Chapter 22 Gauss’s Law

• Gauss’s Law is a relationship between the field at all the points on the surface and the total charge enclosed within the surface.

• Gauss’s Law is part of the key to using symmetry considerations to simplify electric-field calculations.

• In Chapter 21 we asked the question “Given a charge distribution, what is the electric field produced by that distribution at point P?”

• The answer from Chapter 21 is: The total field at point P is the vector sum of the fields due to all the point charges.

Chapter 22 1

Introduction– The movement of electrons

can be shocking (pardon the pun).

– If you look at the girl’s hair (figure to the right), you’ll see the electrons coating each individual hair fiber and then repelling each other.

– Gauss imagined a flow through a surface placed around a charge and then considered outcomes that we will study in Chapter 22.

Chapter 22 3

Twice the charge, twice as many lines penetrating closed surface

The charge within the box determines the magnitude and

direction of the Electric field outside the box.

Chapter 22 4

The net electric flux through the surface of the box is directly proportional to the magnitude of the net charge enclosed in the box.

Electric Flux is the “Flow” of the Electric Field in or out of an enclosed surface

Chapter 22 5

Gauss’s Law – Qualitative statementThe net electric flux (number of lines) penetrating a closed surface is directly proportional to the total electric charge enclosed by the closed surface and is independent of the size of the enclosed surface.

Electric Flux = The net sum of the product of the perpendicular component of E and the surface area

Chapter 22 6

The Electric Flux is the scalar product of E and A

If the surface (A) is not flat and/or the electric field (E) is not uniform, A is divided intosmall elements dA and integrated over the entire surface (A).

En

Et

E

cosEEn

AEEAAEnE

cos

EAE 0cos EAE

Chapter 22 7

Electric Flux Through a disk - Example 22-1 (p.756)

E = 2 x 103 N/C; the disk of radius 0.1m is at an angle. (a) Disk angle = 30o; electric flux E through the disk? Answer: E = 54 N-m2/C (b) Disk angle = 90o ; (the disk parallel to E ) electric flux E through the disk? Answer: E = 0 N-m2/C (c) Disk angle = 0o; (the disk is normal to E) electric flux E through the disk? Answer: E = 63 N-m2/C

Chapter 22 8

Electric Flux through a sphere - Example 22-3 (p.757)

q = 3 µC, r = 0.2 mDetermine electric flux through the sphere dAEEdAdAEAdEE cos

204 r

qE

24 rAdA EAE

Setup

CmNxxEAE /.104.3)20.0)(4)(1075.6( 2529

CNxxxr

qE /1075.6)2.0/()103)(109(

45269

20

Execute

Evaluate The is independent of the radius. Larger r produces a smaller E and a larger A

Chapter 22 9

0

22

0

)4(4

1

q

rr

qEAE

204

1

r

qE

(22-6)

0q

E (where E is the total electric flux penetrating surface A)

Gauss’s Law for a point charge in a sphere

Chapter 22 10

Mathematical Representation of Gauss’s Law - General Case

Figure 22-13

cosEE

AdEdAEdAEd E

cos

Chapter 22 11

Excess charge on a solid conductor resides entirely on the conductor’s surface

Qencl = 0

Therefore:E and E penetrating gaussian surface = 0

Total electric flux penetratingimaginary gaussian surface

This means that the electric field (E) inside thecharged solid conductor is zero.

Applications of Gauss’s Law

Note: A Gaussian surface does not have to be a physical surface

Chapter 22 12

Problem 22.8

Chapter 22 13

Electric Field (E) outside a Charged Solid Conductor Sphere (Ex 22-5, p.762)

Figure 22-18

dAEdAEAdE cos

0encl

E

QAdE

00

2 )4(

qQrEdAE encl

0

2 )4(

qrE

204

1

r

qE

(r R)

q(imaginary)

Chapter 22 14

Electric field of a line charge (Example 22.6, p.763)Problem: Electric charge is distributed uniformly along an infinite long, thin wire. The charge per unit length is λ (assume positive). Find the electric field.

Solution: The field lines are radial and lie in a plane perpendicular to the wire. The field magnitude depends only on the radial distance from the wire.Execute: No flux out the ends. All flux out the walls of the cylinder, EA

lQencl

0

00

)2)((

)2)((

l

rlE

rlElQencl

E

rE

02

1

)2)(( rlEE Total flux: Total enclosed charge:

Solve for E:

From Gauss’s law:

Identify: The system has cylindrical symmetry. The field must point away from the + charge

Evaluation: Not valid for short wire. For a short wire use the method of Example 21.11(p.731).

note: independent of length

Chapter 22 15

Field between oppositely charged parallel conducting plates (Ex 22-8)

00021 22

EEE

Two charged parallel plates . Charge per unit area is +σ and -σ Identify: Opposite charges attract. “Fringing” on the edges of plates. Neglect for large plates. Setup: Exploit symmetry. S1, S2, S3, S4 are cylinders with end area AExecute: Consider S1, flux through the end is EA, net charge enclosed is σATherefore from Gauss’s Law, the total flux = enclosed charge/Єo

E=0E=0

0A

EA0

ESolve for E:

Evaluate: The same result was obtained with superposition of fields . Example 21.13, p.732

There are practical applications– Figure 22.17 treats excess charge as residing on the surface of a

conductor.– Consider Example 22.5.– Figure 22.18 illustrates Example 22.5.

The field of a line or plane of charge– Consider Example 22.6 and Figure 22.19.– See also Example 22.7 and Figure 22.20.

The field of a uniformly charged sphere– Consider Example 22.9.

– Figure 22.22 illustrates the example.– Follow Example 22.10.

Charges on conductors– The electric field within a charged conductor may be found.– Consider Figure 22.23.– Follow Example 22.11 and Figure 22.24.

Chapter 22 20

Excess charge on solid conductors are located on the surface

The electric field at every point within a conductor is zero and any excess charge on a solid conductor is located entirely on the surface.

Experimental tests of Gauss’s Law– Regard Figure 22.25. – A metal container on an insulating stand.

The Van de Graaff generator– The source of all the static on the child’s hair in our introduction.– Consider Figure 22.27 below.

Chapter 22 23

6 in spark

R =15.0 cm radius sphere

Prob. 22.17On a Humid day E=2.00 x 104 N/C will produce a one inch spark

(a)Using Gauss’ Law what is the charge stored on the surface of the sphere.

(b)Assume all charge is concentrated in the sphere center, us e Coulomb’s law to calculated the electric field.

Note : P0 = Є0

Identify: The electric field required to produce a spark 6 in. long is 6 times as strong as the field needed to produce a spark 1 in. long.

Set Up:  Gauss’s law,

By Gauss’s Law solve for the enclosed charge: 0q EAP

P0 = Є0

the electric field is the same as for a point-charge, 20

1

4

qE

r

P

Execute: (a) The electric field for 6-inch sparks is six times the electric field of a one inch spark

4 56 2.00 10  N/C 1.20 10  N/CE

The charge to produce this field is 2 12 2 2 5 2 70 0 (4 ) (8.85 10  C /N m )(1.20 10  N/C)(4 )(0.15 m) 3.00 10  Cq EA E r P P

(b) Using Coulomb’s law gives

79 2 2 5

2

3.00 10  C(9.00 10  N m /C ) 1.20 10  N/C

(0.150 m)E

Evaluate: It takes only about to produce this field 0.3 C

A Faraday cage blocks flow– Refer to Figure 22.28 below.– Science-fiction movies always place alien transmitters in these

to prevent them from calling for help.– Follow Examples 22.12 and 22.13.

Chapter 22 25

Chapter-22 Important Equations

Chapter 22 26

Chapter-22 Important Concepts

Chapter 22 27

Chapter-22 Important concepts and results

A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it.

Compared to the electric flux through surface #1, the flux through surface #2 is

A. greater. B. the same.

C. less, but not zero. D. zero.

E. not enough information given to decide

Q22.1

+q

Gaussian surface #1

Gaussian surface #2

A solid spherical conductor has a spherical cavity in its interior. The cavity is not centered on the center of the conductor.

If a positive charge is placed on the conductor, the electric field in the cavity

A. points generally toward theouter surface of the conductor.

B. points generally away from the outer surface of the conductor.

C. is zero.

D. not enough information given to decide

Q22.4Conductor

Cavity