AnnouncementsCareers in Physics Event:Dr. Jeffrey Phillips from EPRI will discuss working in the field of energy production. 11 AM today Olin Lounge Seminar:Dr. Jeffrey Phillips, Electric Power Research InstituteIntegrated Coal Gasification Combined Cycle Power Plants : A Cleaner Coal-to-Electricity Option 4:00 PM Olin 101 Thursday

Pick up the handout

Announcements II

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Semester Quiz I is on 02/03/05.Lectures 1-5; Ch 3,22-24One {letter-sized} page of notes allowed

Hints: Review {and master} HW, and reading quizzesStudy lecture notes, and do optional problems

for extra practice

Electric Flux and Gauss’ Law

E

En

cosE E dA E dA

•The electric flux through a closed surface is given by Gauss’ Law•Usually you can pick your surface so that the integration doesn’t need to be done•We must distinguish between field and flux!

4E eE dA k q

Special optional Readings

For those of you who are interested, “Div, Grad, Curl and all that” in Ch 2 has a more quantitative analysis of Gauss’ law.

The handout is from the beginning of that chapter and has a more quantitative treatment of unit normals and surface integrals than we did in class. You will not need to know this for the course, butit might aid your understanding.

Gauss’s Law

charge -q

charge 2q

charge q

charge -2q

Which other charge(s) do we have to include in a gaussian surface so as to contain the +2q charge and have flux equal to:Zero ?+3q/0?

-2q/0 ?

IMPOSSIBLE

A) -2q C) q B) -q D)

impossible

Today’s Reading Quiz

Consider Gauss's law:  Which of the following is true?

E must be the electric field due to the enclosed chargeIf a charge is placed outside the surface, then it cannot affect E on the surfaceOn the surface E is everywhere parallel to dAIf q = 0 then E = 0 everywhere on the Gaussian surfaceIf the charge inside consists of an electric dipole, then the integral is zero.

charge -q

charge 2qcharge q

charge -2q

Example

•What is the flux through the first surface?

                                      

Figure 24-29.

•What is the flux through the second surface?

•What is the flux through the third surface?

•What is the flux through the fourthsurface?

•What is the flux through the fifth surface?

A) q/o D) 2q/o B) -q/o C) 0

Applying Gauss’s Law•Can be used to determine total flux through a surface in simple cases•Must have a great deal of symmetry to use easily

1 2 3 1 20

E S S S E E

q

charge q

Charge in an infinite triangular channelWhat is flux out of one side?

3 0S

03S

q

Applying Gauss’s Law

•Infinite cylinder radius R charge density •What is the electric field inside and outside the cylinder?

L

R

•Electric Field will point directly out from the axis

r

•Draw a cylinder with the desired radius inside the cylindrical charge

AE 2 rLE0

V

2

0

r L

0

inq

02

rE

Applying Gauss’s Law

•Infinite cylinder radius R charge density •What is the electric field inside and outside the cylinder?

L

R

•Electric Field will point directly out from the center

r

•Draw a cylinder with the desired radius outside the cylindrical charge

AE 2 rLE0

V

2

0

R L

0

inq

2

02

RE

r

Applying Gauss’s Law

•Sphere radius R charge density . What is E-field inside?

RSphere volume:

V = 4a3/3

Sphere area: A = 4a2

•Draw a Gaussian surface inside the sphere of radius r

r

What is the magnitude of the electric field inside the sphere at radius r?

A) R3/30r2

B) r2/30RC) R/30 D) r/30

AE 24 r E

0

V

3

0

4

3

r

0

inq

03

rE

Conductors in Equilbirum•A conductor has charges that can move freely•In equilibrium the charges are not moving•Therefore, there are no electric fields in a conductor in equilibrium

F qE ma

0 0inq E dA

= 0

•The interior of a conductor never has any charge in it•Charge on a conductor is always on the surface

= 0

Electric Fields near Conductors•No electric field inside the conductor•Electric field outside cannot be tangential – must be perpendicular

Surface charge

0 0 AE 0

A

0

inq

Area A

•Add a gaussian pillbox that penetrates the surface

0

ˆE n

•Electric field points directly out from (or in to) conductor

Conductors shield chargesNo net charge

Charge q•What is electric field outside the spherical conductor?

•Draw a Gaussian surface•No electric field – no charge•Inner charge is hidden – except

Charge -q •Charge +q on outside to compensate•Charge distributed uniformlyCharge +q

20

ˆ

4

qrE

r

Conductors shield charges

Charge q

•Charge is induced on the inner and outer surfaces of the conducting shell, which hides the interior charge•How would the situations change if the charge inside were on a conductor?

Charge -q

Charge +q

Example

•What is the charge enclosed on the wire?

•We know all the physical dimensions!

•We know how the field lines point out from a wire•We know the field on the interior of the cylinder

Suppose that the radius of the central wire is 26 µm, the radius of the cylinder 1.3 cm, and the length of the tube 16 cm. If the electric field at the cylinder's inner wall is 2.7 x 104 N/C

Example

•EA=q/o=E(2rl)+0+0

•What if we know the charge density, , and not the field?

•EA=q/o=E(2rl)+0+0=l/o

Suppose that the radius of the central wire is 26 µm, the radius of the cylinder 1.3 cm, and the length of the tube 16 cm. If the electric field at the cylinder's inner wall is 2.7 x 104 N/C

Practice Problem I

A cube with 1.40 m edges is oriented as shown in the figure

Suppose there is a charge situated in the middle ofthe cube.• What is the magnitude of the flux through the whole

cube?• What is the magnitude of the flux through any one

side?

A) q/o D) q/6o B) q/4o C) 0

Practice Problem II

A cube with 1.40 m edges is oriented as shown in the figure

Suppose the cube sits in a uniform electric field of 10i ?• What is the magnitude of the flux through the whole

cube?• What is the magnitude of the flux through the top

side?• How many sides have nonzero flux?

A) q/o D) q/6o B) q/4o C) 0

A) 2 D) 1 B) 4C) 0