Download Chapter 24: Gauss’ Law

Chapter 24: Gauss’ Law
Gauss’ Law relates the net amount of electric charge enclosed by a surface to the electric field on
that surface.
For certain situations Gauss’ Law provides an easier way to calculate the electric field than the
integration methods discussed in chapter 23. These cases most always involve a situation with a
high degree of symmetry, e.g. a sphere or cylinder.
To be specific, Gauss’ Law is: ε0Φ=qenc
e0 is our friend the permittivity constant=8.85x10-12C2/(N-m2).
qenc is the net electric charge enclosed by our surface.
This is the algebraic sum of all enclosed charges and can be +, -, or zero.
Φ is the electric flux through the surface that encloses our charge(s).
A few words about the “surface”:
1) it is not a real surface, i.e. something we can touch.
2) We try to choose the appropriate surface to simplify the problem.
3) The surface must fully surround the charge(s), i.e. it must be CLOSED.
Karl Friedrich Gauss 1777-1855.
A true math-physics genius.
Example: a rectangular box with 6 sides gives a closed surface. If we take the top of the box off it is no longer a closed surface.
4) We call it a “Gaussian surface”.
The electric flux through the Gaussian surface is:
Φ = ∫ E ⋅ dA
The integral contains the
dot product of 2 vectors.
The ∫ symbol means that the integration is taken over the entire closed surface.
Flux is a scalar even though E and dA are vectors. Flux has units Nm2/C
R. Kass
P132 Sp04 Ch24
1
Electric Flux Calculations
Φ = ∫ E ⋅ dA
So, how do we calculate the electric flux? This integral looks worse than chapter 23’s!
Let’s calculate the electric flux through the box in the figure.
Since the box is closed the Gaussian surface is the 6 box sides. We really have 6 integrals to worry about!!
Φ = ∫ E ⋅ dA = ∫ E ⋅ dA + ∫ E ⋅ dA + ∫ E ⋅ dA + ∫ E ⋅ dA + ∫ E ⋅ dA + ∫ E ⋅ d A
1
2
3
4
5
6
Ex: for surface #1 dA points in the +z direction. For #6 dA points in the –z direction.
1
1
4
4
3
5 1
For this example the electric field is uniform along the +z-axis.
The direction of dA is always normal to a surface and points outward.
∫ E ⋅ dA = ∫ EdA cos(0) = EA
∫ E ⋅ dA = ∫ EdA cos(270) = 0
6
2
4
∫ E ⋅ dA = ∫ EdA cos(90) = 0 ∫ E ⋅ dA = ∫ EdA cos(90) = 0
∫ E ⋅ dA = ∫ EdA cos(270) = 0 ∫ E ⋅ dA = ∫ EdA cos(180) = − EA
2
5
2
5
3
HRW fig. CP1
3
6
6
So, the flux through the Gaussian surface is: Φ=EA+0+0+0+0-EA=0,
The total flux is zero even though the flux through sides 1 and 6 is non-zero.
According to Gauss’ Law there is zero charge enclosed in the box.
This should make sense since if we had enclosed a positive charge in the box we would have electric field
lines pointing outward on each of the 6 surfaces. Likewise if we had a negative charge in the box we would
have field lines pointing inward on each of the 6 surfaces. The only way to have an electric field that points
only in the +z direction is to have the charge(s) outside the box.
R. Kass
P132 Sp04 Ch24
2
Electric Flux Calculations continued
What would the flux be if I rotated the box 450 around the y-axis but kept
the E field along the z axis?
E
E
6
The blue dashed
Looking down on the box.
2
6
arrows represent the
The y axis is coming out
x
normals to the sides.
x
of the page. Sides 3&4 are
5
2
top and bottom.
1
5
1
z
z
In the original orientation sides 2, 3 ,4, and 5 were perpendicular to E so their flux was zero.
In the rotated orientation sides 3 and 4 are still perpendicular but sides 2 and 5 are no longer.
∫ E ⋅ dA = ∫ EdA cos(315) = 0.707 EA
∫ E ⋅ dA = ∫ EdA cos(45) = 0.707 EA ∫ E ⋅ dA = ∫ EdA cos(90) = 0
∫ E ⋅ dA = ∫ EdA cos(270) = 0
∫ E ⋅ dA = ∫ EdA cos(225) = −0.707 EA ∫ E ⋅ dA = ∫ EdA cos(135) = −0.707 EA
1
1
4
4
2
5
2
5
3
6
3
6
The total flux is still zero!
In fact, for any shape closed surface the total flux will be zero if there is no net charge inside.
E
R. Kass
P132 Sp04 Ch24
3
Gauss’ Law meets Coulomb’s Law
HRW fig. 24-8
Consider the situation where we have a single positive charge (q) in space.
Let’s use Gauss’ Law to calculate its electric field (E).
In order to actually do this problem we need to assume the following:
1) the electric field of a positive point charge points radially outward.
2) the electric field of a point charge varies only with the distance (r) from the charge.
What should we use for the Gaussian surface?? Let’s start with a sphere!
We draw an imaginary sphere with the charge at the center like the figure on the right.
Suppose we look at an infinitesimal area (dA) of the sphere.
The normal to this surface will always point radially away from the center of the sphere.
Therefore using assumptions 1) E and dA are parallel.
Φ = ∫ E ⋅ dA = ∫ EdA cos(θ ) = ∫ EdA cos(0) = ∫ EdA
By assumption 2) E has the same value everywhere on the sphere.
Therefore we can move E outside the integral.
Φ = ∫ E ⋅ dA = ∫ EdA cos(0) = ∫ EdA = E ∫ dA
This integral is just the surface area of a sphere: A=4πr2.
If we had used a cube instead of
a sphere for our Gaussian surface
we would not be able to take E
outside the integral. Since the
distance to a cube face is not constant
assumption 2 says E will vary too.
Also, E and dA would no longer be
parallel everywhere on the Gaussian
surface. Thus cosθ would vary on the
Gaussian surface too. The integration
becomes much more complicated!
Φ = ∫ E ⋅ dA = ∫ EdA cos(0) = ∫ EdA = E ∫ dA = 4πr 2 E
Gauss’ Law says: ε0Φ=ε04πr2E= qenc=q
R. Kass
bit of algebra
P132 Sp04 Ch24
E=
q
4πε 0 r
2
Coulomb’s Law!
4
Gauss’ Law and Conductors
Previously, when we discussed the properties of conductors we noted the following properties:
1) The electric field inside of a conductor is zero.
2) Any excess charge on an isolated conductor resides entirely on its outside surface.
Using Gauss’ law and 1) we can prove 2)!
Assume we have a conductor with excess charge.
cross section of conductor
Gauss’ laws says:
∫ E ⋅ dA = q
enc
/ε0
cross section of conductor
with Gaussian surface (dashed)
just inside the conductor.
Since our Gaussian surface is inside the conductor by 1) the electric field, E, is zero
everywhere on the Gaussian surface.
∫ E ⋅ dA = 0 ⇒ q
enc
=0
Therefore any charge inside the Gaussian surface must be zero too.
Since we can move the Gaussian surface arbitrarily close to the conductor’s surface
the charge must lie entirely on the outer surface of the conductor! Try proving this without using
Gauss’ law!!
R. Kass
P132 Sp04 Ch24
5
Gauss’ Law and Conductors continued
Let’s consider the case where we have a spherical metal shell with a point charge +q inside:
The metal shell does not have any excess charge.
The point charge is inside the shell, but not at its center.
metal
-- -
- - -
-
--+q
rinner ----- - router
What does Gauss’ law say for the 3 regions:?
a) r < rinner
b) rinner < r < router
c) r > router
For each case let’s pick a spherical Gaussian surface
centered
on our metal shell.
∫
For case a) we have: E ⋅ dA = q enc / ε 0 = q / ε 0
However, since +q is not at the center of the shell
there is no symmetry that makes the integral easy to do
so Gauss’ law is not much help here.
For case b) the Gaussian surface is everywhere inside the conductor
so E is zero everywhere on the Gaussian surface.
∫ E ⋅ dA = 0 ⇒ q
enc
=0
Since qenc=0 there must be an induced charge = -q on the inner wall to cancel +q!
The induced charge will not be uniformly distributed if +q is not at the center.
R. Kass
P132 Sp04 Ch24
6
Gauss’ Law and Conductors continued continued
+
+
+
+
+
metal
+
In region c) we have included the whole metal
shell and therefore have:
- - --
+
+
+
+
+
router
r
+
+
+
enc
/ε0 = q /ε0
Unlike the negative induced charge on the inner
shell the induced charge, +q, on the outer shell is
uniformly distributed over the surface. This is
+
because the electric field in the region r < rinner
does not make it to the outer surface, i.e. the outer
+ surface is shielded from this field. Therefore the
electric field outside the metal shell at a radius r
+
will have spherical symmetry and we find:
+
--+q
rinner ----- - -
+
∫ E ⋅ dA = q
+
+
+
∫ E ⋅ dA = EA = E (4πr
2
E (4πr 2 ) = q / ε 0 ⇒ E =
)
q
remember:
r ≥ router
4πε 0 r 2
This electric field is the same as that of an isolated point charge!
R. Kass
P132 Sp04 Ch24
7
Gauss’ Law and a non-Conducting Sphere
We just studied a conducting sphere. What about a non-conducting sphere with charge q
distributed uniformly inside its volume?
We can apply Gauss’ law to a hollow non-conducting sphere
to prove the two shells theorems we saw previously:
charge uniformly
rinner
router
distributed in this region
1) a shell of uniform charge attracts or repels a charged particle that is
outside the shell as if all the shell’s charge were concentrated at its
center.
2) a shell of uniform charge exerts no electrostatic force on a charged
particle located inside the shell.
Actually we proved 1) on the previous page! Just put the
Gaussian surface outside the sphere at r ≥ router
2
E
⋅
d
A
⇒
EA
=
E
(
4
π
r
)
∫
E (4πr 2 ) = q / ε 0 ⇒ E =
q
4πε 0 r 2
And so the force that a point particle with charge Q would feel at ≥ router is:
F = QE =
Qq
4πε 0 r 2
To prove 2) we put a Gaussian surface in the hollow region r < rinner. Since there is no charge
in this region there is no electric field here either! So, if we put a charge Q in this region it
will feel zero force.
note: Q will generate an electric field, but Q does not feel its own electric field so there still will be no force on Q.
R. Kass
P132 Sp04 Ch24
8
Gauss’ Law and a non-Conducting Sphere continued
How about a solid non-conducting sphere with total charge=q uniformly spread out through
its volume?
Here we are considering a sphere with radius =a. The volume of a sphere with this radius is:
V=(4/3)πa3
By uniformly spread throughout the volume we mean:
The charge enclosed by a sphere of radius r The charge enclosed by the whole sphere
q
=
=
4 3
The volume of a sphere of radius r
The volume of the whole sphere
πa
3
First, take a Gaussian surface inside the sphere r < a.
∫ E ⋅ dA =
r
a
q enc
ε0
q enc
q
qr 3
=
⇒ q enc = 3
4 3 4 3
a
πr
πa
3
3
Since we have spherical symmetry we can write: ∫ E ⋅ dA = EA = E (4πr 2 )
qr 3
E (4πr ) =
ε 0a3
2
The sphere has total charge q
uniformly distributed throughout
its volume.
R. Kass
E=
qr
4πε 0 a 3
P132 Sp04 Ch24
9
Gauss’ Law and a non-Conducting Sphere continued continued
Finally, lets do the case where r > a. Draw a Gaussian surface outside
the sphere.
r
∫ E ⋅ dA =
q enc
ε0
=
q
ε0
Since we have spherical symmetry we can write:
a
E (4πr 2 ) =
E=
The sphere has total charge q
uniformly distributed throughout
its volume.
r≤a
To recap:
r>a
R. Kass
E=
q
ε0
q
4πε 0 r 2
qr
4πε 0 a
q
E=
4πε 0 r 2
3
“point charge” again!
grows linearly with r.
decreases like r-2.
P132 Sp04 Ch24
10
Gauss’ Law, a non-conducting Sphere with a
non-Uniform Charge Distribution
In order to use Gauss’ law we always need to know how much charge is enclosed in our surface.
If we have a volume charge density, ρ, then the amount of charge dq in a small volume, dV, is:
dq=ρdV
Let’s assume that ρ is only a function distance from the center of the sphere: ρ=ρ(r).
Therefore the amount of charge enclosed by a spherical shell with inner radius, rinner &
router
outer radius, router is:
qenc =
∫ ρ (r )dV
rinner
So, we can rewrite Gauss’ law as:
ε 0 ∫ E ⋅ dA = qenc =
router
∫ ρ (r )dV
rinner
If ρ is only a function of r, the electric field is always radial and E and dA are parallel.
For a spherical Gaussian surface with radius, r we have:
ε 0 ∫ E ⋅ dA = ε 0 E ∫ dA =ε 0 E 4πr 2
Let’s do prob. 41 of ch. 24. Here we have a non-conducting sphere of radius R with ρ=ρsr/R.
Calculate the electric field inside the sphere at radius r. The total charge on the sphere is Q.
We put the Gaussian surface at r. The inner radius is r=0 since we have a solid sphere.
router
q enc
R. Kass
4πρ s r 3
πρ s r 4
r
2
= ∫ ρ (r )dV = ∫ ρ s (4πr dr ) =
r dr =
∫
R
R
R
rinner
0
0
r
P132 Sp04 Ch24
11
Gauss’ Law, a non-Conducting Sphere and a
non-Uniform Charge Distribution continued
router
q enc
r
4
πρ
πρ s r 4
r
2
3
s
= ∫ ρ ( r ) dV = ∫ ρ s ( 4πr dr ) =
r dr =
∫
R
R
R 0
rinner
0
r
How does this
differ from the uniform
ρ example?
We have to evaluate the electric field on the Gaussian surface:
ε 0 ∫ E ⋅ dA = ε 0 E ∫ dA =ε 0 E 4πr 2
This part is the same as the uniform ρ example!
Putting these two equations together we get: Eε 0 4πr =
2
πρ s r 4
R
ρ
r2
s
Solving for the electric field we get: E =
ε 0 4R
We can write E in terms of the total charge in the sphere Q using:
4πρ s 3
Q
r
3
r
dr
=
R
⇒
=
πρ
ρ
Q = ∫ ρ ( r )dV = ∫ ρ s (4πr 2 dr ) =
s
s
∫
R
R
πR 3
0
0
0
R
R
R
Qr 2
E=
4πε 0 R 4
How does this differ from the problem with uniform volume charge density?
What is the electric field for r>R? (ans: same as that of a point charge)
R. Kass
P132 Sp04 Ch24
12