Chapter 21 Electric Charge and Electric Fieldpeople.virginia.edu/~ben/2415141/Lecture_3.pdf · 22-2...

Copyright © 2009 Pearson Education, Inc.

Chapter 21Electric Charge and

Electric Field


21-11 Electric Dipoles

An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance . The dipole moment, p = Q , points from the negative to the positive charge.

ℓℓℓℓ

ℓℓℓℓ��

p

↓↓↓↓�

2H Op


21-11 Electric DipolesAn electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:

So how was I able to move the 2x4 using the electric field of the rod?

= − ⋅= − ⋅= − ⋅= − ⋅��

U p E (cf. Chapt. 10)


21-11 Electric DipolesRecall: The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r3

dependence:

( ) ( )3

2ˆ ˆ [along axis of ]

� � �kpE r rx x p

r= = ±


Along p axis

( ) ( )( )

( )( )

( ) ( ) ( )

( )

( ) ( ) ( ) ( )

2 2

2 2 2

3 32 2 3

ˆ

ˆ ˆ ˆˆ 1 / 2 1 / 2

/ 2 / 2

Recall:

1 1 if 1

Then:

ˆ ˆ 2 2 2 2ˆ ˆ ˆ1 1

�ℓ

� �ℓ ℓ

ℓ ℓ

≪

�� ℓ ℓ ℓ

n

p Q x

Q x Q x kQ xE r rx k r r

rr r

n

kQ x kQ x kp kp kpE r rx x x

r r r r r r r r

ε ε ε

− −

≡

± − ± ± = = + = − − + − +

+ ≈ +

± ± = = + − − = = ± → =


21-11 Electric DipolesExample:

The dipole moment of a water molecule is 6.3 x 10-30 C·m. A sample contains 10 21

water molecules, with their dipole moments all oriented in the direction of an electric field of 2.5 x 10 5 N/C. How much work is required to rotate the dipoles from this orientation ( θ = 0ο) to one in which all moments are perpendicular to the field (θ = 90ο)?


Solution:

( ) ( )( )( )( )21 30 5 3

cos

cos90 cos0

10 6.3 10 2.5 10 1.6 10

final initialext E E E

E

W U U U

u p E pE

W NpE NpE npE

J

θ

− −

= ∆ = −

= − ⋅ = −

→ = − − − =

= × × = ×

� �

��


Photocopy machine:

• drum is charged positively

• image is focused on drum

• only black areas stay charged and therefore attract toner particles

• image is transferred to paper and sealed by heat

21-13 Photocopy Machines and Computer Printers Use

Electrostatics



Electrostatics


Laser printer is similar, except a computer controls the laser intensity to form the image on the drum.


Electrostatics


• Two kinds of electric charge – positive and negative.

• Charge is conserved.

• Charge on electron:

e = 1.602 x 10-19 C.

• Conductors: electrons free to move.

• Insulators: nonconductors.

Summary of Chapter 21


• Charge is quantized in units of e.

• Objects can be charged by conduction or induction.

• Coulomb’s law:

•Electric field is force per unit charge:


1 2 1 212 12 122 2

12 0 12

12 2 20

1ˆ ˆ

4

1where 8.85 10 /

4

Q Q Q QF k r r

r r

C N mk

πε

επ

−

= =

= = × ⋅

�


• Electric field of a point charge:

• Electric field can be represented by electric

field lines.

• Static electric field inside conductor is zero; surface field is perpendicular to surface.


2

2

/ˆ ˆ for single point charge

F Qq r QE k r k r

q q r= = =�

�


Chapter 22Gauss’s Law


• Electric Flux

• Gauss’s Law

• Applications of Gauss’s Law

• Experimental Basis of Gauss’s and Coulomb’s Laws

Units of Chapter 22


Electric flux:

Electric flux through an area is proportional to the total number of field lines crossing the area.

22-1 Electric Flux

[[[[ ]]]] [[[[ ]]]][[[[ ]]]]

[[[[ ]]]]

2

2

0 20

Recall:

= == == == =

= → == → == → == → = ⋅⋅⋅⋅

E

E

NE A m

C

C C

N mε

ε

Φ

Φ


22-1 Electric FluxExample 22-1: Electric flux.

Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.


Flux through a closed surface:

22-1 Electric Flux


22-2 Gauss’s Law

1

2

3

4

0

0

⋅ ∝ + = +⋅ ∝ + = +⋅ ∝ + = +⋅ ∝ + = +

⋅ ∝ − = −⋅ ∝ − = −⋅ ∝ − = −⋅ ∝ − = −

⋅ =⋅ =⋅ =⋅ =

⋅ =⋅ =⋅ =⋅ =

∫∫∫∫

∫∫∫∫

∫∫∫∫

∫∫∫∫

��

��

��

��

�

�

�

�

S

S

S

S

E dA Q Q

E dA Q Q

E dA

E dA

κ

κ


22-2 Gauss’s Law

1

2

3

4

0

0

0

0

++++⋅ =⋅ =⋅ =⋅ =

−−−−⋅ =⋅ =⋅ =⋅ =

⋅ =⋅ =⋅ =⋅ =

⋅ =⋅ =⋅ =⋅ =

∫∫∫∫

∫∫∫∫

∫∫∫∫

∫∫∫∫

��

��

��

��

�

�

�

�

S

S

S

S

QE dA

QE dA

E dA

E dA

ε

ε

0

1≡≡≡≡κε


The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law:

This can be used to find the electric field in situations with a high degree of symmetry .

22-2 Gauss’s Law


22-2 Gauss’s LawFor a point charge,

Therefore,

Solving for E gives the result we expect from Coulomb’s law:

Spherical symmetry


22-2 Gauss’s LawUsing Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives:

Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.


22-2 Gauss’s Law

Finally, if a gaussian surface encloses several point charges, the superposition principle shows that:

Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

←←←←


22-2 Gauss’s Law

(((( ))))

1 2 3

0 0

9

12

2

3.1 5.9 3.1 10

8.85 10

667

−−−−

−−−−

+ ++ ++ ++ +⋅ = =⋅ = =⋅ = =⋅ = =

− − ×− − ×− − ×− − ×====

××××= ⋅= ⋅= ⋅= ⋅

∫∫∫∫��

�encl

S

q q q qE dA

N m C

ε ε

The figure shows five charged clumps of plastic and an electrically neutral coin as well as a Gaussian surface S. What is the flux through S if the charges are: ?1 4 2 5 33.1 , 5.9 , 3.1= = + = = − = −= = + = = − = −= = + = = − = −= = + = = − = −q q nC q q nC q nC


22-2 Gauss’s Law

A charge of 1.8 nC is placed at the center of a cube 3 cm on an edge. What is the electric flux through each face?


22-2 Gauss’s Law

A charge of 1.8 nC is placed at the center of a cube 3 cm on an edge. What is the electric flux through each face?

(((( ))))

0

92

120

6

1.8 1034

6 6 8.85 10

−−−−

−−−−

⋅ = ⋅ =⋅ = ⋅ =⋅ = ⋅ =⋅ = ⋅ =

××××→ ⋅ = = = ⋅→ ⋅ = = = ⋅→ ⋅ = = = ⋅→ ⋅ = = = ⋅××××

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

� ��

��

�Cube Face

Face

QE dA E dA

QE dA N m C

ε

ε


22-3 Applications of Gauss’s LawExample 22-3: Spherical conductor.

A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Spherical symmetry


22-3 Applications of Gauss’s LawExample 22-4: Solid sphere of charge.

An electric charge Qis distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere ( r < r0).

Spherical symmetry


22-3 Applications of Gauss’s LawExample 22-5: Nonuniformly charged solid sphere.

Suppose the charge density of a solid sphere is given by ρE = αr2,where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere. Spherical symmetry


22-3 Applications of Gauss’s Law

Example 22-6: Long uniform line of charge.

A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends.

Cylindrical symmetry


22-3 Applications of Gauss’s LawExample 22-7: Infinite plane of charge.

Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane. Planar, Cylindrical,

& Mirror Symmetry


22-3 Applications of Gauss’s LawExample 22-8: Electric field near any conducting surface.

Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by

E = σ/ε0

where σ is the surface charge density on the conductor’s surface at that point.

Approximate Planar & Cylindrical Symmetry


22-3 Applications of Gauss’s LawConceptual Example 22-9: Conductor with charge inside a cavity.

Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a pointcharge + q. What can

you say about the charges on the inner and outer surfaces of the conductor?


22-3 Applications of Gauss’s LawProcedure for Gauss’s law problems:

1. Identify the symmetry , and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field).

2. Draw the surface.

3. Use the symmetry to find the direction of E.

4. Evaluate the flux by integrating.

5. Calculate the enclosed charge.

6. Solve for the field.

��

E


22-4 Experimental Basis of Gauss’s and Coulomb’s Laws

In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero.


• Electric flux:

• Gauss’s law:

• Gauss’s law can be used to calculate the field in situations with a high degree of symmetry.

• Gauss’s law applies in all situations.

• Follows from Coulomb’s Law


(((( ))))21

r↑↑↑↑


Questions?

Chapter 21 Electric Charge and Electric Fieldpeople.virginia.edu/~ben/2415141/Lecture_3.pdf · 22-2...

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