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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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