Download Chapter 22 Gauss`s Law 1 Charge and Electric Flux

Chapter 22
Gauss’s Law
1
Charge and Electric Flux
In this chapter we will define the flux of an electric field through “open” and
“closed” surfaces. The flux through a closed surface will be directly proportional
to the “net charge” enclosed.
Figure 1: This is Fig. 22.2 show a positive flux (i.e., flow) out of the enclosed volume due to positive
charges, and the negative flux, or flow, into the enclosed volume due to negative charge.
Figure 2: This is Fig. 22.3 showing three possible configurations where the net flux is zero.
1
2
Calculating Electric Flux
There is an analogy called the Fluid-Flow analogy that helps describe the concept
of flux.
For example, the volume flow rate through an area A is dV /dt (m3 /s) and can be
described by the following equation
dV
~
= ~v · A
dt
(1)
Figure 3: This is Fig. 22.5 showing how the flux can change for constant velocity ~v and constant
~ by changing the orientation of the area vector A
~ with respect to the velocity vector ~v .
area A
2
2.1
Flux of a Uniform Electric Field
In a similar fashion, we can describe the flux ΦE of the electric field through an
~
area A:
~ ·A
~
ΦE = E
(2)
ΦE = E A cos θ
(3)
~ = An̂.
where the area vector is defined as A
2.2
Flux of a Nonuniform Electric Field
~ is not uniform?
How can we write the electric flux if the electric field E
Z
ΦE =
Z
E cos φ dA =
~ · dA
~
E
(4)
Examples
• Flux through a Cube:
• Flux through a Sphere:
3
Gauss’s law
Gauss’s law provides a different way to express the relationship between electric
charge and the electric field. Let’s investigate a “point charge” inside a spherical
surface. If a charge q is located at the center of a sphere of radius R, we know the
electric field at the surface will be:
E=
1 q
4πo R2
Knowing the electric field on the surface of the sphere, we can calculate the electric
flux ΦE .
3
1 q
q
2
×
4πR
=
4πo R2
o
~ ·A
~ =
ΦE = E
(5)
Figure 4: This is Fig. 22.11 showing how the flux remains constant for concentric spheres of radii
R and 2R.
3.1
Point Charge inside a Nonspherical Closed Surface
Likewise, it can be shown, by using vector calculus, that the flux through a nonspherical surface is the same as a spherical. That is, the electric flux is:
I
ΦE =
~ · dA
~ = q
E
o
S
For a closed surface enclosing no charge:
I
ΦE =
~ · dA
~ = 0
E
S
4
(6)
Figure 5: This is Fig. 22.13 showing how a charge “outside’ a closed surface generates a net flux of
zero.
3.2
4
General Form of Gauss’s law
Applications of Gauss’s Law
Gauss’s law can be used to calculate the electric field for charge distributions possessing a high degree of symmetry.
4.1
Field of a Charged Conducting Sphere
Figure 6: This is Fig. 22.18 showing the electric field as a function of radius E(r) for a charged
conducting sphere
5
4.2
Field of a Uniform Line Charge
Figure 7: This is Fig. 22.19 showing how a charge “outside’ a closed surface generates a net flux of
zero.
~ barrel · A
~ barrel + 2 × E
~ endcaps · A
~ endcaps = Ebarrel 2πr` = qenclosed
ΦE = E
o
E(r) =
4.3
λ 1
2πo r
(7)
Field of an Infinite Plane Sheet of Charge
Figure 8: This is Fig. 22.20 showing how Gauss’s law is applied to determine the electric field
passing through the end caps of the barrel “cutting through” an infinitely large plane containing a
surface charge density σ.
6
In this case, the electric field passing through the end caps is not zero while the
electric field passing through the barrel is zero.
~ barrel · A
~ barrel + 2 × E
~ endcaps · A
~ endcaps = 2 Eendcaps πr2 = qenclosed
ΦE = E
o
Eendcap =
q
enclosed
π r2
1
σ
=
2 o
2 o
(8)
This is the same electric field we obtained in the previous chapter for an infinitely
large sheet with a surface charge density σ.
4.4
Field Between Oppositely Charged Parallel Conducting Plates
Figure 9: This is Fig. 22.21 showing the electric field between oppositely charged parallel conducting
plates.
7
4.5
Field of a Uniformly Charged Sphere
Figure 10: This is Fig. 22.22 showing how Gauss’s law is applied to a uniformly charged sphere.
The electric field inside and outside the sphere are plotted as a function of r.
8
5
Charges on Conductors
When a charge Q is deposited on a conductor, charge is immediately redistributed
~ = ~0.
such that the electric field in the conducting material is E
Figure 11: This is Fig. 22.23 showing how Gauss’s law can be applied to a charged conductor. The
electric field inside the conducting material of the conductor is zero. This, along with Gauss’s law,
one can determine the distribution of charges in conductors with hollow cavities.
Here is a conceptual example:
Figure 12: This is Fig. 22.24 and shows how +7 nC of charge is redistributed in a hollow charged
conductor when −5 nC of charged is placed inside the hollow region of the cavity.
9
5.1
Testing of Gauss’s Law Experimentally
Because of Gauss’s law, the electric field inside a hollow conductor must always be
~ = ~0. Also notice the the electric field lines must enter and exit normal to the
E
conducting surface.
Figure 13: This is Fig. 22.27 showing the principle of a Faraday cage.
5.2
Field at the Surface of a Conductor
~ must be perpendicular to the
Figure 14: This is Fig. 22.28 illustrating how the electric field E
surface once the charges have come to rest.
E⊥ =
σ
o
E-field near the surface of a conductor
10
(9)
5.3
Electric Field at the surface of a conducting sphere
If a charge q is placed on a conducting sphere of radius R, then the electric field
near the surface will be:
E=
1 q
4πo R2
Likewise, the surface charge density will be:
σ =
q
charge
=
area
4πR2
Conclusion: The smaller the radius of the sphere, the larger the surface charge
~
density, and the larger the electric field E.
E =
11
σ
o