Chapter 22- Gauss's Law and Electric Flux (Exam 2, Physics 1320)

Chapter 22: Gauss's Law

Introduction and Learning Objectives

This chapter introduces Gauss's Law, a fundamental principle in electromagnetism, and the concept of electric flux. By the end of this chapter, students should be able to:

• Determine the amount of charge within a closed surface by examining the electric field on the surface.

• Define electric flux and calculate it for various scenarios.

• Explain how Gauss's law relates electric flux through a closed surface to the charge enclosed by the surface.

• Apply Gauss's law to calculate the electric field due to symmetric charge distributions.

• Describe the location of charge on a charged conductor.

Electric Charge and Electric Flux

Definition of Electric Flux

Electric flux ($\Phi_E$) quantifies the amount of electric field passing through a given area. It is a useful concept for understanding how electric fields interact with surfaces, especially in the context of Gauss's law.

• Area ($A$): The surface through which the electric field passes. This area can be real or imaginary.

• Closed surface: A surface that completely encloses a volume (e.g., a sphere or box).

• Normal: Refers to a direction perpendicular to the surface area.

• Imaginary surfaces are often used to calculate the electric field of a charge distribution or to find the amount of charge enclosed within a volume.

Formula for Electric Flux (Uniform Field):

$\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta$

• $\vec{E}$: Electric field vector

• $\vec{A}$: Area vector (perpendicular to the surface)

• $\theta$: Angle between $\vec{E}$ and $\vec{A}$

• SI unit: $\text{N} \cdot \text{m}^2 / \text{C}$

Electric Flux Through a Closed Surface

The net electric flux through a closed surface depends on the net charge enclosed:

• Positive charge inside the surface: Outward flux (field lines exit the surface).

• Negative charge inside the surface: Inward flux (field lines enter the surface).

• No net charge inside the surface: Zero net flux (inward and outward fluxes cancel).

• Charges outside the surface: Do not contribute to net electric flux through the surface.

Example: If a box encloses a $+q$ charge, the electric field lines point outward, resulting in positive flux. If it encloses a $-q$ charge, the field lines point inward, resulting in negative flux. If both $+q$ and $-q$ are enclosed, their fluxes cancel.

Qualitative Statement of Gauss's Law

Gauss's Law Overview

Gauss's Law relates the net electric flux through a closed surface to the net charge enclosed by that surface. It is independent of the size or shape of the surface, provided it is closed.

• The net electric flux is directly proportional to the net amount of charge enclosed.

• Charges outside the surface do not affect the net flux.

Qualitative Statement: The net electric flux through a closed surface depends only on the net charge enclosed within the surface.

Mathematical Formulation of Electric Flux

Uniform Electric Field

For a uniform electric field passing through a flat surface:

$\Phi_E = EA \cos \theta$

• If $\theta = 0^\circ$ (field perpendicular to surface), $\Phi_E = EA$

• If $\theta = 90^\circ$ (field parallel to surface), $\Phi_E = 0$

Nonuniform Electric Field

For a nonuniform electric field, the electric flux is calculated using a surface integral:

$\Phi_E = \int \vec{E} \cdot d\vec{A} = \int E \cos \theta \, dA$

• This integral sums the contributions of the electric field over each infinitesimal area element $dA$.

Gauss's Law: Mathematical Statement

Integral Form of Gauss's Law

Gauss's law is expressed mathematically as:

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$

• $\oint$ denotes a surface integral over a closed surface.

• $Q_{\text{enc}}$: Net charge enclosed by the surface.

• $\varepsilon_0$: Permittivity of free space ($\varepsilon_0 \approx 8.85 \times 10^{-12} \text{C}^2/\text{N} \cdot \text{m}^2$)

Applications of Gauss's Law

Gauss's law can be used for any charge distribution and any closed surface. It is most useful when the system has sufficient symmetry (spherical, cylindrical, or planar) to simplify the calculation of the electric field.

• If the charge distribution is known and symmetric, Gauss's law can be used to find the electric field.

• If the electric field is known, Gauss's law can be used to find the charge distribution.

Conductors and Excess Charge

Charge Distribution on Conductors

When excess charge is placed on a solid conductor at rest, the charge resides entirely on the surface. This is because, in electrostatics, charges are not moving, and any internal electric field would cause movement. Thus, the interior of a conductor is field-free, and excess charge is found only on the surface.

• If the conductor contains a cavity, the situation changes and must be analyzed using Gauss's law.

Choosing a Gaussian Surface

Symmetry and Surface Selection

The choice of Gaussian surface depends on the symmetry of the problem:

• Spherical symmetry: Use a concentric spherical surface.

• Cylindrical symmetry: Use a coaxial cylindrical surface with flat ends perpendicular to the axis.

• Planar symmetry: Use a cylindrical surface ("pillbox") with flat ends parallel to the plane, or a cube.

To evaluate the electric field at a specific point, the Gaussian surface must include that point.

Examples and Applications

Example 1: Point Charge Enclosed by a Sphere

A charge $q = +3.00 \mu\text{C}$ is surrounded by an imaginary sphere of radius $r$ centered on the charge. The electric flux through the sphere is:

$\Phi_E = \frac{q}{\varepsilon_0}$

Example 2: Charged Conducting Sphere

Find the electric field inside, on the surface of, and outside a charged conducting sphere of radius $R$ with charge $Q$ placed on its surface:

• Inside ($r < R$): $E = 0$

• On the surface ($r = R$): $E = \frac{Q}{4\pi\varepsilon_0 R^2}$

• Outside ($r > R$): $E = \frac{Q}{4\pi\varepsilon_0 r^2}$

Example 3: Infinitely Long Thin Wire

For a wire with charge per unit length $\lambda$, the electric field at a distance $r$ from the wire is:

$E = \frac{\lambda}{2\pi\varepsilon_0 r}$

Example 4: Infinite Sheet of Charge

The electric field a distance $d$ away from a thin, flat, infinite sheet of surface charge density $\sigma$ is:

$E = \frac{\sigma}{2\varepsilon_0}$

Example 5: Two Oppositely Charged Plates

The electric field between two oppositely charged conducting plates is:

$E = \frac{\sigma}{\varepsilon_0}$

Example 6: Uniformly Charged Insulating Sphere

For a sphere of radius $R$ with total charge $Q$ distributed uniformly throughout its volume, the electric field at a distance $r$ from the center ($0 < r < R$) is:

$E = \frac{Q}{4\pi\varepsilon_0 R^3} r$

At the surface ($r = R$):

$E = \frac{Q}{4\pi\varepsilon_0 R^2}$

Outside ($r > R$):

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

Summary Table: Electric Field for Common Charge Distributions

Charge Distribution	Electric Field ($E$)	Notes
Point charge	$E = \frac{q}{4\pi\varepsilon_0 r^2}$	Outside the charge
Conducting sphere	$E = 0$ (inside), $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ (outside)	Charge resides on surface
Infinite line of charge	$E = \frac{\lambda}{2\pi\varepsilon_0 r}$	At distance $r$ from wire
Infinite sheet of charge	$E = \frac{\sigma}{2\varepsilon_0}$	On either side of sheet
Uniformly charged sphere	$E = \frac{Q}{4\pi\varepsilon_0 R^3} r$ (inside), $E = \frac{Q}{4\pi\varepsilon_0 r^2}$ (outside)	Charge distributed throughout volume

Key Points to Remember

• Electric flux measures the "flow" of electric field through a surface.

• Gauss's law links electric flux to the net charge enclosed by a surface.

• Symmetry is crucial for simplifying calculations using Gauss's law.

• Excess charge on conductors resides on the surface.

• Choosing the correct Gaussian surface is essential for problem-solving.

Additional info: These notes expand on the brief points and diagrams in the original slides, providing full academic context, definitions, and formulas for Physics 1320 students.