Chapter 22
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Chapter 22. Gauss’s Law. Goals for Chapter 22. To use the electric field at a surface to determine the charge within the surface To learn the meaning of electric flux and how to calculate it

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Chapter 22

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Chapter 22

Chapter 22

Gauss’s Law


Goals for chapter 22

Goals for Chapter 22

  • To use the electric field at a surface to determine the charge within the surface

  • To learn the meaning of electric flux and how to calculate it

  • To learn the relationship between the electric flux through a surface and the charge within the surface


Goals for chapter 221

Goals for Chapter 22

  • To use Gauss’s law to calculate electric fields

    • Recognizing symmetry

    • Setting up two-dimensional surface integrals

  • To learn where the charge on a conductor is located


Charge and electric flux

Charge and electric flux

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


Charge and electric flux1

Charge and electric flux

  • Positive charge within the box produces outward electric flux through the surface of the box. More charge = more flux!


Charge and electric flux2

Charge and electric flux

  • Negative charge produces inward flux.


Charge and electric flux3

Charge and electric flux

  • More negative charge – more inward flux!


Zero net charge inside a box

Zero net charge inside a box

  • Three cases of zero net charge inside a box

  • No net electric flux through surface of box!


Zero net charge inside a box1

Zero net charge inside a box

  • Three cases of zero net charge inside a box

  • No net electric flux through surface of box!


Zero net charge inside a box2

Zero net charge inside a box

  • Three cases of zero net charge inside a box

  • No net electric flux through surface of box!


Zero net charge inside a box3

Zero net charge inside a box

  • Three cases of zero net charge inside a box

  • No net electric flux through surface of box!


What affects the flux through a box

What affects the flux through a box?

  • Doubling charge within box doubles flux.


What affects the flux through a box1

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

Uniform E fields and Units of Electric Flux

  • F= E·A = EA cos(q°)

  • [F] = N/C ·m2 = Nm2/C

For a UNIFORM E field (in space)


Calculating electric flux in uniform fields

Calculating electric flux in uniform fields


Calculating electric flux in uniform fields1

Calculating electric flux in uniform fields


Calculating electric flux in uniform fields2

Calculating electric flux in uniform fields


Example 22 1 electric flux through a disk

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?


Example 22 1 electric flux through a disk1

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?

F = E·A = EA cos(30°)

A = pr2 = 0.0314 m2

F =54 Nm2/C


Example 22 1 electric flux through a disk2

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? What if n is perpendicular to E? Parallel to E?


Electric flux through a cube

Electric flux through a cube

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


Electric flux through a sphere

Electric flux through a sphere

F = ∫ E·dA


Electric flux through a sphere1

Electric flux through a sphere

  • = ∫ E·dA

    E = kq/r2

    = 1/(4pe0) q/r2 and is parallel to dAeverywhere on the surface

  • = ∫ E·dA = E ∫dA = EA


Electric flux through a sphere2

Electric flux through a sphere

  • = ∫ 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

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S


Gauss law1

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

Electric Flux is produced by charge in space


Gauss law2

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

You integrate over a CLOSED surface (two dimensions!)


Gauss law3

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

E field is a VECTOR


Gauss law4

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

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


Gauss law5

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

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


Gauss law6

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

Dot product is a scalar: E·dA =

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


Gauss law7

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

The TOTAL amount of charge…


Gauss law8

Gauss’ Law

  • = ∫ E·dA =qenc

… but ONLY charge inside S counts!

e0

S


Gauss law9

Gauss’ Law

  • = ∫ E·dA =qenc

e0

S

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


Why is gauss law useful

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

Gauss’ Law for Spherical Surface…

  • Flux through sphere is independent of size of sphere

  • Flux depends only on charge inside.

  • F= ∫ E·dA = +q/e0


Point charge inside a nonspherical surface

Point charge inside a nonspherical surface

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


Positive and negative flux

Positive and negative flux

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


Conceptual example 22 4

Conceptual Example 22.4

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


Conceptual example 22 41

Conceptual Example 22.4

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

FA = +q/e0


Conceptual example 22 42

Conceptual Example 22.4

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

FB = -q/e0

FA = +q/e0


Conceptual example 22 43

Conceptual Example 22.4

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

FB = -q/e0

FA = +q/e0

FC = 0 !


Conceptual example 22 44

Conceptual Example 22.4

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

FB = -q/e0

FA = +q/e0

FC = 0 !

FD = 0 !!


Applications of gauss s law

Applications of Gauss’s law

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

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


Applications of gauss s law1

Applications of Gauss’s law

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

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

    WHY ???


Applications of gauss s law2

Applications of Gauss’s law

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

  • 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!


Applications of gauss s law3

Applications of Gauss’s law

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


Example 22 6 field of a line charge

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

·E = ?

·E = ?

Charge/meter = l

·E = ?

·E = ?


Example 22 6 field of a line charge1

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

Charge/meter = l

·E = ?

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

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


Example 22 6 field of a line charge2

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

Imagine closed cylindricalGaussian Surface around the wire a distance r away…


Example 22 6 field of a line charge3

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

  • Look at∫ E·dA; remember CLOSED surface!

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


Example 22 6 field of a line charge4

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

  • E is orthogonal to dA at the end caps.

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


Example 22 6 field of a line charge5

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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


Example 22 6 field of a line charge6

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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

  • = ∫ E·dA= ∫ EdA


Example 22 6 field of a line charge7

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

dA

rdq

dl

dq

r

  • The integration over area is two-dimensional

  • |dA| = (rdq) dl


Example 22 6 field of a line charge8

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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

  • = ∫ E·dA = ∫ ∫ E(rdq)dl


Example 22 6 field of a line charge9

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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

  • = ∫ E·dA = ∫ ∫ (E) · (rdq)dl = E ∫ ∫ rdqdl


Example 22 6 field of a line charge10

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

  • Limits of integration?

    • dqgoes from 0 to 2p

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

  • = ∫ E·dA = E ∫ ∫ rdqdl


Example 22 6 field of a line charge11

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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


Example 22 6 field of a line charge12

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

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

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


Example 22 6 field of a line charge13

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

How much charge enclosed in the closed surface?


Example 22 6 field of a line charge14

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

How much charge enclosed in the closed surface?

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


Example 22 6 field of a line charge15

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

So… q/e0 = (l l) /e0Gauss’ Law gives us the flux F= E(2pr) l = q/e0 = (l l) /e0


Example 22 6 field of a line charge16

Example 22.6: Field of a line charge

  • E around an infinite positive wire of charge density l?

Don’t forget E is a vector!

And… E = (l l) /e0 r= (l / 2pr e0)r

(2pr) l

Unit vector radially out from line


Field of a sheet of charge

Field of a sheet of charge

  • Example 22.7 for an infinite plane sheet of charge?


Field of a sheet of charge1

Field of a 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 = sA

Where s is the charge/area of the sheet


Field of a sheet of charge2

Field of a 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 ?

  • Flux through entire closed surface of the cylinder?

  • = ∫ E·dA= E1A (left) + E2A (right) (only)

  • Why no flux through the cylinder?


Field of a sheet of charge3

Field of a 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 ?

  • Flux through entire closed surface of the cylinder?

  • = ∫ E·dA = Qenc / e0= 2EA = Qenc / e0 = sA / e0

  • So E =s / 2e0 for infinite sheet


Ex 22 8 field between two parallel conducting plates

Ex. 22.8 Field between two parallel conducting plates

  • E field between oppositely charged parallel conducting plates.


Field between two parallel conducting plates

Field between two parallel conducting plates

  • Superposition of TWO E fields, from each plate…


Field between two parallel conducting plates1

Field between two parallel conducting plates

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

For EACH sheet,

E =s / 2e0

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

Enet =2(s / 2e0) = s / e0


Ex 22 9 a uniformly charged insulating sphere

Ex. 22.9 - A uniformly charged insulating sphere

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


Ex 22 9 a uniformly charged insulating sphere1

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

A uniformly charged insulating sphere

  • E field insidea sphere uniformly filled with charge.

Find Qenclosed to start!

Start with charge density function r

r = Qtotal/Volume (if uniform)

NB!!

Could be non-uniform!r(r) = kr2


A uniformly charged insulating sphere1

A uniformly charged insulating sphere

  • E field insidea sphere uniformly filled with charge.

Find Qenclosed in sphere of radius r <R?

Qenclosed = (4/3 pr3 ) x r


A uniformly charged insulating sphere2

A uniformly charged insulating sphere

  • E field insidea sphere uniformly filled with charge.

  • Find Flux through surface?

  • E field uniform, constant at any r

  • E is parallel to dA at surfaceso…

    • F= ∫ E·dA = E(4pr2)


A uniformly charged insulating sphere3

A uniformly charged insulating sphere

  • E field insidea sphere uniformly filled with charge.

  • F= ∫ E·dA = E(4pr2) = (4/3 pr3 ) x r

  • And

  • E = rr/3e0 (for r < R)

  • or

  • E = Qr/4pR3e0 (for r < R)


  • Ex 22 9 a uniformly charged insulating sphere2

    Ex. 22.9 - A uniformly charged insulating sphere

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


    Charges on conductors with cavities

    Charges on conductors with cavities

    • E = 0 inside conductor…


    Charges on conductors with cavities1

    Charges on conductors with cavities

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


    Charges on conductors with cavities2

    Charges on conductors with cavities

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


    A conductor with a cavity

    A conductor with a cavity

    • Conceptual Example 22.11


    Electrostatic shielding

    Electrostatic shielding

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


    Electrostatic shielding1

    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


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