lecture04 electric flux, gauss' law; conductors and electric fields

8/3/2019 Lecture04 Electric Flux, Gauss' Law; Conductors and Electric Fields

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adapted from http://www.nearingzero.net(nz128)


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Todays agenda:

Announcements.

Electric flux.You must be able to calculate the electric flux through a surface.

Gauss Law.You must be able to use Gauss Law to calculate the electric field of a high-symmetrycharge distribution.

Cultural enlightenment time.You must be culturally enlightened by this lecture.

Conductors in electrostatic equilibrium.You must be able to use Gauss law to draw conclusions about the behavior of chargedparticles on, and electric fields in, conductors in electrostatic equilibrium.


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Announcements

y Exam 1 is Tuesday, September 20, 5:00-6:15 pm.

If you are participating in an official Missouri S&T event or in aclass that offers no makeups, a sponsor may administer the testand provide test security. See Taking a Test at a Different

Time/Place on this web page.

In case of a officially-scheduled course conflict, there is an optionto take the test from 5:30-6:45 on Test Day. Follow theinstructions in paragraph 3 of the above link.

If one of the above two circumstances applies to you, I will needa memo from you (handwritten is OK), signed by you and thefaculty sponsor or course instructor, no later than end of the 1:00lecture on the Wednesday before the exam (Sep. 14 for Exam 1).

E-mails from both you and your sponsor/instructor are also OK.


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Announcements

E-mail me by the 5:00 pm Friday, September 2 if you have atime conflict for Exam 1.

Please provide these details: (1) dates of conflict, (2) nature ofconflict, (3) name of responsible person (other courseinstructor, job supervisor, etc.).

There are no guarantees, but hopefully the other person involved and I can work out something

agreeable.


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Announcements (continued)

According to Testing Center guidelines: students areresponsible for making appointments a minimum of sevenbusiness days prior to date that the test will be

administered. (Both you and I must make an appointment.)

y The Missouri S&TTesting Center provides accommodations forstudents with special needs or disabilities.

To ensure that we meet these guidelines, I need to beinformed no later than Friday, September 9, that you willrequire Testing Center accommodations for Exam 1. I will then

request the necessary accommodations.

Also provide me with your accommodation letter and MissouriS&T e-mail address by the end of the 1:00 Wednesday lectureprior to the test (September 14 for Exam 1).


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Homework problem 22.24 (due tomorrow):

Assume that the charge distribution on the insulator isspherically symmetric and concentric with the sphericalcavity.

Homework for Tuesday, September 6:

21: 46, 55, 90ab, 96; 22: 37, 43


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If you are going to miss recitation, let your recitationinstructor know, but that will not necessarily excuse youfrom boardwork and will not excuse you from turning inhomework.

Find a way to get your homework to your recitationinstructor. Have a friend take it. Scan and e-mail it.Photograph and e-mail it (keep file sizes down!). Take apicture with your cell phone camera and e-mail it. (Clearthis with your recitation instructor so he/she

knows you didnt take a photo of someone elseswork.)


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Announcements

Know your recitation section letter!

The walk-in tutors in 208 Norwood are available from 7-9 pmMonday through Thursday.


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Todays agenda:

Announcements.






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Gauss Law

Electric Flux

We have used electric field lines to visualize electric fields andindicate their strength.

We are now going to count* thenumber of electric field lines passing

through a surface, and use thiscount to determine the electric field.

E

*There are 3 kinds of people in this world: those who can count, and those who cant.


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The electric flux passing through a surface is the number ofelectric field lines that pass through it.

Because electric field lines are drawnarbitrarily, we quantify electric flux

like this: *E=EA, except that

If the surface is tilted, fewer lines cutthe surface.

EA

Later well learn about magnetic flux, which is

why I will use the subscript E on electric flux.

E

U


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E

U

U A

The amount of surface perpendicular

to the electric field is AcosU.

AEffective = A cos U so *E = EAEffective = EA cosU.

We define A to be a vector having amagnitude equal to the area of thesurface, in a direction normal to thesurface.

Therefore, the amount of surface area effectively cutthrough by the electric field is AcosU.

Remember the dot product from Physics 23?E

E A* ! &&


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If the electric field is not uniform, or the surface is not flat

divide the surface intoinfinitesimal surfaceelements and add theflux through each

dAE

i

E i iA 0

i

lim E A( p

* ! (&&

E E dA* !

&&

(A

Remember, the direction of dAis normal to the surface.


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If the surface is closed (completely encloses a volume)

E

we count* lines goingout as positive and linesgoing in as negative

E E dA* ! &&dA

a surface integral, therefore adouble integral

*There are 10 kinds of people in this world: those who can count in binary, and those who cant.

For a closed surface, dA is normalto the surface and always points

away from the inside.


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What the *!@* is this thing?

Nothing to panic about!

The circle just reminds you

to integrate over a closedsurface.


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Question: you gave me five different equations for electric flux.Which one do I need to use?

E E dA* ! &&

E E dA* ! &&

EE A* !

&&

EEAcos* ! U

EEA* !

Answer: use the simplest (easiest!) one that works.

Flat surface, E `` A, E constant over surface. Easy!

Flat surface, E not`` A, E constant over surface.

Flat surface, E not`` A, E constant over surface.

Surface not flat, E not uniform. Avoid, if possible.

Closed surface. Most general. Most complex.

If the surface is closed, you may be able to break it up into

simple segments and still use *E=EA

for each segment.


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Example: Calculate the electric flux through a cylinder with itsaxis parallel to the electric field direction.

E

To be worked at the blackboard


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Possible Homework Hint! (not applicable every semester)

EE A* !

&&

If you know E and A for each surface enclosing a volume, thesimplest way to calculate the total *E might be to calculate

for each surface, and add up all the *Es.


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Todays agenda:

Announcements.






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If there were a + charge inside the cylinder, there would bemore lines going out than in.

If there were a - charge inside the cylinder, there would bemore lines going in than out

which leads us to

E+-


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Gauss Law

Mathematically*, we express the idea of the last two slides as

enclosedE

o

qE dA* ! !

I&&

Gauss LawAlways true, not always useful.

We will find that Gauss law gives a simple way to calculateelectric fields for charge distributions that exhibit a high degreeof symmetry

and save more complex charge distributions for advanced

classes.

*Mathematics is the Queen of the Sciences.Karl Gauss


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To see how this works, lets do an example.

Example: use Gauss Law to calculate the electric field from anisolated point charge q.

To apply Gauss Law, we construct a Gaussian Surface

enclosing the charge.

The Gaussian surface must have the same symmetry as thecharge distribution.

For this example, choose for our Gaussian surface a sphere ofradius r, with the point charge at the center.

Ill work the rest of the example on the blackboard.


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Homework Hint!

For tomorrows homework, you may not apply the equation forthe electric field of a point charge

to a distribution of charges. Instead, use Gauss Law. Later Iwill give you permission to use the point charge equation forcertain specific charge distributions.

2

k qE

r!


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Strategy for Solving Gauss Law Problems

y Evaluate the surface integral (electric flux).

y Determine the charge inside the Gaussian surface.

y Solve for E.

If I were Dr. Bieniek, this would be a litany.

y Select a Gaussian surface with symmetry that matches thecharge distribution.

Use symmetry to determine the direction of on the Gaussian surface.

You want to be constant in magnitude and everywhere perpendicular

to the surface, so that

or else everywhere parallel to the surface so that .

E

&

E dA E dA !

&&

E dA 0 !&&

E&


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Quiz time (maybe for points, maybe just for practice!)


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Example: calculate the electric field outside a long cylinder offinite radius R with a uniform volume charge density V spread

throughout the volume of the cylinder.

The cylinder being long and the radius finite are codewords that tell you to neglect end effects from the cylinder(i.e., assume it is infinitely long).

Know how to interpret code words when exam time comes!

Lets go through this a step at a time (work to be shown at board; skip to here).

y Select a Gaussian surface with symmetry that matches thecharge distribution.

Pick a cylinder of length L and radius r, concentric with thecylinder of the problem.


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Already done!

y Use symmetry to determine the direction of E on the Gaussiansurface.

Electric field points radially away from cylinder, and magnitudedoes not depend on direction.

y Evaluate the surface integral (electric flux).

Surface integral is just E times the curved area.

y Draw the Gaussian surface so that at all points on theGaussian surface either or .E dA E dA !

&&E dA 0 !

&&


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y Determine the charge inside the Gaussian surface.

The charge inside is just the volume of charged cylinder insidethe Gaussian surface, times the charge per volume.

y Solve for E.

2

0

RE

2 r

V!

I

More examples at end of lecture, if time permits.


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Example: use Gauss Law to calculate the electric field due to along line of charge, with linear charge density P.

Example: use Gauss Law to calculate the electric field due toan infinite sheet of charge, with surface charge density W.

These are easy using Gauss Law (remember what a pain theywere in the previous chapter). Study these examples and othersin your text!

sheet

0

E .2

W!

I

line

0

E .2 r

P!TI


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Todays agenda:

Announcements.






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Cultural Enlightenment* Time

The top 5 reasons why we make you learn Gauss Law:

5. You can solve (high-symmetry) problems with it.

4. Its good for you. Its fun! What more can you ask!

3. Its easy. Smart physicists go for the easy solutions.

2. IfI had to learn it, you do too.

And the number one reason

will take a couple of slides to present

*This will not be tested on the exam.


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You have learned the integral form of Gauss Law:

This will not be tested on the exam.

enclosed

o

q

E dA ! I

&&

This equation can also be written in differential form:

0

E V !I

& &

a 3-dimensional derivative operator

Now you can see we are on the trail of something Really Big


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This will not be tested on the exam.

The Missouri S&T Society of Physics Students T-Shirt!


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Todays agenda:

Announcements.






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Homework hints buried in the next 3 pages!

Conductors in Electrostatic Equilibrium

Electrostatic equilibrium means there is no net motion of tnecharges inside the conductor.

The electric field inside the conductor must be zero.

Any excess charge must reside on the outside surface of the

conductor.

If this were not the case, charges would move.

Apply Gauss law to a Gaussian surface just inside theconductor surface. The electric field is zero, so the net chargeinside the Gaussian surface is zero.


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The electric field just outside a charged conductor must beperpendicular to the conductors surface.

Otherwise, the component of theelectric field parallel to the surface

would cause charges to accelerate.

The magnitude of the electric field just outside a charged

conductor is equal to `W/I0, where `W` is the magnitude of thelocal surface charge density.

A simple application Gauss Law, which I will show if timepermits.


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If there is an empty nonconducting cavity inside the conductor,Gauss Law tells us there is no net charge on the interiorsurface of the conductor.

If there is a nonconducting cavity inside the conductor, with anet charge inside the cavity, Gauss Law tells us there is anequal and opposite induced charge on the interior surface ofthe conductor.

Construct a Gaussian surface that includes the inner surface ofthe conductor. The electric field at the Gaussian surface is zero,so no electric flux passes through the Gaussian surface. GaussLaw says the charge inside must be zero. The conductor doesnot have to be symmetric, as shown.

+QConstruct a Gaussian surface that includes the inner surface ofthe conductor. The electric field at the Gaussian surface is zero,so no electric flux passes through the Gaussian surface. GaussLaw says the charge inside must be zero.

Construct a Gaussian surface that includes the inner surface ofthe conductor. The electric field at the Gaussian surface is zero,so no electric flux passes through the Gaussian surface. GaussLaw says the charge inside must be zero. There must be a Qon the inner surface. The conductor does not have to besymmetric, as shown.

-Q


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Demo: Professor Tries to Avoid

Debilitating Electrical Shock WhileDemonstrating Van de Graaff Generator

http://en.wikipedia.org/wiki/Van_de_Graaff_generator


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Example: a conducting spherical shell of inner radius a andouter radius b with a net charge -Q is centered on point charge

+2Q. Use Gausss law to show that there is a net charge of-2Q on the inner surface of the shell, and a net charge of +Qon the outer surface of the shell.

a

b

-Q

+2Q

enclosed

o

qE dA !

I&&

Thanks to Dr. Waddill for the use of the picture.


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Example (if time permits): an insulating sphere of radius a hasa uniform charge density and a total positive charge Q.

Calculate the electric field at a point inside the sphere

a

Q

r

Thanks again to Dr. Waddill for the use of the picture.

enclosed

o

qE dA ! I&&


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Because there may be no time to work the previous twoexamples in class, I will put Dr. Waddills lecture on Gauss Law

from a couple of years ago on the Physics 24 web site. Pleaseclick on the word lecture in the previous sentence toview/download the lecture.

lecture04 electric flux, gauss' law; conductors and electric fields

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