Chapter 22

Gauss’s Law

Goals for Chapter 22

• To use the electric field at a surface to determine the charge within the surface.

• To learn the meaning of electric flux and how to calculate it.

• To learn the relationship between the electric flux through a surface and the charge within the surface.

• To use Gauss’s law to calculate electric fields.

• To learn where the charge on a conductor is located.

Introduction• Would the girl’s hair

stand on end if she were inside the metal sphere?

• Symmetry properties play an important role in physics.

• Gauss’s law will allow us to do electric-field calculations using symmetry principles.

The Big Picture

 Gauss' law for electricity

 Gauss' law for magnetism

 Faraday's law of induction

 Ampere's law

Lorentz Force law

Introduction

Zero net charge inside a box

Charge and electric flux• Positive charge within the box produces outward

electric flux through the surface of the box, and negative charge produces inward flux.

What affects the flux through a box?

A HUMOROUS CHARGE

• A Buddhist monk approaches a hot dog vendor and tells him: "Make me one with everything".

The monk then gives the vendor a $5 bill for the $2 hot dog. After waiting for a few moments, he asks for change.

• The Buddhist hot dog vendor replies, " ah, but change comes from within”.

Electric FluxThe magnitude is “A” and the direction is

directed perpendicular to the area like a force normal.

A

E

Flux ( or FLOW) is a general term associated with a FIELD that is bound by a certain AREA. So ELECTRIC FLUX is any AREA that has a ELECTRIC FIELD passing through it.

We generally define an AREA vector as one that is perpendicular to the surface of the material.

Therefore, you can see in the figure that the AREA vector and the Electric Field vector are PARALLEL. This then produces a DOT PRODUCT between the 2 variables that then define flux.

Calculating electric flux

Electric flux through a diskA disk of radiusis oriented with its normal unit vector n atto a uniform electric fieldof magnitude. Since this isn’t a closed surface, it has no “inside” or “outside”. That’s why we have to specify the direction of. (a) What is the electric flux through the disk? (b) What is the flux through the disk if it is turned so thatis perpendicular to? (c) What is the flux through the disk if is parallel to?

Electric flux through a cube

The First Surface is:

Electric flux through a cube

The First Surface is:

Gauss’s Law• Gauss’s law is an alternative to

Coulomb’s law and is completely equivalent to it.

• Carl Friedrich Gauss formulated this law. We will use the form:

• Given the Electric field we can find the charge

• Given the Charge we can find the electric field!

"This is Gauss's Law. It is very powerful...but it doesn't work in

general”

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 /𝜖0

Electric flux through a sphere• State Gauss’s Law.

• Derive Coulomb's Electric

• Field.

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 /𝜖0

Point charge centered in a spherical surface• The flux through the sphere

is independent of the size of the sphere and depends only on the charge inside.

• The area goes as 4R2 • But the Field goes as 1/ 4R2 • So the product of A and E is

a constant =

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 /𝜖0

Gauss’s Law most general form (multiple charges)

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 /𝜖0

Rules for using Gauss’s Law

1. Chose a Surface such that the electric field vector and the area vector are either parallel or anti-parallel.

2. Chose a surface such that the electric field is constant everywhere on the surface.

3. Make sure you know which physical quantitiy you wish to calculate.

1. The electric field given an charge.

2. The charge given an electric field.

Positive and negative flux

Use the electric field to get flux

Derive the Coulomb field!

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 /𝜖0

Φ𝐸=𝐸 ∙ A⃗= 𝑞4𝜋 𝜖0𝑟

2 × 4𝜋𝑟 2=qenc

𝜖0

Φ𝐸=𝐸 ∙ A⃗

Positive and negative flux Φ𝐸=𝐸 ∙ A⃗

General form of Gauss’s law

A point charge outside a closed surface that encloses NO CHARGE. If an electric field line from the external charge enters

the surface at one point, it must leave at another!

Φ𝐸=𝐸 ∙ A⃗=𝑞𝑒𝑛𝑐

𝜖0

Φ𝐸=𝐸 ∙ A⃗=0

General form of Gauss’s law• What are the next Fluxes in each region?

General form of Gauss’s law• What are the next Fluxes in each region?

Demo – The floating metal!

With a cage and

Without a cage

End Lecture 1

Applications of Gauss’s law• Under electrostatic conditions, any excess charge on a conductor

resides entirely on its surface.

Applications of Gauss’s lawWe place a total positive chargeon a solid conducting sphere with radius . Find the electric field vector at ANY point inside or outside the sphere.

Field of a line charge

Field of a sheet of charge (pill box)The field follows from the fact that there are two surfaces and one charge density

Field between two parallel conducting plates

Charges on conductors

Net enclosed q must be zero!!

A conductor with a cavity

Careful with the concept of NET charge. This is the total charge on the metal at the start!

A solid conductor with a cavity carries a total charge of +7nC. A charge of -5nC is placed in the cavity. How much charge is on each surface (inner and outer).

Multiple Conductors Example

Multiple Conductors Example

Multiple Conductors Example

Multiple Conductors Example

A conductor with a cavity

Answer:

• NO: First understand why there is no net charge.

• Second: understand that the +Q that allows the outside to be at Qnet = 0 will be cancelled by the electrons from the ground plane.

• Third Since the outside is held at Q = 0 it must be true the no net Q will be enclosed…therefore no outer filed.

Testing Gauss’s law experimentally• Faraday’s ice pail experiment, which confirmed the validity of

Gauss’s law.

The Van de Graaff generator• The Van de Graaff generator, shown below, operates on the same

principle as in Faraday’s ice pail experiment.

Electrostatic shielding

• A conducting box (a Faraday cage) in an electric field shields the interior from the field.

Charge anyone?

Field at the surface of a conductor• The electric field at the

surface of any conductor is always perpendicular to the surface and has magnitude /0.

• Show that Coulombs law give the same result

Example: a non-conducting Sphere with a uniform Charge distribution

• What is the force on a charge just outside the shell?

• What is the force on a charge just inside the shell?

Summary and notes