Gauss's Law Complete Guide: Electric Flux, Applications & Solved Problems | HRK Physics

Chapter 29: Gauss's Law

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

Gauss Law Electromagnetism Electric Flux Reading Time: 25 min

📋 Table of Contents

• 1. Introduction to Gauss's Law
• 2. Electric Flux
• 3. Electric Flux through Irregular Shapes
• 4. Statement of Gauss's Law
• 5. Differential Form of Gauss's Law
• 6. Integral Form of Gauss's Law
• 7. Applications of Gauss's Law
• 8. Electric Field due to Infinite Line of Charge
• 9. Electric Field due to Infinite Sheet of Charge
• 10. Electric Field due to Spherical Shell
• 11. Electric Field due to Solid Sphere
• 12. Conductors and Gauss's Law
• 13. Deduction of Coulomb's Law
• 14. Solved Problems
• Frequently Asked Questions

Introduction to Gauss's Law

⚡ Gauss's Law vs Coulomb's Law

Coulomb's law can always be used to calculate the electric field intensity \( \vec{E} \) for any discrete or continuous charge distribution of charges at rest. The sums or integrals might be complicated (and a computer might be needed to evaluate them numerically), but resulting electric field intensity \( \vec{E} \) can always be found.

In this chapter, we discuss an alternative to Coulomb's law, called Gauss's law, that provides a more useful and instructive approach to calculating the electric field in the situations having certain symmetries.

🔬 When to Use Gauss's Law

The number of situations that can directly be analyzed using Gauss's law is small, but those cases can be done with extraordinary ease. Although Gauss's law and Coulomb's law gives identical results in the cases in which both can be used, Gauss's law is considered a more fundamental equation than Coulomb's law.

It is fair to say that while Coulomb's law provides workhorse of electrostatics, Gauss's law provides the insight.

Electric Flux

📏 Definition of Electric Flux

The number of electric lines of force passing normally through a certain area is called the electric flux. It is measured by the product of area and the component of electric field intensity normal to the area. It is denoted by the symbol \( \Phi_e \).

Electric Flux Calculation

Consider a surface '\( S \)' placed in a uniform electric field of intensity '\( \vec{E} \)'. Let '\( A \)' be the area of the surface. The component of \( \vec{E} \) normal to the area \( A \) is \( E \cos \theta \).

Electric Flux Formula

The electric flux through the surface \( S \) is given by:

\[ \Phi_e = A \left( E \cos \theta \right) \]

\[ = EA \cos \theta \]

\[ = \vec{E} \cdot \vec{A} \]

Interpretation

Thus the electric flux is the scalar product of electric field intensity and the vector area. The SI unit of the electric flux is \( \frac{Nm^2}{C} \).

Electric Flux through Irregular Shapes

🔍 Flux Calculation for Complex Surfaces

Consider an object of irregular shape placed in a non-uniform electric field. We want to find out the expression of electric flux through this irregular shaped object.

Step 1: Divide the Surface

We divide the surface into n number of small patches having area \( \Delta A_1, \Delta A_2, \Delta A_3, \ldots, \Delta A_n \).

Step 2: Field Components

Let \( \vec{E}_1, \vec{E}_2, \ldots, \vec{E}_n \) are the electric field intensities which makes angle \( \theta_1, \theta_2, \theta_3, \ldots, \theta_n \) with the normal to the area elements \( \Delta A_1, \Delta A_2, \Delta A_3, \ldots, \Delta A_n \), respectively.

Step 3: Total Flux

If \( \Phi_1, \Phi_2, \Phi_3, \ldots, \Phi_n \) be the electric flux through \( \Delta A_1, \Delta A_2, \Delta A_3, \ldots, \Delta A_n \), then the total electric flux \( \Phi_e \) will be:

\[ \Phi_e = \Phi_1 + \Phi_2 + \Phi_3 + \ldots + \Phi_n \]

\[ = E_1 (\Delta A_1 \cos \theta_1) + E_2 (\Delta A_2 \cos \theta_2) + E_3 (\Delta A_3 \cos \theta_3) + ...... + E_n (\Delta A_n \cos \theta_n) \]

\[ = E_1 \Delta A_1 \cos \theta_1 + E_2 \Delta A_2 \cos \theta_2 + E_3 \Delta A_3 \cos \theta_3 + ...... + E_n \Delta A_n \cos \theta_n \]

\[ = E_1 \Delta A_1 + E_2 \Delta A_2 + E_3 \Delta A_3 + ...... + E_n \Delta A_n \]

Step 4: Summation Form

\[ \Phi_e = \sum_{i=1}^n E_i \Delta A_i \]

Step 5: Integral Form

When \( n \to \infty \), or \( \Delta A \to 0 \), then the sigma is replaced by the surface integral:

\[ \Phi_e = \int_S E_i dA \]

By convention, the outward flux is taken as positive and inward flux is taken as negative.

Statement of Gauss's Law

📜 Formal Statement

The total electric flux through any closed surface is \( \frac{1}{\varepsilon_0} \) times the total charge enclosed by the surface.

Mathematical Formulation

The Gauss's law gives the relation between total flux and total charge enclosed by the surface:

\[ \Phi_e = \frac{q}{\varepsilon_0} \quad \text{(1)} \]

where \( q \) is the net charge enclosed by the surface.

Flux Integral

Also, from the definition of flux:

\[ \Phi_e = \oint E_i dA \quad \text{(2)} \]

Final Form

Comparing (1) and (2), we have:

\[ \oint E_i dA = \frac{q}{\varepsilon_0} \]

Thus we can describe the Gauss's law as: The surface normal integral of electric field intensity is equal to \( \frac{1}{\varepsilon_0} \) times the total charge enclosed by the surface.

Problem 5: Flux through Cubical Surface

Question: A point charge of 1.84 \( \mu C \) is at center of cubical Gaussian surface of 55 cm edge. Find flux through the surface.

Solution:

Given: \( q = 1.84 \mu C = 1.84 \times 10^{-6} C \)

Applying Gauss law:

\[ \Phi_e = \frac{q}{\varepsilon_0} \]

\[ = \frac{1.84 \times 10^{-6}}{8.85 \times 10^{-12}} \]

\[ = 2.07 \times 10^5 \frac{N - m^2}{C} \]

Differential Form of Gauss's Law

📐 From Integral to Differential Form

If the charge is distributed into a volume having uniform volume charge density 'ρ', then charge enclosed q by Gaussian surface is described by expression:

\[ q = \int_v ρ \, dv \]

Step 1: Start with Gauss's Law

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

\[ \Rightarrow \oint \vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0} \int_v ρ \, dv \quad \text{(1)} \]

Step 2: Apply Divergence Theorem

By Gauss's Divergence theorem:

\[ \oint \vec{E} \cdot d\vec{A} = \int_v \text{div} \, E \, dv \quad \text{(2)} \]

Step 3: Compare Equations

Comparing (1) and (2), we have:

\[ \int_v \text{div} \, E \, dv = \frac{1}{\epsilon_0} \int_v ρ \, dv \]

\[ \Rightarrow \int_v \text{div} \, E \, dv - \frac{1}{\epsilon_0} \int_v ρ \, dv = 0 \]

\[ \Rightarrow \int_v \left( \text{div} \, E - \frac{1}{\epsilon_0} ρ \right) dv = 0 \]

Step 4: Final Differential Form

\[ \text{div} \, E - \frac{1}{\epsilon_0} ρ = 0 \]

\[ \Rightarrow \text{div} \, E = \frac{1}{\epsilon_0} ρ \]

This is differential form of Gauss's law.

Integral Form of Gauss's Law

📊 Integral Formulations

By Gauss's law:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

where 'q' is the total charge enclosed.

Volume Charge Distribution

If the charge is uniformly distributed into a volume having charge density 'ρ', then:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0} \int_v ρ \, dv \quad \text{(1)} \]

Surface Charge Distribution

If the charge is uniformly distributed over a surface having a surface charge density 'σ', then:

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} = \frac{1}{\epsilon_0} \int_s σ \, dA \quad \text{(2)} \]

📝 Summary

Equation (1) and (2) are the integral form of Gauss's law for volume and surface charge distributions respectively.

Applications of Gauss's Law

🔧 When to Apply Gauss's Law

Gauss's law can be used to calculate the electric field intensity due to certain charge distributions if the charge distribution has the greater symmetry.

Infinite Line of Charge

Cylindrical symmetry with uniform linear charge density

Infinite Sheet of Charge

Planar symmetry with uniform surface charge density

Spherical Shell

Spherical symmetry with uniform surface charge density

Solid Sphere

Spherical symmetry with uniform volume charge density

Electric Field due to Infinite Line of Charge

📏 Setup and Symmetry

Consider a section of infinite line of charge having uniform linear charge density '\( \lambda \)'. We want to find out electric field intensity at any point '\( P \)' which is at distance '\( r \)' from the wire.

Step 1: Choose Gaussian Surface

We choose a cylindrical Gaussian surface of radius '\( r \)' and length '\( L \)' such that the point '\( P \)' lies on the surface of cylinder.

Step 2: Flux Calculation

The electric flux through the Gaussian surface is:

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} \]

\[ = \int_{\text{curved}} \vec{E} \cdot d\vec{A} + \int_{\text{top}} \vec{E} \cdot d\vec{A} + \int_{\text{bottom}} \vec{E} \cdot d\vec{A} \]

\[ = E \cdot 2\pi r L + 0 + 0 \]

\[ = E \cdot 2\pi r L \]

Step 3: Charge Enclosed

The charge enclosed by the Gaussian surface is:

\[ q = \lambda L \]

Step 4: Apply Gauss's Law

\[ \Phi_e = \frac{q}{\epsilon_0} \]

\[ E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0} \]

\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]

📐 Vector Form

In vector form:

\[ \vec{E} = \frac{\lambda}{2\pi \epsilon_0 r} \hat{r} \]

where \( \hat{r} \) is the unit vector directed radially outward from the line charge.

Electric Field due to Infinite Sheet of Charge

📄 Setup and Symmetry

Consider an infinite sheet of charge having uniform surface charge density '\( \sigma \)'. We want to find out electric field intensity at any point '\( P \)' which is at distance '\( r \)' from the sheet.

Step 1: Choose Gaussian Surface

We choose a cylindrical Gaussian surface (pillbox) of cross-sectional area '\( A \)' such that the point '\( P \)' lies on the surface of cylinder.

Step 2: Flux Calculation

The electric flux through the Gaussian surface is:

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} \]

\[ = \int_{\text{left}} \vec{E} \cdot d\vec{A} + \int_{\text{right}} \vec{E} \cdot d\vec{A} + \int_{\text{curved}} \vec{E} \cdot d\vec{A} \]

\[ = EA + EA + 0 \]

\[ = 2EA \]

Step 3: Charge Enclosed

The charge enclosed by the Gaussian surface is:

\[ q = \sigma A \]

Step 4: Apply Gauss's Law

\[ \Phi_e = \frac{q}{\epsilon_0} \]

\[ 2EA = \frac{\sigma A}{\epsilon_0} \]

\[ E = \frac{\sigma}{2\epsilon_0} \]

📐 Vector Form

In vector form:

\[ \vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n} \]

where \( \hat{n} \) is the unit vector normal to the sheet.

Electric Field due to Spherical Shell

🔘 Setup and Symmetry

Consider a thin spherical shell of radius '\( R \)' having uniform surface charge density '\( \sigma \)'. We want to find out electric field intensity at any point '\( P \)' which is at distance '\( r \)' from the center of the shell.

Case 1: Point Outside the Shell (r > R)

We choose a spherical Gaussian surface of radius '\( r \)' concentric with the shell.

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 \]

\[ q = \sigma \cdot 4\pi R^2 \]

\[ E \cdot 4\pi r^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0} \]

\[ E = \frac{\sigma R^2}{\epsilon_0 r^2} \]

Since \( \sigma = \frac{q}{4\pi R^2} \), we get:

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

Case 2: Point Inside the Shell (r < R)

We choose a spherical Gaussian surface of radius '\( r \)' inside the shell.

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 \]

\[ q = 0 \]

\[ E \cdot 4\pi r^2 = 0 \]

\[ E = 0 \]

📊 Summary

For a charged spherical shell:

• Outside (r > R): \( E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \)
• Inside (r < R): \( E = 0 \)

Electric Field due to Solid Sphere

🔵 Setup and Symmetry

Consider a solid sphere of radius '\( R \)' having uniform volume charge density '\( \rho \)'. We want to find out electric field intensity at any point '\( P \)' which is at distance '\( r \)' from the center of the sphere.

Case 1: Point Outside the Sphere (r > R)

We choose a spherical Gaussian surface of radius '\( r \)' concentric with the sphere.

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 \]

\[ q = \rho \cdot \frac{4}{3}\pi R^3 \]

\[ E \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi R^3}{\epsilon_0} \]

\[ E = \frac{\rho R^3}{3\epsilon_0 r^2} \]

Since \( \rho = \frac{q}{\frac{4}{3}\pi R^3} \), we get:

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

Case 2: Point Inside the Sphere (r < R)

We choose a spherical Gaussian surface of radius '\( r \)' inside the sphere.

\[ \Phi_e = \oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 \]

\[ q = \rho \cdot \frac{4}{3}\pi r^3 \]

\[ E \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\epsilon_0} \]

\[ E = \frac{\rho r}{3\epsilon_0} \]

Since \( \rho = \frac{q}{\frac{4}{3}\pi R^3} \), we get:

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

📊 Summary

For a uniformly charged solid sphere:

• Outside (r > R): \( E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \)
• Inside (r < R): \( E = \frac{1}{4\pi\epsilon_0} \frac{q r}{R^3} \)

Conductors and Gauss's Law

⚡ Properties of Conductors

A conductor is a material in which some of the electrons are free to move. These free electrons are not bound to any particular atom. In electrostatic conditions, the following properties hold for conductors:

Property 1

The electric field is zero everywhere inside the conductor.

\[ E_{\text{inside}} = 0 \]

Property 2

Any net charge on an isolated conductor resides entirely on its surface.

Property 3

The electric field just outside a charged conductor is perpendicular to the surface and has magnitude \( \frac{\sigma}{\epsilon_0} \).

\[ E_{\text{surface}} = \frac{\sigma}{\epsilon_0} \]

🔍 Proof Using Gauss's Law

Consider a Gaussian surface just inside the conductor surface. Since \( E = 0 \) inside the conductor, the flux through this surface is zero. By Gauss's law, the net charge enclosed must be zero. Therefore, any excess charge must reside on the surface.

Deduction of Coulomb's Law

🔄 Gauss's Law to Coulomb's Law

We can deduce Coulomb's law from Gauss's law. Consider a point charge '\( q \)' placed at the origin. We want to find the electric field at a point '\( P \)' at distance '\( r \)' from the charge.

Step 1: Choose Gaussian Surface

We choose a spherical Gaussian surface of radius '\( r \)' centered at the charge.

Step 2: Apply Gauss's Law

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \]

Step 3: Calculate Flux

Due to spherical symmetry, \( \vec{E} \) is radial and constant in magnitude over the Gaussian surface:

\[ \oint \vec{E} \cdot d\vec{A} = E \oint dA = E \cdot 4\pi r^2 \]

Step 4: Equate and Solve

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

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

Step 5: Force on Test Charge

The force on a test charge \( q_0 \) placed at point P is:

\[ F = q_0 E = \frac{1}{4\pi\epsilon_0} \frac{q q_0}{r^2} \]

This is Coulomb's law.

Solved Problems

Problem 16: Electric Field of a Solid Sphere

Question: A solid non-conducting sphere of radius R has a non-uniform charge distribution of volume charge density \( \rho = \rho_0 r \), where \( \rho_0 \) is a constant and r is the distance from the center of sphere. Show that:

(a) The total charge on the sphere is \( Q = \pi \rho_0 R^4 \)

(b) The electric field inside the sphere has a magnitude given by \( E = \frac{Q r^2}{4\pi\epsilon_0 R^4} \)

Solution:

(a) Total charge:

\[ Q = \int \rho dV \]

\[ = \int_0^R (\rho_0 r) (4\pi r^2 dr) \]

\[ = 4\pi \rho_0 \int_0^R r^3 dr \]

\[ = 4\pi \rho_0 \left[ \frac{r^4}{4} \right]_0^R \]

\[ = \pi \rho_0 R^4 \]

(b) Electric field inside:

Using Gauss's law: \( \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} \)

\[ E \cdot 4\pi r^2 = \frac{1}{\epsilon_0} \int_0^r \rho dV \]

\[ = \frac{1}{\epsilon_0} \int_0^r (\rho_0 r') (4\pi r'^2 dr') \]

\[ = \frac{4\pi \rho_0}{\epsilon_0} \int_0^r r'^3 dr' \]

\[ = \frac{4\pi \rho_0}{\epsilon_0} \left[ \frac{r'^4}{4} \right]_0^r \]

\[ = \frac{\pi \rho_0 r^4}{\epsilon_0} \]

\[ E = \frac{\rho_0 r^2}{4\epsilon_0} \]

Substitute \( \rho_0 = \frac{Q}{\pi R^4} \):

\[ E = \frac{Q r^2}{4\pi\epsilon_0 R^4} \]

Problem 28: Electric Field of a Cylinder

Question: A long solid non-conducting cylinder of radius R has a non-uniform volume charge density given by \( \rho_0 r \) for \( r \leq R \), where \( \rho_0 \) is a constant and r is the distance from the axis of cylinder. Find the electric field at a point inside the cylinder.

Solution:

Choose a cylindrical Gaussian surface of radius r and length L:

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} \]

\[ E \cdot 2\pi r L = \frac{1}{\epsilon_0} \int_0^r \rho dV \]

\[ = \frac{1}{\epsilon_0} \int_0^r (\rho_0 r') (2\pi r' L dr') \]

\[ = \frac{2\pi L \rho_0}{\epsilon_0} \int_0^r r'^2 dr' \]

\[ = \frac{2\pi L \rho_0}{\epsilon_0} \left[ \frac{r'^3}{3} \right]_0^r \]

\[ = \frac{2\pi L \rho_0 r^3}{3\epsilon_0} \]

\[ E = \frac{\rho_0 r^2}{3\epsilon_0} \]

Frequently Asked Questions

❓ What is the main advantage of using Gauss's law over Coulomb's law?

Gauss's law provides a much simpler way to calculate electric fields for charge distributions with high symmetry (spherical, cylindrical, or planar). While Coulomb's law can always be used, the calculations can become very complex for symmetric distributions, whereas Gauss's law provides elegant solutions with minimal computation.

❓ Why must the Gaussian surface be closed?

A closed surface is necessary because Gauss's law relates the total flux through the entire surface to the total charge enclosed. If the surface weren't closed, we couldn't properly account for all the field lines entering and leaving the region containing the charge.

❓ Can Gauss's law be used for any charge distribution?

While Gauss's law is always true, it's only practically useful for calculating electric fields when the charge distribution has sufficient symmetry. For asymmetric distributions, the integral in Gauss's law becomes difficult to evaluate, and Coulomb's law (or numerical methods) must be used instead.

❓ Why is the electric field inside a conductor zero in electrostatic conditions?

In a conductor, charges are free to move. If there were an electric field inside, it would exert forces on the charges, causing them to move until they redistribute themselves in such a way that the field inside becomes zero. This redistribution happens almost instantaneously in good conductors.

📚 Continue Your Physics Journey

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

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These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of Gauss's Law based on Halliday, Resnick and Krane

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