chapter 22

Copyright © 2012 Pearson Education Inc – Modified by Scott Hildreth, Chabot College 2016.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 22

Gauss’s Law

Goals for Chapter 22

• Use electric field at a surface to determine charge within the surface

• Learn meaning of electric flux & how to calculate

• Learn relationship between electric flux through a surface & charge within the surface

Goals for Chapter 22

• Use Gauss’s law to calculate electric fields

• Recognizing symmetry

• Setting up two-dimensional surface integrals

• Learn where charge on a conductor is located

Charge and electric flux

• Positive charge within the box produces outward electric flux through the surface of the box.


• Positive charge within the box produces outward electric flux through the surface of the box.

More charge = more flux!


• Negative charge produces inward flux.

• Flux can be negative!!


• More negative charge – more inward flux!

Zero net charge inside a box

• Three cases of zero net charge inside a box

• No net electric flux through surface of box!

1




2




3




1 2 3

What affects the flux through a box?

• Doubling charge within box doubles flux.

What affects the flux through a box?

• Doubling charge within box doubles flux.

• Doubling size of box does NOT change flux.

Uniform E fields and Units of Electric Flux

= E·A = EA cos(°)

C · m2 = Nm2/C

For a UNIFORM E field (in space)

Calculating electric flux in uniform fields

Example 22.1 - Electric flux through a disk

Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux?


Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux?

= E·A = EA cos(30°)

A = r2 = 0.0314 m2

=54 Nm2/C


Disk of radius 0.10 m, magnitude E of 2.0 x 103 N/C. What is the flux if n is perpendicular to E?


Disk of radius 0.10 m, magnitude E of 2.0 x 103 N/C. What is the flux if n is Parallel to E?

Electric flux through a cube

• An imaginary cube of side L is in a region of uniform E. Find the flux through each side…



• Start with easy ones!

• 3 & 4 = ?

• 5 & 6 = ?

• 3,4,5,6 = 0!



• Which will be positive?

• 2 = POSITIVE + EL2

• 1 = NEGATIVE - EL2




• 2 = POSITIVE + EL2

• 1 = NEGATIVE - EL2

• NET flux = 0

• No charge inside!!




• 2,4 = POSITIVE +

• 2 = +EL2 cos• 4 = +EL2 cos(90-

• 1,3 = NEGATIVE –

Electric flux through a sphere

+r

q

Consider flux through a sphere of radius r around a charge of +q….


+r

q

E varies in direction everywhere on the surface, but not magnitude

Normal to surface varies in direction everywhere on the surface


Imagine a small segment of area dA; what is flux through that?


= ∫ E·dA

Now what is flux across entire surface?


= ∫ E·dA

E = kq/r2

= 1/(40) q/r2

and is parallel to dAeverywhere on the surface

= ∫ E·dA = E ∫dA = EA


= ∫ E·dA

E = kq/r2 and is parallel to dAeverywhere on the surface

= ∫ E·dA = E ∫dA = EA

For q = +3.0nC, flux through sphere of radius r=.20 m?

Gauss’ Law

= ∫ E·dA =qenc0

S

Gauss’ Law

= ∫ E·dA =qenc0

Electric Flux is produced by charge in space

S

Gauss’ Law

= ∫ E·dA =qenc0

You integrate over a CLOSED surface (two dimensions!)

S

Gauss’ Law

= ∫ E·dA =qenc0

E field is a VECTOR

S

Gauss’ Law

= ∫ E·dA =qenc0

Infinitesimal area element dA is also a vector; this is what you sum

S

Gauss’ Law

= ∫ E·dA =qenc0

Dot product tells you to find the part of E that is PARALLEL to dA at that point (perpendicular to the surface)

S

Gauss’ Law

= ∫ E·dA =qenc0

Dot product is a scalar: E·dA =

ExdAx + EydAy + EzdAz = |E||dA|cos

S

Gauss’ Law

= ∫ E·dA =qenc0

The TOTAL amount of charge…

S

Gauss’ Law

= ∫ E·dA =qenc0

… but ONLY charge inside S counts!

S

Gauss’ Law

= ∫ E·dA =qenc0

The electrical permittivity of free space, through which the field is acting.

S

Why is Gauss’ Law Useful?

• Given info about a distribution of electric charge, find the flux through a surface enclosing that charge.

•Given info about the flux through a closed surface, find the total charge enclosed by that surface.

•For highly symmetric distributions, find the E field itself rather than just the flux.

Gauss’ Law for Spherical Surface…

• Flux through sphere is independent of size of sphere

• Flux depends only on charge inside.

= ∫ E·dA = +q/0

Point charge inside a nonspherical surface

As before, flux is independent of surface & depends only on charge inside.

Positive and negative flux

• Flux is positive if enclosed charge is positive, & negative if charge is negative.

Conceptual Example 22.4

• What is the flux through the surfaces A, B, C, and D?



A = +q/0



A = +q/0B = -q/0



A = +q/0B = -q/0

C = 0 !



A = +q/0B = -q/0

C = 0 !

D = 0 !!

Applications of Gauss’s law

• Recall from Chapter 21…

Under electrostatic conditions, E field inside a conductor is 0!

WHY?????

E = 0 inside!


• Under electrostatic conditions, E field inside a conductor is 0!

• Assume the opposite! IF E field inside a conductoris not zero, then …

– E field WILL act on free charges in conductor

– Those charges WILL move in response to the force of the field

– They STOP moving when net force = 0

– Which means IF static, THEN no field inside conductor!


•

WHY ???

• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface.


• Consider Gaussian surface inside conductor!


• Under electrostatic conditions, field outside ANY spherical conductor looks just like a point charge!

Example 22.6: Field of a line charge

• E around an infinite positive wire of charge density ?

·E = ?

·E = ?

·E = ?

·E = ?

Charge/meter =



·E = ?

Charge/meter =

• You know charge, you WANT E field (Gauss’ Law!)

• Choose Gaussian Surface with symmetry to match charge distribution to make calculating ∫ E·dA easy!



Imagine closed cylindrical Gaussian Surface around the wire a distance r away…



• Look at ∫ E·dA; remember CLOSED surface means you sum flux over ALL sides!

• Three components: the cylindrical side, and the two ends. Need flux across each!



•E is orthogonal to dA at the end caps.

•E is parallel (radially outwards) to dA on cylinder



•E is constant in value everywhere on the cylinder at a distance r from the wire!



•E is parallel to dA everywhere on the cylinder, so E ◦ dA = EdA

= ∫ E·dA= ∫ EdA



• Integration over curved cylindrical side is two-dimensional |dA| = (rd dl

dl

rd

rd

dA




= ∫ E·dA = ∫ ∫ E(rd)dl




= ∫ E·dA = ∫ ∫ (E) · (rd)dl = E ∫ ∫ rddl



•Limits of integration?

• dgoes from 0 to 2

• dl goes from 0 to l (length of cylinder)

= ∫ E·dA = E ∫ ∫ rddl



Surface Area of cylinder (but not end caps, since net flux there = 0




= ∫ E·dA = (E) x (Surface Area!) = E(2r)l



How much charge enclosed in the closed surface?



How much charge enclosed in the closed surface?

Q(enclosed) = (charge density) x (length) = l



• So… q/0 = (l) /0

• Gauss’ Law gives us the flux

• = E(2r) l = q/0 = (l) /0



And… E = (l) /0 r = ( / 2r 0) r

(2r) l

Don’t forget E

is a vector!

Unit vector radially out from line

Field of a sheet of charge

• Example 22.7 for an infinite plane sheet of charge?


• Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ?

Find Q enclosed to start!

Qenc = A

Where is the charge/area of the sheet



Flux through entire closed surface of the cylinder?

∫ E·dA

= E1A (left) + E2A (right) (only)

Why no flux through the cylinder?



Flux through entire closed surface of the cylinder?

∫ E·dA = Qenc /0

= 2EA = Qenc /0 = A / 0

So

E = / 20 for infinite sheet

Ex. 22.8 Field between two parallel conducting plates

• E field between oppositely charged parallel conducting plates.

Field between two parallel conducting plates

• Superposition of TWO E fields, from each plate…

Field between two parallel conducting plates

• Superposition of 2 fields from infinite sheets with same charge!

For EACH sheet,

E = / 20

And is the same in magnitude, directions of E field from both sheets is the same, so

Enet = 2( / 20) = / 0

Ex. 22.9 - A uniformly charged insulating sphere

• E field both inside and outside a sphere uniformly filled with charge.



• E field outside a sphere uniformly filled with charge? EASY

• Looks like a point=charge field!

A uniformly charged insulating sphere

• E field inside a sphere uniformly filled with charge.

Find Qenclosed to start!

Start with charge density function

Qtotal/Volume (if uniform)

Nota Bene

Density function could be non-uniform! r) r



Find Qenclosed in sphere of radius r <R?

Qenclosed = (4/3 r3 ) x



Find Flux through surface?

•E field uniform, constant at any r

•E is parallel to dA at surface

so…

= ∫ E·dA = E(4r2)



= ∫ E·dA = E(4r2) = (4/3 r3 ) x

And

E = r/30 (for r < R)

or

E = Qr/4R30 (for r < R)

Charges on conductors with cavities

• E = 0 inside conductor…


• Empty cavity inside conductor has no field, nor charge on inner surface


• Isolated charges inside cavity “induce” opposite charge, canceling field inside the conductor!

A conductor with a cavity

• Conceptual Example 22.11

Electrostatic shielding

• A conducting box (a Faraday cage) in an electric field shields the interior from the field.

Electrostatic shielding

• A conducting box (a Faraday cage) in an electric field shields the interior from the field. See http://www.youtube.com/watch?v=FvtfE-ha8dE

http://www.youtube.com/watch?v=FvtfE-ha8dE

chapter 22

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