitesimal segments \(dx\) at position \(x\).
2. Use symmetry: horizontal components of \(\vec{E}\) from symmetrically placed \(dx\) elements cancel.
3. Integrate the vertical components.

Setup:

• Line segment from \(-L/2\) to \(L/2\) along \(x\)-axis.
• Point P at \((0, z)\).
• Differential charge \(dq = \lambda \, dx\).
• Distance from \(dx\) to P: \(r = \sqrt{x^2 + z^2}\).
• Vertical component contribution: \(dE_z = dE \cos\theta\), where \(\cos\theta = z/r\).

Solution:

The total field is purely in the \(z\)-direction (\(\hat{k}\)).

\(dE_z = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \cos\theta = \frac{1}{4\pi\epsilon_0} \frac{\lambda \, dx}{(x^2 + z^2)} \frac{z}{\sqrt{x^2 + z^2}}\)

\(E_z = \frac{\lambda z}{4\pi\epsilon_0} \int_{-L/2}^{L/2} \frac{dx}{(x^2 + z^2)^{3/2}}\)

Due to symmetry, we can integrate from \(0\) to \(L/2\) and multiply by 2:

\(E_z = \frac{2\lambda z}{4\pi\epsilon_0} \int_{0}^{L/2} \frac{dx}{(x^2 + z^2)^{3/2}}\)

This integral evaluates to: \(\frac{1}{z^2} \frac{x}{\sqrt{x^2+z^2}}\)

\(E_z = \frac{2\lambda z}{4\pi\epsilon_0} \left[ \frac{x}{z^2 \sqrt{x^2 + z^2}} \right]_0^{L/2}\)

\(\vec{E}(z) = \frac{1}{4\pi\epsilon_0} \frac{\lambda L}{z \sqrt{z^2 + (L/2)^2}} \hat{k}\)

Significance:

• For \(z \gg L\), this simplifies to \(\vec{E} \approx \frac{1}{4\pi\epsilon_0} \frac{\lambda L}{z^2} \hat{k} = \frac{1}{4\pi\epsilon_0} \frac{Q_{total}}{z^2} \hat{k}\), which is the field of a point charge, as expected.

Electric field of a uniformly charged line segment.

This is a standard example of integrating to find the electric field. The use of symmetry is crucial to simplify the integral. Remember to check limiting cases to build confidence in your result.

Check Your Understanding 5.4

How would the strategy used above change to calculate the electric field at a point a distance z above one end of the finite line segment?

If the point P is above one end, the symmetry argument no longer holds. Both the horizontal (\(x\)) and vertical (\(z\)) components of the field would need to be calculated and integrated separately, and then added vectorially. The limits of integration would be from \(0\) to \(L\) for the wire extending along the \(x\)-axis from \(x=0\) to \(x=L\).

Example: Electric Field of an Infinite Line of Charge

Problem:

Find the electric field a distance \(z\) above the midpoint of an infinite line of charge that carries a uniform line charge density \(\lambda\).

Strategy:

Same as the finite line segment, but with integration limits from \(-\infty\) to \(+\infty\).

Solution:

Using the setup from the finite line segment example, but changing the limits:

\(E_z = \frac{\lambda z}{4\pi\epsilon_0} \int_{-\infty}^{\infty} \frac{dx}{(x^2 + z^2)^{3/2}}\)

This integral evaluates to: \(\frac{2}{z^2}\)

\(\vec{E}(z) = \frac{\lambda z}{4\pi\epsilon_0} \left[ \frac{2}{z^2} \right] \hat{k} = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{z} \hat{k}\)

Or more commonly: \(\vec{E}(z) = \frac{\lambda}{2\pi\epsilon_0 z} \hat{k}\)

Significance:

• The field of an infinite line charge decreases with \(1/z\), not \(1/z^2\). This is an artifact of the infinite length assumption, where “z” effectively becomes the only relevant length scale.
• This result is highly useful for approximating the field of very long, thin wires.

The \(1/z\) dependence is a key characteristic of infinite line charges. This highlights how symmetry and infinite extent can change the functional form of the electric field’s dependence on distance.

Example: Electric Field due to a Ring of Charge

Problem:

A ring of radius \(R\) has a uniform charge density \(\lambda\). Find the electric field at a point on the axis passing through the center of the ring, a distance \(z\) from the center.

Strategy:

1. Divide the ring into infinitesimal arc elements \(dl = R \, d\theta\).
2. Use symmetry: all horizontal components of \(\vec{E}\) cancel. Only vertical components add up.
3. Integrate the vertical components over the entire ring.

Setup:

• Point P at \((0, 0, z)\).
• Ring in \(xy\)-plane, centered at origin.
• Differential charge \(dq = \lambda \, dl = \lambda R \, d\theta\).
• Distance from \(dq\) to P: \(r = \sqrt{R^2 + z^2}\) (constant for all \(dq\)).
• Vertical component contribution: \(dE_z = dE \cos\phi\), where \(\cos\phi = z/r\).

Solution:

\(dE_z = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \cos\phi = \frac{1}{4\pi\epsilon_0} \frac{\lambda R \, d\theta}{R^2 + z^2} \frac{z}{\sqrt{R^2 + z^2}}\)

\(E_z = \int_0^{2\pi} \frac{1}{4\pi\epsilon_0} \frac{\lambda R z}{(R^2 + z^2)^{3/2}} d\theta\)

Since all terms in the integrand are constant with respect to \(\theta\):

\(E_z = \frac{\lambda R z}{4\pi\epsilon_0 (R^2 + z^2)^{3/2}} \int_0^{2\pi} d\theta = \frac{\lambda R z (2\pi)}{4\pi\epsilon_0 (R^2 + z^2)^{3/2}}\)

Let \(Q_{total} = \lambda (2\pi R)\) be the total charge on the ring.

\(\vec{E}(z) = \frac{1}{4\pi\epsilon_0} \frac{Q_{total} z}{(R^2 + z^2)^{3/2}} \hat{k}\)

Significance:

• For \(z \gg R\), this simplifies to \(\vec{E} \approx \frac{1}{4\pi\epsilon_0} \frac{Q_{total}}{z^2} \hat{k}\), the field of a point charge, as expected.

Electric field calculation for a ring of charge.

The ring of charge is another classic problem that demonstrates the power of symmetry. Because every point on the ring is equidistant from P, the integration is particularly straightforward after the initial setup.

Example: The Field of a Disk

Problem:

Find the electric field of a circular thin disk of radius \(R\) and uniform surface charge density \(\sigma\) at a distance \(z\) above the center of the disk.

Strategy:

1. Treat the disk as a collection of concentric rings.
2. For each ring of radius \(r'\) and thickness \(dr'\), the charge is \(dq = \sigma \, dA = \sigma (2\pi r' \, dr')\).
3. Use the result for the electric field of a ring, and integrate over the radius \(r'\) from \(0\) to \(R\).

Setup:

• Differential ring has charge \(dq = \sigma (2\pi r' \, dr')\).
• Field from this ring (from previous example, replacing \(R\) with \(r'\) and \(Q_{total}\) with \(dq\)): \(dE_z = \frac{1}{4\pi\epsilon_0} \frac{(\sigma 2\pi r' dr') z}{(r'^2 + z^2)^{3/2}}\)

Solution:

Integrate \(dE_z\) from \(r'=0\) to \(r'=R\):

\(E_z = \int_0^R \frac{1}{4\pi\epsilon_0} \frac{2\pi\sigma z r'}{(r'^2 + z^2)^{3/2}} dr'\)

\(E_z = \frac{2\pi\sigma z}{4\pi\epsilon_0} \int_0^R (r'^2 + z^2)^{-3/2} r' \, dr'\)

Let \(u = r'^2 + z^2\), then \(du = 2r' dr'\).

\(E_z = \frac{\sigma z}{4\epsilon_0} \int_{z^2}^{R^2+z^2} u^{-3/2} du = \frac{\sigma z}{4\epsilon_0} \left[ -2u^{-1/2} \right]_{z^2}^{R^2+z^2}\)

\(E_z = \frac{\sigma z}{4\epsilon_0} \left[ \frac{-2}{\sqrt{R^2+z^2}} - \frac{-2}{\sqrt{z^2}} \right]\)

\(\vec{E}(z) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2+z^2}} \right) \hat{k}\)

Significance:

• For \(z \gg R\), this approximates to the field of a point charge \(\frac{1}{4\pi\epsilon_0} \frac{\sigma \pi R^2}{z^2} \hat{k}\).
• As \(R \to \infty\) (infinite plane), the field becomes \(\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{k}\). This is a constant field, independent of distance!

Electric field calculation for a uniformly charged disk.

The disk field is a more complex integration problem, building on the ring result. The result for an infinite plane is extremely important: a uniform electric field. This result is used to model parallel-plate capacitors.

Check Your Understanding 5.5

How would the above limit change with a uniformly charged rectangle instead of a disk?

For a uniformly charged rectangle, the symmetry is different. You would integrate over two dimensions (length and width). The limit as the rectangle’s dimensions go to infinity would also result in a constant electric field, similar to the infinite plane, but the exact formula for finite dimensions would be more complex, likely involving inverse tangents.

Example: The Field of Two Infinite Planes

Problem:

Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (\(\sigma\) and \(-\sigma\)).

Strategy:

Use the superposition principle and the known result for a single infinite plane: \(\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n}\), where \(\hat{n}\) is the unit vector perpendicular to the plane, pointing away from it for positive charge.

Solution:

Let the positive plane be at \(x=0\) with charge density \(+\sigma\). Its field points away from it.

Let the negative plane be at \(x=d\) with charge density \(-\sigma\). Its field points toward it.

• Region I (Left of positive plane, \(x<0\)):
  • Field from positive plane: \(-\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Field from negative plane: \(+\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Net field: \(\vec{E}_{\text{total}} = 0\)
• Region II (Between planes, \(0<x<d\)):
  • Field from positive plane: \(+\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Field from negative plane: \(+\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Net field: \(\vec{E}_{\text{total}} = \frac{\sigma}{\epsilon_0} \hat{i}\)
• Region III (Right of negative plane, \(x>d\)):
  • Field from positive plane: \(+\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Field from negative plane: \(-\frac{\sigma}{2\epsilon_0} \hat{i}\)
  • Net field: \(\vec{E}_{\text{total}} = 0\)

Significance:

• This configuration creates a uniform electric field between the plates and zero field outside.
• This is the basis for understanding capacitors.

Electric field from two charged infinite planes.

This is a critical result for understanding capacitors and other electrical devices. The ability to create a uniform electric field is very useful in many applications.

Check Your Understanding 5.6

What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?

If both planes were positively charged with equal densities, the fields would add outside the planes and cancel between them. So, the field would be zero between the planes, and \(E = \sigma/\epsilon_0\) pointing away from the planes in the outer regions.

5.6 Electric Field Lines

Learning Objectives

• Explain the purpose of an electric field diagram
• Describe the relationship between a vector diagram and a field line diagram
• Explain the rules for creating a field diagram and why these rules make physical sense
• Sketch the field of an arbitrary source charge

Electric field lines are a powerful visual tool for understanding the geometry and strength of electric fields. They provide an intuitive way to represent a vector field.

Visualizing Electric Fields

• Purpose: To visualize how source charges alter space, allowing us to estimate, predict, and understand the electric field without complex vector calculations.
• From Vector Diagrams to Field Line Diagrams:
  • Instead of drawing many individual field vectors, we connect them into continuous lines and curves.

Vector field of a positive point charge.

Vector field of a dipole (less clear).

Vector diagrams become cumbersome quickly for multiple charges. Field line diagrams offer a more abstract but much clearer representation.

Interpreting Field Line Diagrams

• Direction of Field:
  • At any point, the electric field vector is tangent to the field line at that point.
  • Arrowheads on lines indicate the direction.
• Magnitude of Field:
  • Indicated by field line density (number of lines per unit area perpendicular to the field).
  • Closely spaced lines = strong field.
  • Widely spaced lines = weak field.

Field line diagrams: (a) positive point charge, (b) dipole.

Field lines indicating varying field magnitude.

These two rules are fundamental. The tangent rule gives direction, and the density rule gives magnitude. Remember that field lines are a conceptual tool, not physical objects.

Rules for Drawing Electric Field Lines

1. Origin/Termination:
   • Originate on positive charges or come from infinity.
   • Terminate on negative charges or extend to infinity.
2. Proportional to Magnitude:
   • The number of lines originating from or terminating on a charge is proportional to its magnitude. (e.g., \(2q\) has twice as many lines as \(q\)).
3. Tangent to Field Vector:
   • At every point, the field vector is tangent to the field line.
4. Density Indicates Magnitude:
   • Field line density is proportional to the magnitude of the field.
5. Never Cross:
   • Field lines can never cross. If they did, it would imply two directions for the field at one point, which is impossible.

These rules ensure that field diagrams accurately represent the electric field. Practice with simulations or drawing tools is highly recommended to develop this skill.

Examples of Field Line Diagrams

Three typical electric field diagrams: (a) A dipole, (b) Two identical charges, (c) Two charges with opposite signs and different magnitudes.

• (a) Dipole (+q, -q):
  • Lines start on +q, end on -q.
  • Denser lines between charges (strong field).
• (b) Two Identical Positive Charges (+q, +q):
  • Lines originate from both +q, extend to infinity.
  • Region between charges has fewer lines (weak field, potentially a null point).
• (c) Two Charges, Opposite Signs, Different Magnitudes (+3q, -q):
  • More lines originate from +3q than terminate on -q.
  • The remaining lines from +3q extend to infinity.
  • Field is generally stronger near the larger charge.

These diagrams illustrate the rules. Notice how the number of lines reflects charge magnitude and how their density shows field strength. The “no crossing” rule is very important conceptually.

5.7 Electric Dipoles

Learning Objectives

• Describe a permanent dipole
• Describe an induced dipole
• Define and calculate an electric dipole moment
• Explain the physical meaning of the dipole moment

We’ve already encountered dipoles when discussing continuous charge distributions. Now, we’ll delve deeper into their properties, especially their behavior in external electric fields.

Permanent Dipoles

• Definition: Two equal and opposite charges separated by a small distance.
  • Examples: Water molecules (\(H_2O\)) are permanent dipoles due to their bent molecular structure and unequal sharing of electrons.
• Behavior in a Uniform Electric Field:
  • Net Force: Zero (forces on +q and -q are equal and opposite).
  • Net Torque (\(\vec{\tau}\)): Non-zero, causing the dipole to rotate.
  • \(\vec{\tau} = \vec{p} \times \vec{E}\)
    • Where \(\vec{p}\) is the electric dipole moment.

Permanent dipoles are common in chemistry, explaining properties like water’s high boiling point. Even though the net force is zero in a uniform field, the torque causes alignment, which is a key concept.

Electric Dipole Moment (\(\vec{p}\))

• Definition: A vector quantity that characterizes a dipole.
  • Magnitude: \(p = qd\) (charge magnitude \(\times\) separation distance).
  • Direction: Points from the negative charge to the positive charge.
• Significance: Its value determines the torque the dipole experiences in an external field.
• Effect of Torque: Aligns the dipole moment (\(\vec{p}\)) parallel to the external electric field (\(\vec{E}\)).

\[\vec{p} \equiv q\vec{d}\] Where \(\vec{d}\) is the displacement vector from \(-q\) to \(+q\).

A dipole in a uniform external electric field, showing forces and dipole moment.

Example: Water molecule

The oxygen atom tends to pull electrons more strongly than hydrogen, creating partial negative charge on oxygen and partial positive charges on hydrogen atoms.

This creates a permanent dipole moment.

The dipole moment vector \(\vec{p}\) is an important concept. It simplifies the description of the torque and potential energy of a dipole in an electric field. The direction from negative to positive is a convention, but it’s consistent.

Induced Dipoles

• Definition: Occurs when a neutral atom or nonpolar molecule is placed in an external electric field.
• Mechanism:
  1. External field exerts oppositely directed forces on the positive nucleus and negative electron cloud.
  2. This causes a slight separation of charge, deforming the atom/molecule.
  3. A temporary induced dipole moment is created.
• Alignment: The induced dipole moment aligns parallel to the external electric field.
• Magnitude: Generally much smaller than permanent dipoles.

A dipole induced in a neutral atom by an external electric field.

Impact on Electric Field:

• Once aligned, the induced dipole’s field opposes the external field inside the dipole charges.
• This effectively decreases the total electric field within the material.
• This effect is crucial for understanding dielectrics and capacitors.

Net electric field is sum of dipole field and external field.

Induced dipoles are why neutral objects can be attracted to charged objects. The internal field reduction is fundamental to how capacitors store charge more efficiently when filled with a dielectric material.

Key Takeaways

• Electric Charge: Fundamental property of matter, existing in two types (positive and negative). Like charges repel, unlike charges attract. Quantized (multiples of \(e\)) and conserved.
• Conductors & Insulators: Materials differ in their ability to allow charge movement due to the presence or absence of free electrons.
• Charging by Induction: Redistribution of charge in a conductor due to a nearby charged object, without direct contact or charge transfer from the inducing object. Can involve grounding.
• Coulomb’s Law: Quantifies the electrostatic force between two point charges: \(F = k_e \frac{|q_1 q_2|}{r^2}\). Force decreases with the square of the distance.
• Superposition Principle: The net force (or electric field) on a charge from multiple source charges is the vector sum of individual forces (or fields).
• Electric Field (\(\vec{E}\)): A vector field representing the force per unit positive charge (\(\vec{E} = \vec{F}/Q\)). It describes how space is modified by source charges, independent of a test charge. Points away from positive charges, towards negative charges.
• Electric Field Lines: Visual representation of electric fields. Tangent to field lines gives direction; density of lines gives magnitude. Lines originate from positive charges and terminate on negative charges, never cross.
• Electric Dipoles: Two equal and opposite charges separated by a distance. Characterized by an electric dipole moment (\(\vec{p} = q\vec{d}\)). Experience a torque (not a net force) in a uniform electric field, causing them to align with the field. Can be permanent or induced.

Key Equations

Equation	Description
\(\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}\)	Coulomb’s Law: Electric force between two point charges. (\(k_e = \frac{1}{4\pi\epsilon_0}\))
\(\vec{E} = \frac{\vec{F}}{Q}\)	Definition of Electric Field: Force per unit test charge.
\(\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{N} \frac{q_i}{r_i^2} \hat{r}_i\)	Electric Field for Discrete Charges: Superposition of fields from multiple point charges.
\(\vec{E}(P) = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r^2} \hat{r}\)	Electric Field for Continuous Charges: General form using infinitesimal charge elements.
\(\vec{p} = q\vec{d}\)	Electric Dipole Moment: Magnitude \(qd\), direction from negative to positive charge.
\(\vec{\tau} = \vec{p} \times \vec{E}\)	Torque on an Electric Dipole: Experienced by a dipole in an external electric field.

Key Terms

Term	Definition
Electric Charge	An intrinsic property of matter that gives rise to electric forces.
Coulomb (C)	The SI unit of electric charge.
Quantized Charge	The principle that electric charge exists only in discrete integer multiples of the elementary charge \(e\).
Elementary Charge (\(e\))	The smallest unit of positive charge, \(1.602 \times 10^{-19}\) C, carried by a proton.
Conductor	A material that allows electric charges to move freely through it.
Insulator	A material that resists the flow of electric charges.
Charging by Induction	A method of charging an object without direct contact by redistributing existing charges.
Electric Field (\(\vec{E}\))	A vector field representing the electric force per unit positive charge at any point in space.
Electric Field Lines	Imaginary lines used to visualize the direction and strength of an electric field.
Electric Dipole	A pair of equal and opposite electric charges separated by a small distance.
Electric Dipole Moment (\(\vec{p}\))	A vector quantity that characterizes an electric dipole, pointing from the negative to the positive charge.
Permittivity of Free Space (\(\epsilon_0\))	A fundamental physical constant representing the ability of a vacuum to permit electric fields.
Superposition Principle	The principle that the net electric force or field at a point is the vector sum of individual forces or fields due to each charge.
Linear Charge Density (\(\lambda\))	Electric charge per unit length.
Surface Charge Density (\(\sigma\))	Electric charge per unit area.
Volume Charge Density (\(\rho\))	Electric charge per unit volume.