Physics 212 Lecture 6, Slide 1 Physics 212 Lecture 6 Today's Concept: Electric Potential Defined in terms of Path Integral of Electric Field

Physics 212 Lecture 6, Slide Physics 212 Lecture 6, Slide 11

Physics 212Physics 212Lecture 6Lecture 6

Today's Concept:Today's Concept:Electric PotentialElectric Potential

Defined in terms of Path Integral of Electric FieldDefined in terms of Path Integral of Electric Field

Physics 212 Lecture 6Physics 212 Lecture 6

MusicWho is the Artist?Who is the Artist?

A)A) John PrineJohn PrineB)B) Little FeatLittle FeatC)C) Taj MahalTaj MahalD)D) Ry CooderRy CooderE)E) Los LobosLos Lobos

Why?Why?

Last time did Buena Vista Social Club (Cuba)Last time did Buena Vista Social Club (Cuba)Ry Cooder was the guy who brought them to our attention in this countryRy Cooder was the guy who brought them to our attention in this country

Also, this album is great…. Also, this album is great….

NOTE: Ellnora starts tonight….. NOTE: Ellnora starts tonight…..


Your CommentsYour Comments

05

““I'm pretty sure my head just exploded. Just... everywhere.”.”

“help! i need somebody help! not just anybody help! you know i need someone HELLPP"““That last one was a doozy. Equipotential lines seem like they hold the key to something, but I don't know what yet.””

“Do we have to be able to use/do problems with gradients?”“Are we going to differentiate the electric potential in three dimensions in order to get the electric field?”

“The calculations with the spherical insulator were hard to follow. And I understand simple ideas, but I hope that the lecture helps me understand more fully.”

Electric potentialElectric potential is related to energy – a key aspect of E’nM. is related to energy – a key aspect of E’nM.We will use We will use electric potentialelectric potential extensively when we talk about circuits. extensively when we talk about circuits.

We really only need to know about derivatives (partial derivatives in a We really only need to know about derivatives (partial derivatives in a few cases).few cases).

See example at end of class.See example at end of class.

Not too bad. After discussion today, I understand Gauss's Law!


BIG IDEABIG IDEA• Last time we defined the electric potential energy of charge q in an electric field:

40

b

a

b

aba ldEqldFU

•The only mention of the particle was through its charge q.

• We can obtain a new quantity, the electric potential, which is a PROPERTY OF THE SPACE, as the potential energy per unit charge.

b

a

baba ldE

qUV

•Note the similarity to the definition of another quantity which is also a PROPERTY OF THE SPACE, the electric field.

qFE


Electric Potential from E fieldElectric Potential from E field• Consider the three points A, B, and C located in a region of constant electric field as shown.

40

• What is the sign of VAC = VC - VA ? (A) VAC < 0 (B) VAC = 0 (C) VAC > 0

• Remember the definition:

C

ACA ldEV

• Choose a path (any will do!)

DD xx

C

D

D

ACA ldEldEV

00 xEldEV

C

DCA

When you integrate 0, you get 0, everywhere.


CheckpointCheckpoint 2 2

08

AB

D

• Remember the definition

B

ABA ldEV

0 BAV V is V is constant !!constant !!

0E

When the electric field is zero in a certain region, the electric potential is surely zero in that region. However it is not certain that the electric potential is zero elsewhere.

Suppose the electric field is zero in a certain region of space. Which of the following statements best describes the electric potential in this region of space?

A. The electric potential is zero everywhere in this regionB. The electric potential is zero at at least one point in this regionC. The electric potential is constant everywhere in this regionD. There is not enough information to distinguish which of the answers is correct

The potential is the integral of E dx. If E is zero then when you integrate you will just get a constant.


E from VE from V• If we can get the potential by integrating the electric field:

40

• We should be able to get the electric field by differentiating the potential??

b

aba ldEV

VE

• In Cartesian coordinates:

xVEx

yVEy

zVEz


CheckpointCheckpoint 1a 1a

08

• How do we get E from V??Look at slopes !!!Look at slopes !!!VE

xVE = -x

““The highest electric potential also means the greatest electric field.““

““The slope is the steepest there““

““The field and the potential are inversely related““

The electric potential in a certain region is plotted in the following graph:

At which point is the magnitude of the E-field greatest?A. B. C. D.


CheckpointCheckpoint 1b 1b

08

ABCD

• How do we get E from V??Look at slopes !!!Look at slopes !!!VE

xVE = -x

“Because B is along the negative slope on the graph.“

“The slope of the electric potential at point C is positive“

The electric potential in a certain region is plotted in the following graph:

At which point is the direction of the E-field along the negative x-axis?A. B. C. D.


EquipotentialsEquipotentials• Equipotentials are the locus of points having the same potential

40

Equipotentials produced by a point

charge

Equipotentials are ALWAYS

perpendicular to the electric field lines

The SPACING of the equipotentials indicates

The STRENGTH of the electric field


CheckpointCheckpoint 3a 3a

08

ABCD

“The field lines are dense at points A, B, and C, which indicates that the field is strong. However the lines are very spread out at D, indicating that the field is weakest there..“

The field-line representation of the E-field in a certain region of space is shown below. The dashed lines represent equipotential lines.

At which point in space is the E-field weakest?A. B. C. D.


CheckpointCheckpoint 3b 3b

08

ABCD

ABCD“The equipotential is weaker in the region between c and d than in the region from a to b.“

“The work is directly proportional to the distance between two charges. The distance between C and D is further away from that of A and B.”

“The charge would have to cross the same number of equipotential lines regardless, so the change in potential energy is the same and therefore the work is the same.“


Compare the work done moving a negative charge from A to B and from C to D. Which move requires more work?A. From A to B B. From C to DC. The same D. Cannot determine without performing calculation


HINTHINT

08

ABCD ELECTRIC FIELD LINES !!

What are these ?

EQUIPOTENTIALS !!• What is the sign of WAC = work done by E field to move negative charge from A to C ? (A) WAC < 0 (B) WAC = 0 (C) WAC > 0

A and C are on the same equipotential WAC = 0 !! Equipotentials are perpendicular to the E field:

No work is done along an equipotential


What are these ?


Checkpoint 3b Checkpoint 3b Again?Again?

08

ABCD

ABCD

• A and C are on the same equipotential • B and D are on the same equipotential• Therefore the potential difference between A and B is the SAME as the potential between C and D


Compare the work done moving a negative charge from A to B and from C to D. Which move requires more work?A. From A to B B. From C to DC. The same D. Cannot determine without performing calculation


Checkpoint 3cCheckpoint 3c

08

ABCD

ABCD

“because the electric field is constant, moving the charge from A to D would require the most work since it has the furthest displacement.”

“The electric field is greater at point D than B.“

“B and D are on the same equipotential line.“


Now compare the work done moving a negative charge from A to B and from A to D. Which move requires more work?A. From A to B B. From A to DC. The same D. Cannot determine without performing calculation


Calculation for PotentialCalculation for PotentialPoint charge q at center of concentric conducting spherical shells of radii a1, a2, a3, and a4. The inner shell is uncharged, but the outer shell carries charge Q.

What is V as a function of r?+q+q

metal

metal

+Q+Q

a1

a2

a3a4

• Conceptual Analysis: – Charges q and Q will create an E field throughout space

0

( )r

r

V r E d

–

• Strategic Analysis: – Spherical symmetry: Use Gauss’ Law to calculate E

everywhere – Integrate E to get V

cross-sectioncross-section

40


Calculation: Quantitative AnalysisCalculation: Quantitative Analysisr > a4 : What is E(r)?

(A) (A) 0 0 (B) (C) (B) (C) +q+q

metal

metal

+Q+Q

a1

a2

a3a4


0

12

Q qr

2

0

14

Qr

(D) (E)(D) (E) 2

0

14

Q qr

2

0

14

Q qr

2

0

4

Q qE r

2

0

14

Q qEr

Why?Why?

0

enclosedQE dA

Gauss’ law:Gauss’ law:

rr


Calculation: Quantitative AnalysisCalculation: Quantitative Analysisa3 < r < a4 : What is E(r)?

(A) (A) 0 0 (B) (C)(B) (C) +q+q

metal

metal

+Q+Q

a1

a2

a3a4


(D) (E)(D) (E) 2

0

14

qr

2

0

14

Q qr

How is this possible???How is this possible??? -q must be induced at r=a-q must be induced at r=a33 surface surface

2

0

14

qr

0

12

qr

rr

Applying Gauss’ law, what is Applying Gauss’ law, what is QQenclosedenclosed for red sphere shown?for red sphere shown?

(A) (A) qq (B) (B) –q–q (C)(C) 00

charge at r=acharge at r=a44 surface = Q+q surface = Q+q

3 2

34qa

4 2

44Q qa


Calculation: Quantitative AnalysisCalculation: Quantitative AnalysisContinue on in….a2 < r < a3 :

+q+q

metal

metal

+Q+Q

a1

a2

a3a4


2

0

14

qEr

rr0E a1 < r < a2 :

r < a1 : 2

0

14

qEr

To find V:To find V:1)1) Choose rChoose r00 such that V(r such that V(r00) = 0 (usual: r) = 0 (usual: r00 = infinity) = infinity)2)2) Integrate !!Integrate !!

r > a4 :0

14

Q qVr

a3 < r < a4 :

0 4

14

Q qVa

0V

0 3

14

Q qVa

(A)

(B)

(C)

4( ) 0V a


Calculation: Quantitative AnalysisCalculation: Quantitative Analysis

+q+q

metal

metal

+Q+Q

a1

a2

a3a4


r > a4 :0

14

Q qVr

a3 < r < a4 : 0 4

14

Q qVa

a2 < r < a3 :

4 3( ) ( ) 0 ( )V r V a V a r

0 4 0 3

1 1( ) 04 4Q q qV r

a r a 0 4 3

1( )4

Q q q qV ra r a

a1 < r < a2 : 0 4 2 3

1( )4

Q q q qV ra a a

0 < r < a1 :0 4 2 3 1

1( )4

Q q q q q qV ra a a r a

SummarySummary

V

0 ra1a2 a3a4

b

aba ldEV

VE

Documents

Physics 212 Lecture 6, Slide 1 Physics 212 Lecture 6 Today's Concept: Electric Potential Defined in terms of Path Integral of Electric Field