Flux andFlux andGauss’s LawGauss’s Law

Spring 2008Spring 2008

Last Time: Definition – Sort of –

Electric Field Lines

DIPOLE FIELD LINKCHARGE

Field Lines Electric Field

Last time we showed that

Ignore the Dashed Line … Remember last time .. the big plane?

00

00

0

0

E=0 0 E=0

We will use this a lot!

NEW RULES (Bill Maher)

Imagine a region of space where the ELECTRIC FIELD LINES HAVE BEEN DRAWN.

The electric field at a point in this region is TANGENT to the Electric Field lines that have been drawn.

If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area. The DENSITY of the lines. We won’t use this much

What would you guess is inside the cube?A. A positive

chargeB. A negative

chargeC. Can’t tellD. An AlienE. A car

What about now?

A. A positive charge

B. A negative charge

C. Can’t tellD. An AlienE. A car

How about this??

A. Positive point chargeB. Negative point chargeC. Large Sheet of chargeD. No chargeE. You can’t tell from this

Which box do you think contains more charge?

A. TopB. BottomC. Can’t tellD. Don’t care

All of the E vectors in the bottom box are twice as large as those coming from the top box. The top box contains a charge Q. How much charge do you think is in the bottom box?

1. Q2. 2Q3. You can’t tell4. Leave me alone.

So far …

The electric field exiting a closed surface seems to be related to the charge inside.

But … what does “exiting a closed surface mean”?

How do we really talk about “the electric field exiting” a surface? How do we define such a concept? CAN we define such a concept?

Mr. Gauss answered the question with..

Another QUESTION:

Solid Surface

Given the electric field at EVERY pointon a closed surface, can we determinethe charges that caused it??

A Question: Given the magnitude and direction of the

Electric Field at a point, can we determine the charge distribution that created the field?

Is it Unique? Question … given the Electric Field at a

number of points, can we determine the charge distribution that caused it? How many points must we know??

Still another question

Given a small area, how can you describe both the area itself and its orientation with a single stroke!

The “Area Vector” Consider a small area. It’s orientation can be described by a vector

NORMAL to the surface. We usually define the unit normal vector n. If the area is FLAT, the area vector is given by

An, where A is the area. A is usually a differential area of a small part of a

general surface that is small enough to be considered flat.

The “normal component” of the ELECTRIC FIELD

E

nEn

nEnEnE

)cos(

)cos(

E

E

n

n

DEFINITION FLUX

E

nEn

)cos(

)(

AE

nEAE

AFlux n

We will be considering CLOSED surfaces

nEnE )cos(nEThe normal vector to a closed surface is DEFINED as positiveif it points OUT of the surface. Remember this definition!

“Element” of Flux of a vector E leaving a surface

dAddalso

dd NORMAL

nEAE

AEAE

For a CLOSED surface:n is a unit OUTWARD pointing vector.

This flux is LEAVING the closed surface.

Definition of TOTAL FLUX through a surface

dA

is surface aLEAVING FieldElectric theofFlux Total

out

surfaced

nE

Flux is

A. A vectorB. A scalerC. A triangle

Visualizing Flux

ndAEflux

n is the OUTWARD pointing unit normal.

Definition: A Gaussian Surface

Any closed surface thatis near some distribution

of charge

Remember

ndAEflux

)cos(nEnE

n E

A

Component of Eperpendicular tosurface.

This is the fluxpassing throughthe surface andn is the OUTWARDpointing unit normalvector!

ExampleCube in a UNIFORM Electric Field

L

E

E is parallel to four of the surfaces of the cube so the flux is zero across thesebecause E is perpendicular to A and the dot product is zero.

Flux is EL2

Total Flux leaving the cube is zero

Flux is -EL2

Note sign

area

Simple Example

0

22

0

20

20

20

44

14

14

1

41

qrrq

ArqdA

rq

dArqdA

Sphere

nE

r

q

Gauss’ Law

n is the OUTWARD pointing unit normal.

0

0

enclosedn

enclosed

qdAE

qndAE

q is the total charge ENCLOSEDby the Gaussian Surface.

Flux is total EXITING theSurface.

Simple ExampleUNIFORM FIELD LIKE BEFORE

E

A AE E

00

qEAEA

No

Enclosed Charge

Line of Charge

L

Q

LQ

lengthcharge

Line of Charge

From SYMMETRY E isRadial and Outward

rk

rrE

hrhE

qdAEn

24

22

2

00

0

0

What is a Cylindrical Surface??

Ponder

Looking at A Cylinder from its END

Circular RectangularDrunk

Infinite Sheet of Charge

cylinderE

h

0

0

2

E

AEAEA

We got this sameresult from thatugly integration!

Materials

Conductors Electrons are free to move. In equilibrium, all charges are a rest. If they are at rest, they aren’t moving! If they aren’t moving, there is no net force on them. If there is no net force on them, the electric field must be

zero.

THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!

More on Conductors

Charge cannot reside in the volume of a conductor because it would repel other charges in the volume which would move and constitute a current. This is not allowed.

Charge can’t “fall out” of a conductor.

Isolated Conductor

Electric Field is ZERO inthe interior of a conductor.

Gauss’ law on surface shownAlso says that the enclosedCharge must be ZERO.

Again, all charge on a Conductor must reside onThe SURFACE.

Charged Conductors

E=0E

- --

-

-

Charge Must reside onthe SURFACE

0

0

E

or

AEA

Very SMALL Gaussian Surface

Charged Isolated Conductor

The ELECTRIC FIELD is normal to the surface outside of the conductor.

The field is given by:

Inside of the isolated conductor, the Electric field is ZERO.

If the electric field had a component parallel to the surface, there would be a current flow!

0

E

Isolated (Charged) Conductor with a HOLE in it.

0

0QdAE

n

Because E=0 everywhereinside the conductor.

So Q (total) =0 inside the holeIncluding the surface.

A Spherical Conducting Shell with

A Charge Inside.

Insulators

In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in

the volume and it can’t “flow” anywhere so it stays there.

You can therefore have a charge density inside an insulator.

You can also have an ELECTRIC FIELD in an insulator as well.

Example – A Spatial Distribution of charge.

Uniform charge density = charge per unit volume

0

0

3

0

2

0

3

1344

rE

rVrE

qdAEn

(Vectors)r EO

A Solid SPHERE

Outside The Charge

r

EO

R

20

0

3

0

2

0

41

344

rQE

or

QRrE

qdAEn

Old Coulomb Law!

Graph

R

E

r

Charged Metal Plate

E is the same in magnitude EVERYWHERE. The direction isdifferent on each side.

E

++++++++

++++++++

E

A

A

Apply Gauss’ Law

++++++++

++++++++

E

A

A

AEAEAEA

Bottom

E

AEAAEA

Top

0

0

0

22

0

Same result!

Negatively ChargedISOLATED Metal Plate

---

E is in opposite direction butSame absolute value as before

Bring the two plates together

A

ee

B

As the plates come together, all charge on B is attractedTo the inside surface while the negative charge pushes theElectrons in A to the outside surface.

This leaves each inner surface charged and the outer surfaceUncharged. The charge density is DOUBLED.

Result is …..

A

ee

B

EE=0

E=0

0

1

0

0

2

E

AEA

VERY POWERFULL IDEA

Superposition The field obtained at a point is equal to the

superposition of the fields caused by each of the charged objects creating the field INDEPENDENTLY.

Problem #1Trick Question

Consider a cube with each edge = 55cm. There is a 1.8 C chargeIn the center of the cube. Calculate the total flux exiting the cube.

CNmq /1003.21085.8108.1 25

12

6

0

NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!

Easy, yes??

Note: the problem is poorly stated in the text.

Consider an isolated conductor with an initial charge of 10 C on theExterior. A charge of +3mC is then added to the center of a cavity.Inside the conductor.

(a) What is the charge on the inside surface of the cavity?(b) What is the final charge on the exterior of the cavity?

+3 C added+10 C initial

Another Problem

m,q both given as is

0

0

2

2

E

AEA

Gauss

GaussianSurface

Charged Sheet

-2

m,q both given as is

mg

qE

T

Free body diagram

02)sin(

)cos(

qqET

mgT

-3

290

0

1003.5)tan(2

2)(

mC

qmg

andmgqTan

Divide

(all given)

A Last ProblemA uniformly charged cylinder.

R

rRE

hRqrhE

Rr

rE

hrrhE

Rr

0

2

0

2

0

0

0

2

2

)()2(

2

)()2(