Electric Field: Complete Guide with Formulas & Problems | Physics by Halliday, Resnick & Krane

Chapter 28: The Electric Field

Complete Physics Notes Based on Halliday, Resnick and Krane - B.Sc. Physics Edition 2015-16

Electric Field Electromagnetism Field Intensity Reading Time: 20 min

📋 Table of Contents

• 1. Introduction to Electric Field
• 2. Electric Field Intensity
• 3. Electric Field Lines
• 4. Electric Field Due to Point Charge
• 5. Electric Field Due to Multiple Charges
• 6. Electric Field Due to a Dipole
• 7. Electric Field Due to Continuous Charge Distributions
• 8. Torque on a Dipole
• 9. Energy of Dipole
• 10. Solved Problems
• Frequently Asked Questions

Introduction to Electric Field

⚡ The Concept of Electric Field

Electric charges interact with each other over vast distances. Electrons or ionized atoms at the furthest reaches of the known universe can exert forces that cause electrons to move on the Earth. How can we explain these interactions? We do so in terms of electric field - the distant charge sets up an electric field, which exists throughout the space between the Earth and is the origin of the field.

In this chapter we consider only the static electric field due to charges at rest.

🔬 Faraday's Field Concept

To describe the mechanism by which one charged particle exerts force on other charged particles, Michael Faraday introduced the concept of electric field.

The region or space around a charge in which it can exert the force of attraction or repulsion on other charged bodies is called electric field.

Electric Field Intensity

📏 Definition of Electric Field Intensity

The electrostatic force on unit positive charge at a specific field point is called the electric field intensity.

To find out electric field intensity, a test charge \( q_0 \) is placed in the electric field at a field point. The electric field intensity \( \vec{E} \) is expressed as:

Electric Field Intensity Formula

\[ \vec{E} = \frac{\vec{F}}{q_0} \]

where \( \vec{F} \) is the electrostatic force on test charge \( q_0 \).

🔍 Important Consideration

The test charge \( q_0 \) should be very small, so that it cannot disturb the field produced by source charge \( q \). Therefore the electric field intensity can be written as:

\[ \vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0} \]

⚖️ Analogy with Gravitational Field

The gravitational field strength \( \vec{g} \) is described as the gravitational force \( \vec{F_g} \) per unit test mass \( m \):

\[ \vec{g} = \frac{\vec{F_g}}{m} \]

In analogy with gravitational field intensity, the electric field intensity \( \vec{E} \) is defined as electrostatic force \( \vec{F} \) per unit test charge \( q_0 \):

\[ \vec{E} = \frac{\vec{F}}{q_0} \]

Sample Problem 1: Balancing Proton Weight

Question: A proton is placed in a uniform electric field. What must be the magnitude and direction of this field if the electrostatic force acting on proton is just to balance its weight?

Solution:

Charge on proton \( q = 1.6 \times 10^{-19} \, C \)

\( = 1.6 \times 10^{-19} \, C \)

Mass of proton \( m = 1.67 \times 10^{-27} \, kg \)

\( = 1.67 \times 10^{-27} \, kg \)

For balance: \( F_e = F_g \)

\( qE = mg \)

\( \Rightarrow E = \frac{mg}{q} \)

\( E = \frac{1.67 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}} \)

\( = 10.02 \times 10^{-8} \, \frac{N}{C} \)

The direction of electric field will be opposite to that of weight, i.e., electric field will be directed upward.

Electric Field Lines

📈 Visual Representation of Electric Field

A visual representation of the electric field can be obtained in terms of electric field lines. Electric field lines can be thought of as a map that provides information about the direction and strength of the electric field at various places. As electric field line provides the information about the electric force exerted on a charge, the lines are commonly called "Lines of Force".

Properties of Electric Field Lines

• Electric field lines originate from positive charges and end on negative charges.
• The tangent to a field line at any point gives the direction of the electric field intensity at that point.
• The lines are closer where the field is strong, the lines are farther apart where the field is weak.
• No two lines cross each other.

Field Line Density

The number of field lines per unit area is proportional to the magnitude of the electric field. This means:

• Closer lines = Stronger field
• Farther lines = Weaker field

Conceptual Q # 4. Explain why electric field lines never cross.

The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions, which is not possible.

Electric Field Due to Point Charge

🔘 Field from a Single Point Charge

Consider a test charge \( q_0 \) placed at point P in the electric field of a point charge \( q \) at a distance \( r \) apart.

Step 1: Electrostatic Force

The electrostatic force \( F \) between \( q \) and \( q_0 \) can be found using Coulomb's law:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{qq_0}{r^2} \]

Step 2: Electric Field Intensity

The electric field intensity \( E \) due to a point charge \( q \) can be obtained by putting the value of electrostatic force in expression of electric field intensity:

\[ E = \frac{F}{q_0} \]

\[ = \frac{\left( \frac{1}{4\pi \epsilon_0} \frac{qq_0}{r^2} \right)}{q_0} \]

Step 3: Final Expression

\[ E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \]

This expression gives the magnitude of electric field intensity due to a point charge \( q \).

Vector Form of Electric Field

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

where \( \hat{r} \) is the unit vector which gives the direction of electric field intensity.

Sample Problem 2: Electric Field in Ionized Helium

Question: In an ionized helium atom (a helium atom in which one of the two electrons has been removed) the electron and nucleus are separated by a distance of 26.5 pm. What electric field due to the nucleus at the location of the electron?

Solution:

Total charge of helium nucleus \( q = 2e \)

\( = 2 \times 1.6 \times 10^{-19} \)

\( = 3.2 \times 10^{-19} \, C \)

Distance \( r = 26.5 \, pm \)

\( = 26.5 \times 10^{-12} \, m \)

\( E = k \frac{q}{r^2} \)

\( = 9 \times 10^9 \times \frac{3.2 \times 10^{-19}}{(26.5 \times 10^{-12})^2} \)

\( = 4.1 \times 10^{12} \, \frac{N}{C} \)

Conceptual Q # 5. Explain what happens to the magnitude of the electric field created by a point charge as r approaches zero.

The electric field around a point charge approaches infinity as r approaches zero.

Electric Field Due to Multiple Charges

🧮 Principle of Superposition

Let \( q_1, q_2, q_3, \ldots, q_n \) be the \( n \) point charges which are at distances \( r_1, r_2, r_3, \ldots, r_n \) respectively from a field point P.

If \( \vec{E_1}, \vec{E_2}, \vec{E_3}, \ldots, \vec{E_n} \) be the electric field intensities at a field point due to the point charges \( q_1, q_2, q_3, \ldots, q_n \) respectively, then the total electric field intensity due to assembly of \( n \) point charges will be:

Superposition Principle Formula

\[ \vec{E} = \vec{E_1} + \vec{E_2} + \vec{E_3} + \ldots + \vec{E_n} \]

where:

\[ \vec{E_1} = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1^2} \hat{r_1} \]

\[ \vec{E_2} = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2^2} \hat{r_2} \]

\[ \vdots \]

\[ \vec{E_n} = \frac{1}{4\pi\epsilon_0} \frac{q_n}{r_n^2} \hat{r_n} \]

General Formula for Multiple Charges

\[ \vec{E} = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{r_i} \]

This equation gives the total electric field intensity due to assembly of \( n \) point charges at a specific field point.

Problem 40: Two Point Charges

Question: Two point charges of magnitude \( q_1 = 2.16 \, \mu C, \, q_2 = 85.3 \, nC \) are 11.7 cm apart. Find magnitude of electric field that each produce on other. Find the magnitude of force on each charge?

Solution:

\( q_1 = 2.16 \, \mu C \)

\( = 2.16 \times 10^{-6} \, C \)

\( q_2 = 85.3 \, nC \)

\( = 85.3 \times 10^{-9} \, C \)

Distance \( r = 11.7 \, cm \)

\( = 11.7 \times 10^{-2} \, m \)

Electric field produce by \( q_1 = E_{12} = k \frac{q_1}{r^2} \)

\( = 9 \times 10^9 \times \frac{2.16 \times 10^{-6}}{(11.7 \times 10^{-2})^2} \)

\( = 0.14 \times 10^7 \frac{N}{C} \)

Electric field produce by \( q_2 = E_{21} = k \frac{q_2}{r^2} \)

\( = 9 \times 10^9 \times \frac{85.3 \times 10^{-9}}{(11.7 \times 10^{-2})^2} \)

\( = 5.62 \times 10^4 \frac{N}{C} \)

Force on charge \( q_1 = F_1 = k \frac{q_1 q_2}{r^2} \)

\( = 9 \times 10^9 \times \frac{2.16 \times 10^{-6} \times 85.3 \times 10^{-9}}{(11.7 \times 10^{-2})^2} \)

\( = 12.139 \times 10^{-2} \, N \)

Force on charge \( q_2 = F_2 = k \frac{q_1 q_2}{r^2} \)

\( = 9 \times 10^9 \times \frac{2.16 \times 10^{-6} \times 85.3 \times 10^{-9}}{(11.7 \times 10^{-2})^2} \)

\( = 12.139 \times 10^{-2} \, N \)

Electric Field Due to a Dipole

🧲 What is an Electric Dipole?

Two equal and opposite charges separated by a small distance form an electrical dipole. Consider two point charges \( +q \) and \( -q \) of equal magnitude lying distance \( d \) apart.

Step 1: Setup and Components

We want to determine the electric field intensity E due to a dipole at point P. The point P is at a distance \( x \) along the perpendicular bisector of the line joining the charges. Let the electric field intensities at point P due to the charges \( +q \) and \( -q \) be \( E_1 \) and \( E_2 \) respectively.

Step 2: Magnitude of Individual Fields

\[ E_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \]

\[ E_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \]

where \( r = \sqrt{x^2 + \left(\frac{d}{2}\right)^2} \)

Step 3: Vector Components

The vertical components of \( E_1 \) and \( E_2 \) cancel each other, while the horizontal components add up:

\[ E = E_1 \cos\theta + E_2 \cos\theta \]

\[ = 2E_1 \cos\theta \]

Step 4: Substitution

\[ E = 2 \times \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \times \frac{d/2}{r} \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{qd}{r^3} \]

Step 5: Final Expression

\[ E = \frac{1}{4\pi\epsilon_0} \frac{qd}{(x^2 + (d/2)^2)^{3/2}} \]

For \( x \gg d \), we can approximate:

\[ E \approx \frac{1}{4\pi\epsilon_0} \frac{qd}{x^3} \]

Electric Dipole Moment

The electric dipole moment \( \vec{p} \) is defined as:

\[ \vec{p} = q \vec{d} \]

where \( \vec{d} \) is the displacement vector from \( -q \) to \( +q \).

📐 Electric Field Along Axial Line

For a point on the axial line of the dipole:

\[ E = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \]

For a point on the equatorial line (perpendicular bisector):

\[ E = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \]

Electric Field Due to Continuous Charge Distributions

📊 Continuous Charge Distributions

In many practical situations, we encounter charges that are distributed continuously over a region of space rather than being concentrated at points. There are three types of continuous charge distributions:

Linear Charge Distribution

When charge is distributed along a line:

\[ \lambda = \frac{dq}{d\ell} \]

where \( \lambda \) is the linear charge density.

Surface Charge Distribution

When charge is distributed over a surface:

\[ \sigma = \frac{dq}{dA} \]

where \( \sigma \) is the surface charge density.

Volume Charge Distribution

When charge is distributed throughout a volume:

\[ \rho = \frac{dq}{dV} \]

where \( \rho \) is the volume charge density.

General Method for Continuous Distributions

To find the electric field due to a continuous charge distribution:

1. Divide the charge distribution into infinitesimal elements \( dq \).
2. Calculate the electric field \( d\vec{E} \) due to one such element at the field point.
3. Integrate \( d\vec{E} \) over the entire charge distribution to find the total field \( \vec{E} \).

General Formula for Continuous Charge

\[ \vec{E} = \int d\vec{E} \]

\[ = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r^2} \hat{r} \]

where the integration is carried out over the entire charge distribution.

Problem 29: Electric Field of a Ring

Question: A thin non-conducting rod of finite length L has a charge q spread uniformly along it. Show that E at point P on the perpendicular bisector is given by:

\[ E = \frac{1}{2\pi\epsilon_0} \frac{q}{y\sqrt{L^2 + 4y^2}} \]

Solution:

Linear charge density \( \lambda = \frac{q}{L} \)

Consider a small element of length \( dx \) at distance \( x \) from center

Charge on element \( dq = \lambda dx \)

Distance from P: \( r = \sqrt{x^2 + y^2} \)

Field due to element: \( dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \)

\( = \frac{1}{4\pi\epsilon_0} \frac{\lambda dx}{x^2 + y^2} \)

Vertical components cancel, horizontal components add: \( dE_x = dE \cos\theta \)

\( = dE \frac{y}{r} \)

\( E = \int_{-L/2}^{L/2} dE_x \)

\( = \frac{1}{4\pi\epsilon_0} \int_{-L/2}^{L/2} \frac{\lambda y dx}{(x^2 + y^2)^{3/2}} \)

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

Using standard integral: \( \int \frac{dx}{(x^2 + y^2)^{3/2}} = \frac{x}{y^2\sqrt{x^2 + y^2}} \)

\( E = \frac{\lambda y}{4\pi\epsilon_0} \left[ \frac{x}{y^2\sqrt{x^2 + y^2}} \right]_{-L/2}^{L/2} \)

\( = \frac{\lambda y}{4\pi\epsilon_0} \left[ \frac{L/2}{y^2\sqrt{(L/2)^2 + y^2}} - \frac{-L/2}{y^2\sqrt{(L/2)^2 + y^2}} \right] \)

\( = \frac{\lambda y}{4\pi\epsilon_0} \times \frac{L}{y^2\sqrt{(L/2)^2 + y^2}} \)

\( = \frac{\lambda L}{4\pi\epsilon_0 y \sqrt{(L/2)^2 + y^2}} \)

Substituting \( \lambda = \frac{q}{L} \): \( E = \frac{q}{4\pi\epsilon_0 y \sqrt{(L/2)^2 + y^2}} \)

\( = \frac{1}{2\pi\epsilon_0} \frac{q}{y\sqrt{L^2 + 4y^2}} \)

Torque on a Dipole

🔄 Dipole in Uniform Electric Field

When an electric dipole is placed in a uniform electric field, the two charges experience equal and opposite forces. These forces form a couple that tends to rotate the dipole to align it with the electric field.

Step 1: Forces on Charges

Force on \( +q \): \( \vec{F}_+ = q\vec{E} \)

Force on \( -q \): \( \vec{F}_- = -q\vec{E} \)

Step 2: Torque Calculation

The torque \( \tau \) about the center of the dipole is:

\[ \tau = \text{Force} \times \text{Perpendicular distance} \]

\[ = qE \times \frac{d}{2} \sin\theta + qE \times \frac{d}{2} \sin\theta \]

Step 3: Final Expression

\[ \tau = qEd \sin\theta \]

In vector form:

\[ \vec{\tau} = \vec{p} \times \vec{E} \]

⚖️ Stable and Unstable Equilibrium

The torque is zero when \( \theta = 0^\circ \) or \( \theta = 180^\circ \):

• At \( \theta = 0^\circ \): Stable equilibrium (dipole aligned with field)
• At \( \theta = 180^\circ \): Unstable equilibrium (dipole anti-aligned with field)

Energy of Dipole

🔋 Potential Energy of Dipole

The potential energy of an electric dipole in an external electric field is defined as the work done in rotating the dipole from a reference position to its current orientation.

Step 1: Work Done

The work done in rotating the dipole through a small angle \( d\theta \) is:

\[ dW = \tau d\theta \]

\[ = pE \sin\theta d\theta \]

Step 2: Total Work

The total work done in rotating the dipole from angle \( \theta_0 \) to \( \theta \) is:

\[ W = \int_{\theta_0}^{\theta} pE \sin\theta d\theta \]

\[ = pE (\cos\theta_0 - \cos\theta) \]

Step 3: Potential Energy

Taking \( \theta_0 = 90^\circ \) as reference (\( U = 0 \) when dipole is perpendicular to field):

\[ U = -pE \cos\theta \]

In vector form:

\[ U = -\vec{p} \cdot \vec{E} \]

📉 Energy Considerations

The potential energy is minimum when \( \theta = 0^\circ \) (stable equilibrium) and maximum when \( \theta = 180^\circ \) (unstable equilibrium).

\[ U_{\text{min}} = -pE \quad (\theta = 0^\circ) \]

\[ U_{\text{max}} = pE \quad (\theta = 180^\circ) \]

Solved Problems

Problem 36: Electric Field of a Ring

Question: A thin glass rod is bent into a semicircle of radius R. A charge +Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half. Find the electric field E at P, the center of the semicircle.

Solution:

Linear charge density \( \lambda = \frac{Q}{\pi R/2} \)

\( = \frac{2Q}{\pi R} \)

Consider a small element of length \( d\ell = R d\theta \)

Charge on element \( dq = \lambda d\ell \)

\( = \lambda R d\theta \)

Field due to element: \( dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{R^2} \)

\( = \frac{1}{4\pi\epsilon_0} \frac{\lambda R d\theta}{R^2} \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} d\theta \)

Components: \( dE_x = dE \cos\theta, \quad dE_y = dE \sin\theta \)

For upper half (\( 0 \to \pi/2 \)): \( E_{x1} = \int_0^{\pi/2} dE \cos\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \int_0^{\pi/2} \cos\theta d\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \)

\( E_{y1} = \int_0^{\pi/2} dE \sin\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \int_0^{\pi/2} \sin\theta d\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \)

For lower half (\( \pi/2 \to \pi \)): \( E_{x2} = \int_{\pi/2}^{\pi} dE \cos\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \int_{\pi/2}^{\pi} \cos\theta d\theta \)

\( = -\frac{\lambda}{4\pi\epsilon_0 R} \)

\( E_{y2} = \int_{\pi/2}^{\pi} dE \sin\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \int_{\pi/2}^{\pi} \sin\theta d\theta \)

\( = \frac{\lambda}{4\pi\epsilon_0 R} \)

Total field: \( E_x = E_{x1} + E_{x2} = 0 \)

\( E_y = E_{y1} + E_{y2} \)

\( = \frac{2\lambda}{4\pi\epsilon_0 R} \)

\( = \frac{\lambda}{2\pi\epsilon_0 R} \)

Substitute \( \lambda = \frac{2Q}{\pi R} \): \( E_y = \frac{2Q}{\pi R} \times \frac{1}{2\pi\epsilon_0 R} \)

\( = \frac{Q}{\pi^2 \epsilon_0 R^2} \)

So \( \vec{E} = \frac{Q}{\pi^2 \epsilon_0 R^2} \hat{j} \)

Problem 38: Electric Field of a Disk

Question: A disk of radius R has a uniform surface charge density \( \sigma \). Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk.

Solution:

Consider a ring of radius r and width dr

Area of ring \( dA = 2\pi r dr \)

Charge on ring \( dq = \sigma dA \)

\( = \sigma 2\pi r dr \)

Field due to ring: \( dE = \frac{1}{4\pi\epsilon_0} \frac{x dq}{(x^2 + r^2)^{3/2}} \)

\( = \frac{1}{4\pi\epsilon_0} \frac{x \sigma 2\pi r dr}{(x^2 + r^2)^{3/2}} \)

Total field: \( E = \int_0^R dE \)

\( = \frac{\sigma x}{2\epsilon_0} \int_0^R \frac{r dr}{(x^2 + r^2)^{3/2}} \)

Let \( u = x^2 + r^2 \), then \( du = 2r dr \Rightarrow r dr = \frac{du}{2} \)

\( E = \frac{\sigma x}{2\epsilon_0} \int_{x^2}^{x^2 + R^2} \frac{du/2}{u^{3/2}} \)

\( = \frac{\sigma x}{4\epsilon_0} \int_{x^2}^{x^2 + R^2} u^{-3/2} du \)

\( E = \frac{\sigma x}{4\epsilon_0} \left[ -2u^{-1/2} \right]_{x^2}^{x^2 + R^2} \)

\( = \frac{\sigma x}{2\epsilon_0} \left[ \frac{1}{x} - \frac{1}{\sqrt{x^2 + R^2}} \right] \)

\( = \frac{\sigma}{2\epsilon_0} \left[ 1 - \frac{x}{\sqrt{x^2 + R^2}} \right] \)

Frequently Asked Questions

❓ What is the physical significance of electric field lines?

Electric field lines provide a visual representation of the electric field. The tangent to a field line at any point gives the direction of the electric field at that point, and the density of field lines indicates the strength of the field.

❓ Why can't electric field lines cross each other?

If electric field lines crossed, it would mean that at the point of intersection, the electric field would have two different directions, which is physically impossible. The electric field at any point has a unique direction and magnitude.

❓ What happens to the electric field as we approach a point charge?

As we approach a point charge, the electric field strength increases and approaches infinity as the distance approaches zero. This is because the electric field due to a point charge follows an inverse square law.

❓ How does the electric field due to a dipole differ from that due to a single charge?

The electric field due to a dipole decreases as 1/r³ at large distances, while the field due to a single point charge decreases as 1/r². This is because the dipole consists of two equal and opposite charges that partially cancel each other's fields at large distances.

📚 Continue Your Physics Journey

Mastering electric fields is fundamental to understanding electromagnetism. These comprehensive notes based on Halliday, Resnick and Krane provide a solid foundation for further studies in physics including Gauss's Law and electric potential.

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© 2025 Physics Education Initiative | Chapter 28: The Electric Field

These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of electric fields based on Halliday, Resnick and Krane

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