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Note 4 Gauss’ Law
Gauss’ Law
Gauss’ law relates the the electric field passing through an imaginary, closed surface to the
charges enclosed by that surface. It is another way to calculate the electric field or force that is
simpler under certain specific circumstances. Otherwise, we would have to add up all of the
contributions from all of the existing charges.
–
+
+
electric fields
–
This imaginary, closed surface is called a gaussian surface. The amount of electric field passing
through a surface is called the electric flux.
–
total +
+ enclosed
charge, qenclosed
–
gaussian
surface
electric flux
Φ
Gauss’ law says that the total electric flux, ɸ, at the gaussian surface, relates to the charges inside
the surface in this way.
Φ=
qenclosed
εo
The constant, εo, is called the permittivity of free space.
2
εo = 8.85 ×10−12 C
N ⋅ m2
page 1
Electric Flux
The electric flux is defined as the amount of electric field passing through a surface area. The
surface area is represented by an area vector. The area vector is perpendicular to the surface,
pointing toward the outside of the surface, and has a magnitude equal to the surface area.
magnitude
of area vector
=
area
area
vector
inside of
gaussian surface
direction
of area vector
=
direction
of the outside
outside of
gaussian surface
When electric fields pass through this area, there is an electric flux.
area vector
Epassing through surface
=
θ
Eparallel to area vector
=
E cosθ
electric field vector
The electric flux around an area is calculated this way.
Φat area A = Eat area AA cos θbetween E and A
The total electric flux through the entire gaussian surface is just the sum of all of the electric field
passing through all of the areas of the surface. Normally, this is difficult to calculate for a random
surface. The most instructive examples of Gauss’ law are applied to situations in which there is a
large degree of symmetry.
page 2
Electric Field Around a Point Charge
A distance r from a point charge forms a sphere. Therefore, we use a spherical gaussian surface.
gaussian surface
r
q
At this gaussian surface, the area vectors look like this.
gaussian surface
q
A
At this gaussian surface, the electric fields look like this.
gaussian surface
q
E
Gauss’ Law says this.
Φ=
qenclosed
ε0
For any patch of the sphere's surface, the electric flux is this.
Φpatch = Eat patch Apatch cos θbetween E and A
For any patch, the area vector and the electric field vector are in the same direction. This means
the cosine is 1. Thus, the electric flux for any patch is this.
Φpatch = Eat patch Apatch
The magnitude of the electric field at this gaussian surface is constant. So are the areas of the
patches.
Φpatch = Eat patch Apatch = constant
page 3
Adding up all of the patches gives the total electric flux. Since the electric field is constant, it
factors our when we add up all of the fluxes from all of the patches.
Φ = ΣΦpatch = EΣApatch = E ⋅ Asurface of
sphere
= E ⋅ 4πr 2
The magnitude of the electric field at a distance r from a point charge is this.
E =k
q
r2
Φ=k
q
⋅ 4πr 2 = 4π ⋅ kq
r2
Finally,
Combining the two sides of Gauss’ law, we find that the value of k is actually this.
q
= 4π ⋅ kq
εo
⇒ k=
1
4πεo
The electric field around a point charge is this.
!
E point charge =
1 q
r̂
4πεo r 2 radially outward
page 4
Electric Field Above an “Infinite” Sheet of Charges
Above a sheet of charges forms another plane. Gauss’ law requires a closed surface so I will use
a cylinder. I will use a circular cylinder since the sheet is the same is all horizontal directions. The
amount of charge in the sheet itself depends on the area covered. Therefore, we use a charge
density to characterize the sheet’s charge.
cylindrical
gaussian surface
end
side
charged sheet
with a
charge density
σ
Using the above gaussian surface, we end up with two sets of surfaces. They are the ends and
the side of the cylinder.
Gauss’ law says that
Φ=
qenclosed
ε0
⇒ Φside + Φend =
qenclosed
ε0
Let’s look at the body of the cylinder first. It turns out that the electric field is always parallel to this
surface. Any charge on the sheet has another exact charge on the sheet that will cancel out the
horizontal component of the electric field from the first charge.
electric field
from
right charge
electric field
from
both charges
electric field
from
left charge
area vector
left charge
right charge
This results in the electric field and the area vector always being perpendicular to each other.
Therefore, the electric flux through the side of the cylinder is zero.
Φside = Eat sideAside cos θbetween E and A = Eat sideAside ⋅ 0 = 0
page 5
Now, let’s look at the ends. The same argument can be made as before. The total electric field
will point up and so does the area vector.
electric field
from
both charges
area
vector
electric field
from
right charge
electric field
from
left charge
right charge
left charge
The electric field at the end points up. The amount of electric field comes from all of the charges
in the sheet but it is constant as the amount of charge is constant. Therefore, the electric flux at
each of the end is the following. The radius of the end is r.
Φend = Eat end ΣAend = Eat end (πr 2 )
Going back to Gauss’ law and with the fact that we have two ends.
2 ⋅ E(πr 2 ) =
qenclosed
ε0
The total charge enclosed in the gaussian surface is just the charge density times the area
covered.
qenclosed = σA = σ(πr 2 )
Finally,
E(2πr 2 ) =
σ(πr 2 )
ε0
⇒ E =
σ
2ε0
Notice that the electric field is independent of the distance from the sheet. This is a consequence
of the fact that sheet is implicitly infinite. For finite sheets, the electric field will decrease as the
distance increases. The infinite sheet results is applicable to finite sheets as long as you are very
close to the finite sheet.
page 6
Electric Field Between Two “Infinite” Sheets of Opposite Charges
This configuration is also called the parallel plate capacitor.
E–
E+
sheet of +charges
E–
E+
sheet of –charges
E–
E+
There are two sources of electric field. Within the gap, they are in the same direction so they add.
Both sheets contribute the same magnitude as we model this as two infinite sheets.
The electric field between the sheets is
E =
σ
εo
Outside the sheets, the electric field is zero.
page 7
Electric Field Around an “Infinite” Wire
A wire has the same symmetry as a cylinder, so I will use a cylindrical gaussian surface. This
time, it will line up along the length of the wire.
gaussian surface
wire of charge density λ
Let’s look at the end first time time. The electric field always points away from the wire. The area
of the end points along the wire. They are always perpendicular to each other so the electric flux
is zero at the ends.
total field
right field
left field
area
vector
left charge
right charge
The electric field still points away from the wire at the side. The area vector also points away from
the wire.
total field
right field
left charge
area
vector
left field
right charge
The electric flux at the side is this. The radius of the gaussian surface is r.
Φside = Eat sideAside = EΣAside = E(2πr ⋅ L)
Going back to Gauss’ law with the charge enclosed as the charge density times the length of wire,
E(2πr ⋅ L) =
λL
εo
⇒ E =
1 λ
2πεo r
This is the electric field a distance r from an infinitely long wire.
page 8
Electric Field Around to a Uniformly Charged Sphere
A uniformly charged sphere look like this. It is a sphere so I will use a spherical gaussian surface.
gaussian surface
r
R
total charge
q
At any point on the gaussian surface, the symmetry of the charges will produce an electric field
that is radial. The area vectors is also radial for a sphere.
total field
right field
left charge
area
vector
left field
right charge
The electric flux is
Φ = EΣApatch = E(4πr 2 )
Gauss’ law says that the electric field at the gaussian surface is this.
E(4πr 2 ) =
q
εo
⇒ E =
1 q
4πεo r 2
This is the same result as a single charge at the center of the sphere.
This is the same argument we would make to show that the mass of the Earth can be view as a
point mass where all of the masses is concentrated at a single point located at the enter of the
Earth.
page 9