Last Lecture

Gauss’s law

Using Gauss’s law for:

spherical symmetry

This lecture

Using Gauss’s law for:

line symmetry

plane symmetry

Conductors in electric fields

Coulomb’s law tutorial:Consider two positively charged particles, one of charge q0 (particle 0) fixed at the origin, and another of charge

q11 (particle 1) fixed on the y-axis at (0, d1, 0).

What is the net force F on particle 0 due to particle 1?  Express your answer (a vector) using any or all of k, q0, q1, d1, x-hat, y-hat, and z-hat.

Answer: F =  

Example of wrong answer:

( )

Part B What is the new net force on particle 0, from particle 1 and particle 2?

Answer: F =

 

Here is an example of a wrong answer - just a careless mistake.

This one presumably meant to put a bracket around the two terms, but didn’t.

Part DWhat is the net force on particle 0 due solely to this charge 3?

Answer: F =

The answer here was close but still missing a factor of ½. The message is that accuracy matters.

These are two-dimensional cross sections through three dimensional closed spheres and a cube.

Which of them has the largest flux through surfaces A to E?

Which has the smallest flux?

A B C D E

Answers: ΦB= ΦE > ΦA= ΦC= ΦD

The flux through a CLOSED surface depends only on the amount of enclosed charge, not the size or shape of the surface.

Example 21.2Field of a hollow spherical sphere

From Gauss’s law, E = 0 inside shell

Example 21.3Field of a point charge within a

shellDone on board

Read TIP: SYMMETRY MATTERS!

Line symmetry

Example 21.4 Field of a line of charge

rE

02

(line of charge)

A section of an infinitely long wire with a uniform linear charge density, .

Find an expression for E at distance r from axis of wire.

Applying Gauss’ Law: cylindrical symmetry

rE

02

(line of charge)

(Compare this result with that obtained using Coulomb’s law in Example 20.7, when wire is infinitely long.)

Applies outside any cylindrical charge distribution

Plane symmetryExample 21.6 Field of an infinite plane sheet of charge

p 357

Gaussian surface (cylinder)

Applying Gauss’ Law: planar symmetryA thin, infinite,

nonconducting sheet with uniform surface

charge density

Find E at distance r from sheet.

0

0

02

enclosedqEA EA

AEA EA

E

(sheet of charge)

02xE

Electric field due to plane of charge is

02xE

02xE

2 22 1

xE k

x R

02xE

By integration methods:

- which reduces to above formula for very large R(see problem 20.69)

CHECKPOINT: The figure shows two large parallel, nonconducting sheets with identical (positive) volume charge density. Rank the four labelled points according to the magnitude of the net electric field there, greatest first.

Answer:

C, D equal

B

A

A

B C D

21.5 Fields of arbitrary charge

distributions

CHECKPOINT: There is a certain net flux I through a Gaussian sphere of radius r enclosing an isolated charged particle. Suppose the Gaussian surface is changed to (a) a larger Gaussian sphere, (b) a Gaussian cube with edge length equal to r, and (c) a Gaussian cube with edge length 2r.

In each case is the net flux through the new Gaussian surface

A. greater thanB. less than, or C. equal to I ?

Answers: all equal as charge enclosed is the same

Find E at point P due to a finite line charge

Charge Q is uniformly distributed on a straight line segment of length L.

Choose axes and draw diagram

Write expression for |dE| due to dq

Find dEy

Integrate to find Ey

(what are the limits?)

Substitute trigonometric formulae

Repeat for Ex

When we must use Coulomb - no symmetry!

True or False?

1. If the net electric flux out of a closed surface is zero, the electric field must be zero everywhere on the surface.

2. If the net electric flux out of a closed surface is zero, the charge density must be zero everywhere inside the surface.

3. The electric field is zero everywhere within the material of a conductor in electrostatic equilibrium.

4. The tangential component of the electric field is zero at all points just outside the surface of a conductor in electrostatic equilibrium.

5. The normal component of the electric field is the same at all points just outside the surface of a conductor in electrostatic equilibrium.

True

True

False

False

False

True A False F