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Lecture 4.1 :Electric Potential

Lecture Outline:Electric Potential Energy

Potential Energy of Point ChargesElectric Potential

Feb. 4, 2014

Textbook Reading:Ch. 28.1 - 28.4

Announcements

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•Exam #1 in class Thu. (Feb. 6) covers Ch. 25-27 material.

‣You are responsible for bringing your own calculator and writing utensils.

‣1 sheet of notes (front and back) allowed.

‣Seating assignment will be posted by entrance doors to Stolkin. You must sit in your assigned seat.

‣Optional review session TONIGHT from 7-8pm in Stolkin. If you attend, please bring your clickers! (Note: I know a few of you have an overlapping PHY222 lab. My apologies...I will try to schedule around this in the future. You can join us late. I actually have the room booked until 9pm and can stay late to cover anything you miss before you join.)

•HW4 due next Tue. (2/11) at 9am on Mastering Physics.

•During recitation on Friday you will go over exam solutions.

From HW3...

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Last Lecture...

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Electric field must be zero at all points within a conductor in electrostatic equilibrium.

Any excess charge has to reside on the outside surface.

From HW3...

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Last Lecture...

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From Recitation 3.2

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27-18 Chapter 27

© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Thus, net 0 ˆ[ /(2 )]E jη= .G

! In region 3,

P P P0 0 0

ˆ ˆ ˆ 4 2 4

E j E j E jη η η′= = − = −

G G G0! ! !

Thus, net 0 ˆ/(2 )E jη= −[ ] .G

! In region 4,

P P P0 0 0

ˆ ˆ ˆ 4 2 4

E j E j E jη η η′= = − =

G G G0! ! !

Thus net 0 N/CE = .GG

27.48. Model: The charge has planar symmetry, so the electric field must point toward or away from the slab. Furthermore, the field strength must be the same at equal distances on either side of the center of the slab. Visualize:

Choose Gaussian surfaces to be cylinders of length 2z centered on the 0z = plane. The ends of the cylinders have area A. Solve: (a) For the Gaussian cylinder inside the slab, with 0< ,z z Gauss’s law is

intop bottom sides 0

QE dA E dA E dA E dA⋅ = ⋅ + ⋅ + ⋅ =G G G GG G G GÑ Ñ Ñ !

The field is parallel to the sides, so the third integral is zero. The field emerges from both ends, so the first two integrals are the same. The charge enclosed is the volume of the cylinder multiplied by the charge density, or

in (2 ).Q V zAρ ρ= = Thus

intop bottom0 0 0

(2 ) 0 2Q zA zE dA E dA EA Eρ ρ= = ⋅ + ⋅ + = � =G GG GÑ Ñ! ! !

The field increases linearly with distance from the center. The sign of ρ determines the direction of the electric field,

so 00

ˆ( ) .zE z z kρ≤ =G

!

(b) The analysis is the same for the cylinder that extends outside the slab, with 0> ,z z except that the enclosed charge 0(2 )Q z Aρ= is that within a cylinder of length 02z rather than 2z. Thus

in 0 0top bottom0 0 0

(2 ) 0 2Q z A zE dA E dA EA Eρ ρ= = ⋅ + ⋅ + = � =G GG G

! ! !Ñ Ñ

The field strength outside the slab is constant, and it matches the result of part (a) at the boundary. The sign of r

determines the direction of the electric field, so 00

0

ˆ( ) zE z z kρ≥ =G

!

∫v

Gauss’s Law 27-19

© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

(c)

27.49. Model: The infinitely wide plane of charge with surface charge density η polarizes the infinitely wide conductor. Visualize:

Because 0E =GG

in the metal there will be an induced charge polarization. The face of the conductor adjacent to the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus have three infinite planes of charge. These are P (top conducting face), P′ (bottom conducting face), and P″ (plane of charge). Solve: Let 1 2 3, , and η η η be the surface charge densities of the three surfaces with 2 < 0.η The electric field due to a plane of charge with surface charge density η is 0/(2 )E η= .! Because the electric field inside a conductor is zero (region 2),

21 2 3P P P 1 2

0 0 0

ˆ ˆ ˆ0 N/C 0 N/C 0 C/m2 2 2

E E E j j jη η η η η η′+ + = � − + + = � − + + =G GG G G

0 ! ! !

We have made the substitution 3 .η η= Also note that the field inside the conductor is downward from planes P and

P′ and upward from P .′′ Because 21 2 0 C/m ,η η+ = because the conductor is neutral, 2 1.η η= − The above equation

becomes 2 1 1

1 1 1 22 20 C/mη η η η η η η− − + = � = � = −

We are now in a position to find electric field in regions 1–4. For region 1,

P P P0 0 0

ˆ ˆ ˆ 4 4 2

E j E j E jη η η′= = − =

G G G0! ! !

The electric field is net P P P 0 ˆ[ /(2 )]E E E E jη′= + + = .G G G G

0 !

In region 2, net 0E =GG

N/C. In region 3,

P P P0 0 0

ˆ ˆ ˆ 4 4 2

E j E j E jη η η′= − = =

G G G0! ! !

Infinite Slab of Charge (27.48)

Electric Potential Energy

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�Emech = �K +�U = 0

So far in this course we have described electric charge in terms of fields and forces. We now need to consider Energy.

W =

� j

i

�F · d�s

�U = Uf � Ui = �W

Reminders from Mechanics:

Conservation of Energy

Work/Potential Energy Relation

Work done by a Force.

Electric Potential Energy

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Potential Energy in a Uniform Gravitational Field.

�Ugrav = mgyf �mgyi

Ugrav = U0 +mgy = mgy

Clicker Question #1

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Two rocks have equal mass. Which has more gravitational potential energy?

A.  Rock A.

B.  Rock B.

C.  They have the same potential energy.

D.  Both have zero potential energy.

Electric Potential Energy

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Consider a positive charge inside a parallel-plate capacitor:

Welec = Felec�rcos 0� = qE|sf � si| = qEsi � qEsf

�Uelec = Uf � Ui = �Welec = qEsf � qEsi

Define s=0 at negative plate, and U0 is potential at s=0.

Electric Potential Energy

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Signs can be confusing here, so remember these pictures:

Clicker Question #2

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Two positive charges are equal. Which has more electric potential energy? A.  Charge A.

B.  Charge B.

C.  They have the same potential energy.

D.  Both have zero potential energy.

Electric Potential Energy

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Example problem: Proton/Electron in a 2.0 cm x 2.0 cm parallel plate capacitor

Clicker Question #3

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A positive charge moves as shown. Its kinetic energy

A.  Increases.

B.  Remains constant.

C.  Decreases.

Potential Energy of Point Charges

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Potential Energy of Point Charges

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Two like-sign point charges launched at each other:

Potential Energy of Point Charges

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Two opposite-sign point charges launched away from each other:

Emech<0 implies a bound system. Can you think of any other bound systems in nature?

Clicker Question #4

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A positive and a negative charge are released from rest in vacuum. They move toward each other. As they do:

A.  A positive potential energy becomes more positive. B.  A positive potential energy becomes less positive.

C.  A negative potential energy becomes more negative. D.  A negative potential energy becomes less negative.

E.  A positive potential energy becomes a negative potential energy.

Potential Energy of a Dipole

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Electric Potential

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Alessandro Volta

We introducted “Electric Field” to indicate an electric charge’s alteration of space. Now we

need a concept of potential energy at all points in space due to a source charge.

V � Uq+sources

q

1 volt = 1 V ≡  1 J/C

Electric Potential:

1.5 V Battery

Electric Potential

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�U = q�V

�V = Potential Difference, or Voltage.

Electric Potential

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What is the speed of a proton that has been accelerated from rest through a potential difference of -1000V?

Reminders

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•Exam #1 on Thursday! Bring note sheet and calculator!

•Review session tonight from 7-8pm.