Lecture Set 3Lecture Set 3Gauss’s LawGauss’s Law

Calendar for the WeekjCalendar for the Weekj

Today (Monday)Today (Monday)– One or two problemsOne or two problems– Introduction to the concept of FLUXIntroduction to the concept of FLUX

Wednesday, FridayWednesday, Friday– Gauss’s Law & some problemsGauss’s Law & some problems

EXAM DATE WILL BE NEXT EXAM DATE WILL BE NEXT WEDNESDAY through GaussWEDNESDAY through Gauss

Protons are projected with an initial speed vi = 9.83 103 m/s into a region where a uniform electric field E = (-720 j) N/C is present, as shown in Figure P23.49. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field in Figure 23.49.

Summary from last week

222

,2

2

2

)()()(

r

rdsk

r

rdAk

r

rdVk

r

Qk

q

General

r

Qk

q

r

qQk

unitjj

jjj

unit

unit

E

rF

EE

rF

E

rF

(Note: I left off the unit vectors in the lastequation set, but be aware that they should

be there.)

VECTOR VECTOR

Electric Field

We will now introduce a convenient way to represent the overall electric field in a region of space.

It is kinda sorta a map of the field strength. We will then introduce a new concept

FLUX We will use this new concept to introduce Gauss’s

Law

Look at the “Field Lines” of an infinite sheet of charge …

How do you do that?

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

00

00

0

0

E=0 0 E=0

NEW RULES (Bill Maher)

Imagine a region of space where the ELECTRIC FIELD LINED 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.

Point Charges

They don’t like each other …

Mr. Gauss …

Recall…

We were given Coulomb’s Law We defined the electric field. Calculated the Electric Field given a distribution

of charges using Coulomb’s Law.

unitVi uniti

r

rdqk

r

qrrE

2210

)(

4

1

(Units: N / C)

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??

Another QUESTION:

Solid Surface

Given the electric field at EVERY point

on a closed surface, can we determinethe charges that caused it??

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 a vector

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

ANOTHER DEFINITION:Element of Flux through a surface

EENORMAL

NORMAL

A E=|ENORMAL| | |A|

(a scalar)

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

dAd

also

d NORMAL

nEAE

AEAE

n is a unit OUTWARD pointing vector.

This flux was LEAVING the closed surface

Definition of TOTAL FLUX through a surface

dA

is surface aLEAVING Field

Electric theofFlux Total

out

surfaced

nE

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

1

4

1

4

1

4

1

qr

r

q

Ar

qdA

r

q

dAr

qdA

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

q

EAEA

No

Enclosed Charge

Line of Charge

L

Q

L

Q

length

charge

Line of Charge

From SYMMETRY E isRadial and Outward

r

k

rrE

hrhE

qdAEn

2

4

2

2

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=0

E

---

-

-

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

0Q

dAEn

Because E=0 everywhereinside the surface.

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

1

3

44

rE

rV

rE

qdAEn

(Vectors)

r EO

A Solid SPHERE

Outside The Charge

r

E

O

R

20

0

3

0

2

0

4

1

3

44

r

QE

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

/1003.21085.8

108.1 2512

6

0

NOTE: INDEPENDENT OF THE SHAPE OF THE SURFACE!

Easy, yes??

Problem #2(15 from text)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 from the book

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(

q

qET

mgT

-3

290

0

1003.5)tan(2

2)(

mC

q

mg

and

mg

qTan

Divide

(all given)

A Last ProblemA uniformly charged cylinder.

R

r

RE

hRqrhE

Rr

rE

hrrhE

Rr

0

2

0

2

0

0

0

2

2

)()2(

2

)()2(