Today in Physics 122: use of Gauss’s Law
 Hints for applying
Gauss’s Law
 Example calculations
for 3-D charge density
distributions
 Perfect conductors
 Fields from charges
distributed on perfect
conductors
 Gauss’s law for gravity
18 September 2012
Frame from a video of a sequence of
three lightning bolts “drawn” to the
lightning rod on top of the Empire
State Building (New York Daily News).
Physics 122, Fall 2012
1
Recap of Gauss’s Law
The flux of E which emerges
from a closed surface S
depends upon how much
charge the volume V
bounded by this surface
contains:
∫ EdA = 4π kQencl.
S
dA
ρ ( r ′)
E
S
dV
where dA is perpendicular to
S and points away from V,
and
Qencl. = ∫ ρ ( r ′ ) dV
V
18 September 2012
Physics 122, Fall 2012
2
How to use Gauss’s Law
Can use when
 one is given a distribution of electric charge, and
 the symmetry and/or extent of the charge distribution
makes it clear what the pattern (direction!) of E is.
To use,
 draw imaginary closed surface (Gaussian surface)
through point at which one wants to know the field.
 The Gaussian surface needs to be drawn so as to make the
unknown E come out of the flux integral: 
∫ EdA = E ∫ dA.
 This usually means the symmetry of surface matches
symmetry of the charge distribution.
 Then calculate Qencl. = ∫ ρ dV and solve for E.
18 September 2012
Physics 122, Fall 2012
3
Example 1: infinite sheet of charge
We did this via Coulomb’s law
last week: what is the electric field a
z
distance z from an infinite plane
which carries a uniform charge per
unit area σ?
First, coordinates and symmetry:
 Let the plane be x-y.
 Because the plane is infinite
and the charge is spread
uniformly, E can’t depend
upon x or y.
 Nor can E point along a
direction other than ±z.
18 September 2012
Physics 122, Fall 2012
E=?
z
y
σ
x
4
Infinite sheet of charge (continued)
 Nor can E be anything but symmetrical about the plane:
the field at –z must be equal and opposite to that at +z.
 So the Gaussian surface has to be symmetrical about the
plane, and have all sides either parallel or perpendicular
to the plane. Any shape with these properties will do.
σ
18 September 2012
Physics 122, Fall 2012
5
Infinite sheet of charge (continued)
 Let’s choose the Gaussian surface
with rectangular sides: let its
dimensions be a×a×2z, and let the
charged plane bisect the a-2z sides.
 Because E has to be ⊥ to the plane
(cosθ = 0), the flux is zero through
all of the vertical, a×z faces.
 And because E is uniform on
the a×a faces and points the same
way as the area vectors (i.e. away
from the charged plane; cosθ = 1),
the flux through each of these faces
is Ea2.
18 September 2012
Physics 122, Fall 2012
E, dA
a
a
z
z
a
σ
a
E, dA
6
Infinite sheet of charge (continued)
 Now we’re ready to use the
formula: the left-hand side is
E, dA
2
=
E

dA
=
E
dA
2
Ea
∫
∫
a
a
z
a×a
and because the Gaussian surface
intercepts an area a×a on the plane,
the right-hand side is
4π kQencl. = 4π kσ a
Ea2 4π kσ a2
So 2=
2
⇒=
E 2π kσ
z
a
σ
a
E, dA
(Note: independent of a and z.)
18 September 2012
Physics 122, Fall 2012
7
Infinite sheet of charge (continued)
 The complete answer is
2π kσ zˆ , z > 0
E=
−2π kσ zˆ , z < 0
the same answer we got (on 6
and 11 September) with
Coulomb’s law, of course.
 Note that the process requires
more spatial/geometrical
reasoning and intuition than
the Coulomb’s-law method,
but much less calculus.
18 September 2012
Physics 122, Fall 2012
E = 2π kσ zˆ
σ
E = −2π kσ zˆ
8
Example 2: infinite coaxial line
y
Consider a long coaxial line: a thin, straight central
wire with charge per unit length, surrounded by a
thin cylinder with inner radius R. The inner and
outer wire have uniform charge per unit length λ
−λ
and -λ, respectively. Calculate E everywhere.
−λ
 Long (≈ infinite): point one of the
coordinate axes along the center.
We’ll use z for this.

λ
 Infinite, uniform λ: E must
R
point ⊥ to z axis, and its value
λ r Gaussian
can’t depend upon the value
r
surface
of z.
z
R
 Two distinct regimes: r < R, ≥ R.
18 September 2012
Physics 122, Fall 2012
x
9
Infinite coaxial line (continued)
 So use a cylindrical Gaussian surface,
length  , radius r, and let r run from zero
to > R.
• Flux through circular ends would be
zero, as E ⊥ z axis (i.e. cosθ = 0).
−λ
• Since radii are ⊥ to circles, cosθ
= 1 for the cylinder walls, and
• the cylindrical symmetry
λ
guarantees that E is
uniform on the cylinder
wall, as it all lies the same
r
distance r away from the
z
R
central wire.
18 September 2012
Physics 122, Fall 2012
y
x

10
Infinite coaxial line (continued)
y
 Now put all this into Gauss’s Law:
∫
∫
E dA E=
=
dA 2π r E
r <R:
cyl. walls
4π kQencl. =
4π kλ 
2π r E = 4π kλ 
−λ
Same result
as on 11 Sept.
E = 2 kλ r
r ≥ R : 4π kQ
=
encl. 4π k ( λ − λ )
=0 ⇒ E =0

λ
With r̂ as a unit vector ⊥ z,
2 kλ rˆ r , r < R
E=
0, r > R
18 September 2012
x
Physics 122, Fall 2012
r
z
R
11
Infinite coaxial line (continued)
Notice how much less painful this calculation
is, than would be the case if we needed to use
Coulomb’s Law.
 Field from outer cylinder would require a
lot of effort in both ranges of r, for a
−λ
disappointingly simple result.
 Instead Gauss’s Law exploits the
symmetry of flux of E for the
λ
symmetric charge
distributions, and indicates
field cancellation in situations
r
in which that might not be
z
R
intuitively obvious.
18 September 2012
Physics 122, Fall 2012
y
x

12
Example 3: uniformly charged sphere
A sphere with radius R has a uniform electric charge per unit
volume ρ. Calculate E everywhere.
 Spherical symmetry strongly
ρ
suggests placing the
coordinate origin in the
sphere’s center, and that it
r
doesn’t matter which way
the axes point.
 Charge density spherically
symmetric: E must point
R
radially, so that E ⊥ dA
(cosθ = 1) on a spherical
Gaussian surface (radius r).
18 September 2012
Physics 122, Fall 2012
13
Uniformly charged sphere (continued)
 E will be uniform on all spheres, for the same reason.
 Again two regimes for the radius of the Gaussian surface:
r < R, ≥ R.
ρ
 Ready to put into Gauss’s Law:
=
E dA E=
dA 4π r 2 E
∫
∫
G. sph.
r
Inside the sphere (r < R):
Qencl. =
∫
ρ dV
R
G. sph.
4 3
πr ρ
= ρ=
VG. sph.
3
18 September 2012
Physics 122, Fall 2012
14
Uniformly charged sphere (continued)
 so, again for inside the sphere (r < R),
4 3
4π r E = 4π k π r ρ
3
4
E = π kρr
3
 Note one important implication of Gauss’s law: the
field at r only depends on the
charge contained within r.
Fields from the charges in the
spherical shell from r to R
cancel out. (!)
2
18 September 2012
Physics 122, Fall 2012
ρ
r
R
15
Uniformly charged sphere (continued)
 Now outside the sphere (r ≥ R):
Qencl. =
∫
ρ dV
whole sph.
Vwhole sph.
= ρ=
4
π R3 ρ
3
4
4π r 2 E = 4π k π R 3 ρ
3
k 4
Q
3
E =
k
=
πρ R
2 3
r
r2
 4π k ρ rrˆ 3 , r < R
∴E =

2
ˆ
r
kQ
r
, r≥R

18 September 2012
Physics 122, Fall 2012
ρ
R
r
16
Aside: singularity of Coulomb’s law
Did you ever worry about the fact that we make field
calculations with point charges and Coulomb’s law all the
time, even though
=
E kQrˆ r 2
→ ∞ as r → 0 ?
Now you can stop worrying. Since it appears that there’s no
such thing as a point charge – even elementary particles act
as if they have nonzero extent – one can be sure that this
singularity is never expressed in nature. A nonzero-size
charge distribution will always lead to finite-strength fields in
their interiors, as we just saw with the uniform sphere:
=
E 4π k ρ rrˆ 3 , r < R
→ 0 as r → 0.
18 September 2012
Physics 122, Fall 2012
17
Example 4: nonuniformly charged sphere
A charge distribution is spherically symmetric and has charge per
unit volume given by
ρ ( r ′ ) = ρ0 e −r ′ R
Calculate the electric field everywhere.
 The setup, and indeed most of the solution, is the same as
that of the previous example: the LHS of Gauss’s law is
still
=
E dA E=
dA 4π r 2 E
∫
and the RHS is still
∫
G. sph.
Qencl. =
∫
ρ dV
G. sph.
• This time ρ doesn’t come out of the integral, though.
18 September 2012
Physics 122, Fall 2012
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Nonuniformly charged sphere (continued)
 As the infinitesimal volume in the
enclosed-charge calculation, take an
infinitesimally thin spherical shell,
with surface area 4πr’2 and thickness
dr’:
r
Qencl. = ρ0 e −r
∫
′R
0
=4πρ0 R
3
0
(
)
r R
 e −u u2 + 2u + 2 
e u du =−

 0
−u 2

3
= 4πρ0 R  2 − e −r

18 September 2012
dr’
4π r ′2 dr ′
r R
∫
r’
2


r
r
R
 2 + 2 + 2 
R

R


Physics 122, Fall 2012
19
How to do that integral
= xy − ∫ ydx.
It’s a simple exercise in integration by parts: ∫ xdy
First, let x =
2 udu , y =
u2 , dy =
e −u du ⇒ dx =
− e −u
b
b
b
−u 2
 −u2 e −u  + 2 e −u udu
e
u
du
=
∫
∫

a
a
a
Now, in last term, let x =
u , dy =
e −u du ⇒ dx =
du , y =
− e −u
b
b
b
b
−u 2
 −u2 e −u  +  −2 ue −u  + 2 e −u du
e
u
du
=
∫
∫

a 
a
a
a
2 −u 
=  −u e

18 September 2012
b
+  −2 ue
a 
−u 
b
+  −2 e
a 
b
(
)
b
 e −u u2 + 2u + 2 
=−
 a
 a 
−u 
Physics 122, Fall 2012
20
Nonuniformly charged sphere (continued)
 So using this enclosed charge in Gauss’s Law gives
4π r 2 E = 4π kQencl.
kQencl.
4π k ρ0 R 3
=
E
=
rˆ
r2
r2

 2 − e −r

R
 r2

r
 2 + 2 + 2   rˆ
R

R


 Although that may have had a non-trivial integral, it was
still doable.
Now imagine trying to do the same calculation with
Coulomb’s law!
18 September 2012
Physics 122, Fall 2012
21
Perfect electrical conductors
So far we’ve been considering charge distributions in which
the charges are stuck in predetermined positions: that is,
perfect insulators.
 In contrast are materials in which the charges are free to
move in response to a field. If charges can move without
any impediment, the materials are called perfect
conductors.
 Perfect conductors are not to be confused with
superconductors, which have other interesting properties
in addition to perfect conductivity.
 Of course most conductors are not perfect, but “perfect” is
a good approximation for many metals if the charge
distribution is static (constant in time).
18 September 2012
Physics 122, Fall 2012
22
Properties of perfect conductors
 Electric charge can only reside on the
surface of a perfect conductor.
• Since like charges repel each other, any
free charges within it will move as far as
they can away from other ones, and the
surface is as far as they are free to go.
 Near a perfectly-conducting surface, E is
perpendicular to that surface.
• If it weren’t, then charge would flow
along the surface until any component of
E  to the surface were cancelled, leaving
only the ⊥ component.
• Aside: This is why lightning rods work.
18 September 2012
Physics 122, Fall 2012
Particularly
large E
23
Example 5: Gauss’s Law and conducting planes
Space at z > 0 is empty, but at z = 0 lies an infinite planar surface
of a conductor, which carries a uniform charge per unit area σ.
Calculate E everywhere.
E, dA
We use the same setup as Example 1.
a
 Because E has to be ⊥ to the plane
a
(cosθ = 0), the flux is still zero
z
through all of the vertical, a×z faces.
σ
 For the upper a×a face, E and dA
still point along +z, so its flux
is still Ea2.
z
a
 But this time the lower a×a face
a
E
=
0
has E = 0: no flux threads it.
18 September 2012
Physics 122, Fall 2012
24
Gauss’s Law and conducting planes (continued)
 So Gauss’s Law for this Gaussian surface becomes
2
E
dA
Ea
=
=
E

dA
∫
∫
a×a
= 4=
π kQencl. 4π kσ a2
 4π kσ zˆ , z > 0
∴E =

0, z ≤ 0
 That is, the field at z > 0 is twice
as strong as in Example 1.
• The charge produces the
same flux as before, but
instead of dividing between
up and down, it all goes up.
18 September 2012
Physics 122, Fall 2012
E = 4π kσ zˆ
σ
E=0
25
Example 6: point charge and conducting shell
A neutral, conducting spherical shell – inner and outer radii R1 and
R2 respectively – has a point charge Q at its center. Calculate the
charge, and charge per unit area on each surface.
r
σ2
 To do this methodically, follow
the setup of Examples 3 and 4.
However, I don’t think you’ll
σ1
r1
mind the following shortcuts:
Q
 Consider a spherical
Gaussian surface with radius
R1
r1 , R1 < r1 < R2 . It lies within
R2
the conductor, so E = 0
everywhere on its surface.
18 September 2012
Physics 122, Fall 2012
26
Point charge and conducting shell (continued)
 For E to be zero in the conductor there must be charge
induced on the inner surface, to cancel the point charge’s
field.
r
σ2
 So Gauss’ Law for this surface
(1) and the volume it encloses is
∫ EdA=
0= 4π kQencl.
1
= 4π k ( Q + Q1 ) ,
and Q1 = −Q
σ1 = −Q
18 September 2012
σ1
Q
r1
R1
R2
4π R12
Physics 122, Fall 2012
27
Point charge and conducting shell (continued)
 But the conductor is neutral, so if a charge –Q is induced
on its inner surface, there must be a charge +Q on its outer
surface.
r
σ2
 And thus the outer-surface charge
density is σ 2 = Q 4π R22 .
 We weren’t asked, but: since the
σ1
r1
outer surface acts as if all of its
Q
charge lies at its center, the
field outside the conductor
R1
2
ˆ
is E = kQr r , just as if the
R2
conductor weren’t there at all.
18 September 2012
Physics 122, Fall 2012
28
Your turn.
Suppose that the center of the inner surface and the point
charge were displaced from the center of the outer surface, as
shown.
σ2
Which quantities differ from their
E (r )
values in the previous example?
(Check all that apply.)
σ1
o Q1
Q
o σ
1
o Q2
o σ2
o E(r)

o None of these has changed.
18 September 2012
Physics 122, Fall 2012
29
Your turn again.
Now suppose the inner and outer surfaces of the shell still
have the same dimensions, but the point charge and the
centers of both surfaces are all displaced from each other.
σ2
Which quantities differ from their
E (r )
values in Example 6?
(Check all that apply.)
Q
o Q1

o σ1
o
o
o
o
σ1
Q2
σ2
E(r)
None of these has changed.
18 September 2012
Physics 122, Fall 2012
30
Gauss’s Law and gravity
Gauss’s Law works for gravity too. If one repeats the
reasoning which leads from Coulomb’s to Gauss’s law (13
September, esp. pp. 23-29), and substitutes masses and
gravitational field g for charges and electric field E:
kqQ
GmM
FQ→q =
rˆ
FM →m = −
rˆ
r2
r2
=
F qE
=
⇒ F mg
kQ
GM
E=
rˆ
g= −
rˆ
2
2
r
r
one obtains the gravitational analogue of Gauss’s Law:
4π kQencl.
∫ EdA =
18 September 2012
⇒
−4π GMencl.
∫ g dA =
Physics 122, Fall 2012
31
Gauss’s Law and gravity (continued)
 Since most physicists and many engineers never deal with
gravity, their lives are not drastically improved by
Gauss’s Law for gravity.
 Astrophysicists and geophysicists use it, though,
wherever gravity is the dominant force and symmetrical
but nonuniform mass densities are encountered.
Milky Way mass model: Tom Jarrett, Caltech
18 September 2012
Physics 122, Fall 2012
32