Transcript Gauss` Law - Chabot College

Chapter 22
Gauss’s Law
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc – Modified by Scott Hildreth, Chabot College 2016.
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!
Charge and electric flux
• Negative charge produces inward flux.
• Flux can be negative!!
Charge and electric flux
• 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
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.
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
For a UNIFORM E field (in space)
F= E·A = EA cos(q°)
[F]
= N/C · m2 = Nm2/C
Calculating electric flux in uniform fields
Calculating electric flux in uniform fields
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?
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 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!
• f3 & f4 = ?
• f5 & f6 = ?
• F3,4,5,6 = 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?
• f2 = POSITIVE + EL2
• F1 = 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?
• f2 = POSITIVE + EL2
• F1 = 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?
• f2,4 = POSITIVE +
• F2 = +EL2 cosq
• F4 = +EL2 cos(90-q)
• F1,3 = NEGATIVE –
Electric flux through a sphere
Consider flux through a sphere of radius r around
a charge of +q….
r
+
q
Electric flux through a sphere
E varies in direction
everywhere on the
surface, but not
magnitude
Normal to surface
varies in direction
everywhere on the
surface
r
+
q
Electric flux through a sphere
Imagine a small segment of area dA; what is flux
through that?
Electric flux through a sphere
Now what is flux across entire surface?
F = ∫ E·dA
Electric flux through a sphere
F= ∫ E·dA
E = kq/r2
= 1/(4pe )
0
q/r2
and is parallel to dA
everywhere on the surface
F= ∫ E·dA = E ∫dA = EA
Electric flux through a sphere
F = ∫ E·dA
E = kq/r2 and is parallel to dA
everywhere on the surface
F = ∫ E·dA = E ∫dA = EA
For q = +3.0nC, flux through sphere of radius r=.20 m?
Gauss’ Law
F= ∫ E·dA =qenc
S
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
Electric Flux is produced by charge
in space
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
You integrate over a CLOSED
surface (two dimensions!)
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
E field is a VECTOR
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
Infinitesimal area element dA is
also a vector; this is what you sum
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
Dot product tells you to find the part
of E that is PARALLEL to dA at that
point (perpendicular to the surface)
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
Dot product is a scalar: E·dA =
ExdAx + EydAy + EzdAz =
|E||dA|cosq
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
The TOTAL amount of charge…
e0
Gauss’ Law
… but ONLY charge inside S
counts!
F= ∫ E·dA =qenc
S
e0
Gauss’ Law
F= ∫ E·dA =qenc
S
The electrical permittivity of free space,
through which the field is acting.
e0
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.
 F= ∫ E·dA = +q/e0
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?
Conceptual Example 22.4
• What is the flux through the surfaces A, B, C, and D?
FA = +q/e0
Conceptual Example 22.4
• What is the flux through the surfaces A, B, C, and D?
FB = -q/e0
FA = +q/e0
Conceptual Example 22.4
• What is the flux through the surfaces A, B, C, and D?
FB = -q/e0
FA = +q/e0
FC = 0 !
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
• 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 conductor
is 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
• Under electrostatic conditions, any excess charge on a
conductor resides entirely on its surface.
WHY ???
Applications of Gauss’s law
• Consider Gaussian surface inside conductor!
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 l?
·E = ?
·E = ?
Charge/meter = l
·E = ?
·E = ?
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 charge
• E around an infinite positive wire of charge density l?
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 l?
• 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 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 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 charge
• E around an infinite positive wire of charge density l?
•E is parallel to dA everywhere on the cylinder, so
E ◦ dA = EdA
F = ∫ E·dA= ∫ EdA
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density l?
• Integration over curved cylindrical side is twodimensional
|dA| = (rdq) dl
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!
F = ∫ E·dA = ∫ ∫ E(rdq)dl
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!
F = ∫ E·dA = ∫ ∫ (E) · (rdq)dl = E ∫ ∫ rdqdl
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density l?
•Limits of integration?
• dq goes from 0 to 2p
• dl goes from 0 to l (length of cylinder)
F = ∫ E·dA = E ∫ ∫ rdqdl
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 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!
F = ∫ E·dA = (E) x (Surface Area!) = E(2pr)l
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 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 charge
• E around an infinite positive wire of charge density l?
• So… q/e0 = (l l) /e0
•
Gauss’ Law gives us the flux
• F = E(2pr) l = q/e0 = (l l) /e0
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density l?
And… E =
(l l) /e0 r
= (l / 2pr e0) r
Don’t
forget E
is a
vector!
(2pr) l
Unit vector
radially out
from line
Field of a sheet of charge
• Example 22.7 for an infinite plane sheet of charge?
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 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?
F = ∫ 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?
F = ∫ 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
• 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 = 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
• 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 r
r = Qtotal/Volume (if uniform)
Nota Bene! NB
Density function could be non-uniform!
r(r) = kr2
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 pr3 ) x r
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…
F= ∫ E·dA = E(4pr2)
A uniformly charged insulating sphere
• E field inside a 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 sphere
• E field both inside and outside a sphere uniformly filled with
charge.
Charges on conductors with cavities
• E = 0 inside conductor…
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!
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