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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

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)





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 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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