physics 212 - university of illinoisphysics 212 lecture 3, slide 12 “the flux is directly...

Physics 212 Lecture 3, Slide 1

Physics 212Lecture 3

Today's Concepts:

Electric Flux and Field Lines

Gauss’s Law

Music

Who is the Artist?

A) Professor LonghairB) Henry ButlerC) David EganD) Dr. JohnE) Allen Toussaint

WHY?

Mardi Gras is 4 weeks from today

Physics 212 Lecture 13


Our Comments

• SmartPhysics scores will not be visible in the main gradebook for a few weeks

• You will need to understand integrals in this course!!!

• Forces and Fields are Vectors

• Always Draw a Picture First… What do the Forces/Fields Look Like?

Prelecture

Infinite line of charge

E

Lecture

Finite line of charge

(constant λ)

E

Worked Example

Finite line of charge

(non-constant λ)

E

Homework

Arc of charge

(constant λ)

E2 2

0

1ˆ ˆ

4

dq dqE k r r

r rπε= =∫ ∫�

WORKS FOR ALL !!

The “1/4 shell problem” is not

special !!

It’s the same as all the others!


Electric Field Lines

Direction & Density of Lines represent

Direction & Magnitude of E

Point Charge:Direction is radialDensity α 1/R2

07


Electric Field Lines

Dipole Charge Distribution:Direction & Density

much more interesting…

simulation08


Preflight 3

Checkpoint 3.1

simulation09

The following three questions pertain to the electric field

lines due to the two charges shown. Compare the

magnitude of the two charges

A. |Q1| > |Q2| B. |Q1| = |Q2| C. |Q1| < |Q2|

D. There isn’t enough information to determine the relative

magnitude of the charges

“more field lines emanating from the charge 1, so must be greater in magnitude.”


Checkpoint 3.3

10simulation

What do we know about the signs of the charges from

looking at the picture?

A. Q1 and Q2 have the same sign

B. Q1 and Q2 have the opposite sign

C. There isn’t enough information to determine the relative

signs of the charges

“The field lines leave one and go to the other..”


Preflight 3

Checkpoint 3.5

12

Compare the magnitude of the electric field at point A and

B.

A. |EA| > |EB| B. |EA| = |EB| C. |EA| < |EB|

D. There isn’t enough information to determine the relative

magnitude of the electric field at points A and B

“Since the electric field lines are denser near

point B the magnitude of the electric field is greater at point B than at point A.”


Point Charges

+2q

-q

What charges are inside the red circle?

+Q +Q

-Q

+2Q

-Q -2Q

+Q

-Q

A B C D E

“Telling the difference between positive and negative charges while looking at field lines.

Does field line density from a certain charge give information about the sign of the charge?”

13


Which of the following field line pictures best represents the electric field from two charges that have the same sign but different magnitudes?

A B

C D

simulation15


Electric Flux “Counts Field Lines”

S

S

E dAΦ = ⋅∫��

Flux through surface S

Integral of on surface S

E dA⋅��

“I still feel a little unclear about flux integrals. Can you go over some more example problems?”

18


“The flux is directly proportional to

the charge enclosed by the

Gaussian surface, therefore Case

1 will have 2 times the flux of of

Case 2 because it has two charges enclosed, versus 1 in Case 2. “

“The flux is the measure of field

lines passing through the surface.

Once calculated, it can be shown

that the surface area of both cylinders are the same.”

Checkpoint 1

(A)Φ1=2Φ2

Φ1=Φ2

(B)Φ1=1/2Φ2

(C)none(D)

TAKE s TO BE RADIUS !

L/2

An infinitely long charged rod has uniform

charge density λ and passes through a

cylinder (gray). The cylinder in Case 2 has

twice the radius and half the length compared

with the cylinder in Case 1.

23


Checkpoint 1

(A)Φ1=2Φ2

Φ1=Φ2

(B)Φ1=1/2Φ2

(C)none(D)

TAKE s TO BE RADIUS !

L/2

An infinitely long charged rod has uniform

charge density l and passes through a cylinder

(gray). The cylinder in Case 2 has twice the

radius and half the length compared with the

cylinder in Case 1.

Definition of Flux:

∫ ⋅≡Φsurface

AdE��

E constant on barrel of cylinderE perpendicular to barrel surface(E parallel to dA)

Case 1

sE

0

12πε

λ=

LsA )2(1 π= 0

1ε

λL=Φ

Case 2

)2(2 0

2s

Eπε

λ=

sLLsA ππ 22/))2(2(2 ==0

2

)2/(

ε

λ L=Φ

RESULT: GAUSS’ LAWΦ proportional to charge enclosed !

“The flux is the measure of field

lines passing through the surface.

Once calculated, it can be shown

that the surface area of both cylinders are the same.”

WHAT IS WRONG WITH THIS REASONING??

THE E FIELD AT THE BARREL SURFACE

IS NOT THE SAME IN THE TWO CASES !!

barrelbarrel

EAAdE =∫=Φ�

26


Direction Matters:

0S

S

E dAΦ = ⋅ >∫��

dA

dA

dA dA

dA

E

E

E

E

E

E

EE

E

For a closed surface, A points outward

29


Direction Matters:

0S

S

E dAΦ = ⋅ <∫��

dA

dA

dA dA

dA

E

E

E

E

E

E

EE

E

For a closed surface, A points outward

30


Trapezoid in Constant Field

0ˆE E x=

�

Label faces:1: x = 0

2: z = +a3: x = +a

4: slanted

0E�

y

z

Define Φn = Flux through Face n

A

B

C

Φ1 < 0

Φ1 = 0

Φ1 > 0

A

B

C

Φ2 < 0

Φ2 = 0

Φ2 > 0

A

B

C

Φ3 < 0

Φ3 = 0

Φ3 > 0

A

B

C

Φ4 < 0

Φ4 = 0

Φ4 > 0

x

1 2

3

dA

E

0<⋅ AdE��

31


Trapezoid in Constant Field + Q0ˆE E x=

�

Label faces:1: x = 0

2: z = +a3: x = +a

Define Φn = Flux through Face nΦ = Flux through Trapezoid

A

B

C

Φ1 increases

Φ1 decreases

Φ1 remains same

Add a charge +Q at (-a,a/2,a/2)

+Q

How does Flux change?

A

B

C

Φ increases

Φ decreases

Φ remains same

A

B

C

Φ3 increases

Φ3 decreases

Φ3 remains same

x

y

z

0E�

x

1 2

3

36


Q

Gauss Law

dA

dA

dA dA

dA

E

E

E

E

E

E

EE

E

o

enclosed

surfaceclosed

SQ

AdEε

=∫ ⋅≡Φ��

41


Checkpoint 2.3

(A)

Φ increases(B)

Φ decreases

(C)

Φ stays same

43

“The flux throughout the entire

surface does not change as the

charge moves, for it is still in the

sphere. However, if it was moved

outside of the sphere, the fluxes

would be different because one

would be zero!“


“charge is closer, field is larger, more

lines through a so increase in flux.“

“as long as the charge is within the

shell, the flux will remain the same”

Checkpoint 2.1

(A)

dΦA increases

dΦB decreases

(B)

dΦA decreases

dΦB increases

(C)

dΦA stays same

dΦB stays same

44


The total flux is the same in both cases (just the total number of lines)

The flux through the right (left) hemisphere is smaller (bigger) for case 2.

1 2

Think of it this way:

45


Things to notice about Gauss’s Law

If Qenclosed is the same, the flux has to

be the same, which means that the

integral must yield the same result

for any surface.

o

enclosed

surfaceclosed

SQ

AdEε

=∫ ⋅≡Φ��

47


Things to notice about Gauss’s Law

In cases of high symmetry it may be possible to bring E outside

the integral. In these cases we can solve Gauss Law for E

So - if we can figure out Qenclosed and the area of the

surface A, then we know E !

This is the topic of the next lecture…

“Gausses law and what concepts are most important that we know .”

o

enclosed

surfaceclosed

QAdE

ε=∫ ⋅��

o

enclosed

surfaceclosed

QEAAdE

ε==∫ ⋅

��

0

enclosedQE

Aε=

48


Final Thoughts….

• Keep plugging away– Prelecture 4 and Checkpoint 4 due Thursday

– Homework 2 due next Tuesday … 7 questions mostly Gauss’ law – try them today!

50

physics 212 - university of illinoisphysics 212 lecture 3, slide 12 “the flux is directly...

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