Chapter 24

1. Review on Chapter 23

2. From Coulomb's Law to Gauss’s Law

3. Applications of Gauss’s Law

Review on Chapter 23: Coulomb’s Law and the electric field definition

Coulomb’s Law: the force between two point charges

The electric field is defined as

and is represented through field lines. The force a charge experiences in an

electric filed

1 212 122

ˆe

q qk

rF r

oq

FE

e qF E

Two examplesExample 23.9 (page 662) Example 23.10 (page 663)

From Coulomb’s Law to Gauss’s law

Try to calculate the electric field of A point charge An infinitely long straight wire with evenly distributed charge A wire loop A round disk An infinitely large plane A solid sphere with evenly distributed charge

Are there other ways to calculate electric field generated from a charge distribution?

Electric field is generated by source charges, are there ways to connect electric field directly with these source charges?

The answer is YES!

Some preparation:Electric Flux through a perpendicular plane

Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the electric field:

ΦE = EA Compare to a water flux

in a tube: ΦW = –V1A1= V2A2

This sign means water flows into the tube, by convention.

Electric Flux, plane with an angle θ When the field lines make an

angle θ with the direction (i.e., the normal) of the surface, the flux is calculated as:

And the electric field E has to be a constant all over the area A.

Question: when this is not the case, what do you do the get the flux?

AE

cosEAEAE

Review on math: 1. direction of a surface is defined as the

(outwards) normal to that surface.2. Dot product of two vectors.

Electric Flux, General In the more general case,

look at a small area element

In general, this becomes

cosE i i i i iE A θ E A

0

surface

limi

E i iA

E

E A

d

E A

The surface integral means the integral must be evaluated over the surface in question

In general, the value of the flux will depend both on the field pattern and on the surface

When the surface is closed, the direction of the surface (i.e. the normal of it) points outwards.

The unit of electric flux is N.m2/CReview on math: Integral over a surface.

Example 1: flux through a cube of a uniform electric field

The field lines pass through two surfaces perpendicularly and are parallel to the other four surfaces

For side 1, ΦE = -El 2

For side 2, ΦE = El 2

For the other sides,

ΦE = 0 Therefore, Φtotal = 0

Example 2: flux through a sphere with a charge at its center.

From Coulomb’s Law to Gauss’s Law A positive point charge, q, is located at the

center of a sphere of radius r According to Coulomb’s Law, the magnitude

of the electric field everywhere on the surface of the sphere is

The field lines are directed radially outwards and are perpendicular to the surface at every point, so

Combine these two equations, we have

24AE rEdAEEdAdAEd nE

2r

qkE e

0

22

2 444

qqkr

r

qkrE eeE

The Gaussian Surface and Gauss’s Law Closed surfaces of various shapes can surround the

charge Only S1 is spherical The flux through all other surfaces (S2 and S3) are the

same. These surfaces are all called the Gaussian Surface.

Gauss’s Law (Karl Friedrich Gauss, 1777 – 1855): The net flux through any closed surface surrounding a

charge q is given by q/εo and is independent of the shape of that surface

The net electric flux through a closed surface that surrounds no charge is zero

Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as

Gauss’s Law connects electric field with its source charge

0

AE

qdE

0

2121 A)EE(AE

...qq

d...dE

Gauss’s Law – Summary

Gauss’s law states

qin is the net charge inside the Gaussian surface represents the electric field at any point on the

surface is the total electric field at a point in space and may have

contributions from charges both inside and outside of the surface

Although Gauss’s law can, in theory, be solved to find for any charge configuration, in practice it is limited to a few symmetric situations

E

E

E

0

AE

inE

qd

Preview sections and homework 1/27, due 2/3 Preview sections:

Section 24.3 Section 24.4

Homework: Problem 4, page 687. Problem 9, page 687. (optional = do it if you find it fun, or would like to challenge

yourself) Problem 11, page 687. (optional): On an insulating ring of radius R there evenly

distributed 73 point charges, each with a charge Q =+1 μC. The charges are fixed on the ring and cannot move. There is a bug with charge q = -0.1 μC sits at the center of the ring, and enjoys zero net force on it. When one of the charge Q is removed from the ring, what is the net force of the remaining charges exert on the poor bug?