Transcript Gauss`s 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.
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 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
• 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.
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!
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 box
• Three cases of zero net charge inside a box
• No net electric flux through surface of box!
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?
• 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 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
• An imaginary cube of side L is in a region of uniform
E. Find the flux through each side…
Electric flux through a sphere
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.
Q22.1
A spherical Gaussian surface (#1)
encloses and is centered on a point charge
+q. A second spherical Gaussian surface
(#2) of the same size also encloses the
charge but is not centered on it.
Compared to the electric flux
through surface #1, the
flux through surface #2 is
A. greater.
B. B. the same.
Gaussian
surface #1
C. less, but not zero.
D. zero.
E. not enough information given to decide
© 2012 Pearson Education, Inc.
+q
Gaussian
surface #2
A22.1
A spherical Gaussian surface (#1)
encloses and is centered on a point charge
+q. A second spherical Gaussian surface
(#2) of the same size also encloses the
charge but is not centered on it.
Compared to the electric flux
through surface #1, the
flux through surface #2 is
A. greater.
B. B. the same.
Gaussian
surface #1
C. less, but not zero.
D. zero.
E. not enough information given to decide
© 2012 Pearson Education, Inc.
+q
Gaussian
surface #2
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 !!
Q22.2
Two point charges, +q (in
red) and –q (in blue), are
arranged as shown.
Through which closed
surface(s) is the net
electric flux equal to zero?
A. surface A
B. B. surface B
C. surface C
D. surface D
E. both surface C and surface D
© 2012 Pearson Education, Inc.
A22.2
Two point charges, +q (in
red) and –q (in blue), are
arranged as shown.
Through which closed
surface(s) is the net
electric flux equal to zero?
A. surface A
B. B. surface B
C. surface C
D. surface D
E. both surface C and surface D
© 2012 Pearson Education, Inc.
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 law
• Under electrostatic conditions, field outside spherical conductor
looks just like a point charge!
Q22.3
A conducting spherical shell with inner
radius a and outer radius b has a
positive point charge Q located at its
center. The total charge on the shell is –
3Q, and it is insulated from its
surroundings. In the region a < r < b,
A. the electric field points radially outward.
B. the electric field points radially inward.
C. the electric field is zero.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
A22.3
A conducting spherical shell with inner
radius a and outer radius b has a
positive point charge Q located at its
center. The total charge on the shell is –
3Q, and it is insulated from its
surroundings. In the region a < r < b,
A. the electric field points radially outward.
B. the electric field points radially inward.
C. the electric field is zero.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
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?
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 on the cylinder, and at the ends.
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 constant in value everywhere on the cylinder
at a distance r from the wire!
F = ∫ E·dA
Example 22.6: Field of a line charge
• E around an infinite positive wire of charge density l?
•The integration over area is two-dimensional
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?
F= E
𝑙 2p
rdqdl
0 0
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(enc) = (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 = 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
(2pr) l
= (l / 2pr e0) r
Don’t
forget E
is a
vector!
Field of a sheet of charge
• Example 22.7 for an infinite plane sheet of charge?
Field between two parallel conducting plates
• Example 22.8 for field between oppositely charged parallel
conducting plates.
A uniformly charged insulating sphere
• Example 22.9 for 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
Q22.4
A solid spherical conductor has a
spherical cavity in its interior. The
cavity is not centered on the center of
the conductor.
Conductor
Cavity
If a positive charge is placed on the
conductor, the electric field in the cavity
A. points generally toward the
outer surface of the conductor.
B. points generally away from the outer surface of the conductor.
C. is zero.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
A22.4
A solid spherical conductor has a
spherical cavity in its interior. The
cavity is not centered on the center of
the conductor.
Conductor
Cavity
If a positive charge is placed on the
conductor, the electric field in the
cavity
A. points generally toward the
outer surface of the conductor.
B. points generally away from the outer surface of the conductor.
C. is zero.
D. not enough information given to decide
© 2012 Pearson Education, Inc.
Q22.5
There is a negative surface charge density in a certain region on
the surface of a solid conductor.
Just beneath the surface of this region, the electric field
A. points outward, toward the surface of the conductor.
B. points inward, away from the surface of the conductor.
C. points parallel to the surface.
D. is zero.
E. not enough information given to decide
© 2012 Pearson Education, Inc.
A22.5
There is a negative surface charge density in a certain region on
the surface of a solid conductor.
Just beneath the surface of this region, the electric field
A. points outward, toward the surface of the conductor.
B. points inward, away from the surface of the conductor.
C. points parallel to the surface.
D. is zero.
E. not enough information given to decide
© 2012 Pearson Education, Inc.
Q22.6
For which of the following charge distributions would Gauss’s
law not be useful for calculating the electric field?
A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly
distributed over its surface
C. a right circular cylinder of radius R and height h with
charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with
charge uniformly distributed over its surface
E. Gauss’s law would be useful for finding the electric
field in all of these cases.
© 2012 Pearson Education, Inc.
A22.6
For which of the following charge distributions would Gauss’s
law not be useful for calculating the electric field?
A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly
distributed over its surface
C. a right circular cylinder of radius R and height h with
charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with
charge uniformly distributed over its surface
E. Gauss’s law would be useful for finding the electric
field in all of these cases.
© 2012 Pearson Education, Inc.
Testing Gauss’s law experimentally
• Faraday’s icepail experiment confirmed validity of Gauss’s law.
Testing Gauss’s law experimentally
• Faraday’s icepail experiment confirmed validity of Gauss’s law.
Testing Gauss’s law experimentally
• Faraday’s icepail experiment confirmed validity of Gauss’s law.
The Van de Graaff generator
• The Van de Graaff generator, shown in Figure 22.26 below,
operates on the same principle as in Faraday’s icepail experiment.
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