Chapter 22
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Transcript of 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.
Charge and electric flux
• Positive charge within the box produces outward electric flux through the surface of the box.
More charge = more 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
Zero net charge inside a box
• Three cases of zero net charge inside a box
• No net electric flux through surface of box!
2
Zero net charge inside a box
• Three cases of zero net charge inside a box
• No net electric flux through surface of box!
3
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
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)
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 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?
= E·A = EA cos(30°)
A = r2 = 0.0314 m2
=54 Nm2/C
Example 22.1 - Electric flux through a disk
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?
Example 22.1 - Electric flux through a disk
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…
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!
Electric flux through a cube
• An imaginary cube of side L is in a region of uniform E. Find the flux through each side…
• Which will be positive?
• 2 = POSITIVE + EL2
• 1 = NEGATIVE - EL2
Electric flux through a cube
• An imaginary cube of side L is in a region of uniform E. Find the flux through each side…
• Which will be positive?
• 2 = POSITIVE + EL2
• 1 = NEGATIVE - EL2
• NET flux = 0
• No charge inside!!
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 cube
• An imaginary cube of side L is in a region of uniform E. Find the flux through each side…
• Which will be positive?
• 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….
Electric flux through a sphere
+r
q
E varies in direction everywhere on the surface, but not magnitude
Normal to surface varies in direction everywhere on the surface
Electric flux through a sphere
= ∫ E·dA
E = kq/r2
= 1/(40) q/r2
and is parallel to dAeverywhere on the surface
= ∫ E·dA = E ∫dA = EA
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
= ∫ 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
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/0B = -q/0
C = 0 !
Conceptual Example 22.4
• What is the flux through the surfaces A, B, C, and D?
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!
Applications of Gauss’s law
• 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 law
•
WHY ???
• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface.
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
• E around an infinite positive wire of charge density ?
·E = ?
·E = ?
·E = ?
·E = ?
Charge/meter =
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
·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!
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
Imagine closed cylindrical Gaussian Surface around the wire a distance r away…
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
• 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!
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•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 charge
• E around an infinite positive wire of charge density ?
•E is constant in value everywhere on the cylinder at a distance r from the wire!
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•E is parallel to dA everywhere on the cylinder, so E ◦ dA = EdA
= ∫ E·dA= ∫ EdA
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
• Integration over curved cylindrical side is two-dimensional |dA| = (rd dl
dl
rd
rd
dA
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•E is constant in value everywhere on the cylinder at a distance r from the wire!
= ∫ E·dA = ∫ ∫ E(rd)dl
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•E is constant in value everywhere on the cylinder at a distance r from the wire!
= ∫ E·dA = ∫ ∫ (E) · (rd)dl = E ∫ ∫ rddl
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•Limits of integration?
• dgoes from 0 to 2
• dl goes from 0 to l (length of cylinder)
= ∫ E·dA = E ∫ ∫ rddl
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
Surface Area of cylinder (but not end caps, since net flux there = 0
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
•E is constant in value everywhere on the cylinder at a distance r from the wire!
= ∫ E·dA = (E) x (Surface Area!) = E(2r)l
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
How much charge enclosed in the closed surface?
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
How much charge enclosed in the closed surface?
Q(enclosed) = (charge density) x (length) = l
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
• So… q/0 = (l) /0
• Gauss’ Law gives us the flux
• = E(2r) l = q/0 = (l) /0
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density ?
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: 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
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 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 /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 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.
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
A uniformly charged insulating sphere
• E field inside a sphere uniformly filled with charge.
Find Qenclosed in sphere of radius r <R?
Qenclosed = (4/3 r3 ) x
A uniformly charged insulating sphere
• E field inside a sphere uniformly filled with charge.
Find Flux through surface?
•E field uniform, constant at any r
•E is parallel to dA at surface
so…
= ∫ E·dA = E(4r2)
A uniformly charged insulating sphere
• E field inside a sphere uniformly filled with charge.
= ∫ E·dA = E(4r2) = (4/3 r3 ) x
And
E = r/30 (for r < R)
or
E = Qr/4R30 (for r < R)
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
• Empty cavity inside conductor has no field, nor charge on inner surface
Charges on conductors with cavities
• Isolated charges inside cavity “induce” opposite charge, canceling field inside the conductor!
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