EECS 117
Lecture 8: Electrostatic Potential & Gauss Law
Prof. Niknejad
University of California, Berkeley
University of California, Berkeley
EECS 117 Lecture 8 – p. 1/11
Electric Dipole
r+
ẑ
r
+q
r−
ŷ
x̂
−q
An electric dipole is two equal and opposite charges q
separted by a distance d
Consider the potential due to an electric dipole at points
far removed from the dipole r ≫ d
−q
q
1
q
1
+
=
− −
φ(r) =
+
−
+
4πǫr
4πǫr
4πǫ r
r
University of California, Berkeley
EECS 117 Lecture 8 – p. 2/11
Law of Cosines
Vector summation is fundamentally related to a triangle.
Note that we can think of adding two vectors r and ẑd/2
as forming r+ = r + ẑd/2
The length of a vector is given by |a|2 = a · a
We can therefore express r+ in terms of r and d
r
√
d
d
+
+
+
r = r · r = (r + ẑ) · (r + ẑ)
2
2
Or simplifying a bit and recalling that r ≫ d
s
r
2
d
d
d
+
r =r 1+
+ cos θ ≈ r 1 + cos θ
2r
r
r
University of California, Berkeley
EECS 117 Lecture 8 – p. 3/11
Back to Potential Calculation
So we have
1
1
1
1
≈
= q
+
r
r 1 + d cos θ
r
r
and
1
1
≈
+
r
r
d
1−
cos θ
2r
d
1+
cos θ
2r
So that the potential is given by
−qd cos θ
1
1
q
− − =
φ(r) =
+
4πǫ r
r
4πǫr2
Unlike an isolated point charge, the potential drops like
1/r2 rather than 1/r
University of California, Berkeley
EECS 117 Lecture 8 – p. 4/11
Electric Field of a Dipole
The potential calculation was easy since we didn’t have
to deal with vectors
The electric field is simply E = −∇φ
Since our answer is in spherical coordinates, but there
is no ϕ variation due to symmetry, we have
1 ∂φ
2qd cos θ
∂φ
1 qd
+ θ̂
= r̂
∇φ = r̂
+ θ̂
sin θ
3
2
∂r
r ∂θ
4πǫr
r 4πǫr
Thus the electric field is given by
qd r̂2 cos θ + θ̂ sin θ
E=−
3
4πǫr
University of California, Berkeley
EECS 117 Lecture 8 – p. 5/11
Electric Flux Density
It shall be convenient to define a new vector D in terms
of E
D = ǫE
Note that the units are simply C/m2 .
We call this the “flux” density because the amount of it’s
flux crossing any sphere surrounding a point source is
the same
I
Z 2π Z π
q 2
q
D · dS =
2π2 = q
r sin θdθdϕ ==
2
4π
0 4πr
0
S
University of California, Berkeley
EECS 117 Lecture 8 – p. 6/11
Gauss’ Theorem
Because of the 1/r2 dependence of the field, the
integral is a constant
Gauss’ law proves that for any surface (not just a
sphere), the result is identical
I
D · dS = q
S
Furthermore by superposition, the result applies to any
distribution of charge
I
Z
D · dS =
ρdV = qinside
S
University of California, Berkeley
V
EECS 117 Lecture 8 – p. 7/11
Gauss’ Law “Proof”
First of all, does it makes sense? For instance, if we
consider a region with no net charge, then the flux
density crossing the surface is zero. This means that
the flux lines entering the surface from the left equal the
flux lines leaving the surface on the right
We can prove that the flux crossing an infinitesimal
surface of any shape is the same as the flux crossing a
radial cone
Notice that if the surface is tilted relative to the radial
surface by an angle θ, it’s cross-sectional area is larger
by a factor of 1/ cos θ
The flux is therefore a constant
dS ′
cos θ = Dr dS
dΨ = D · dS = Dr dS cos θ = Dr
cos θ
University of California, Berkeley
EECS 117 Lecture 8 – p. 8/11
Application of Gauss’ Law
For any problem with symmetry, it’s easy to calculate
the fields directly using Gauss’ Law
Consider a long (infnite) charged wire. If the charge
density is λC/m, then by symmetry the field is radial
Applying Gauss law to a small concentric cylinder
surrounding the wire
I
D · dS = Dr 2πrdℓ
Since the charge inside the cylinder is simply λdℓ
Dr 2πrdℓ = λdℓ
λ
Dr =
2πr
University of California, Berkeley
EECS 117 Lecture 8 – p. 9/11
Spherical Cloud of Charge
Another easy problem is a cloud of charge Q. Since the
charge density is uniform ρ = Q/V , and V = 43 πa3
(
I
r 3
4
3
r<a
ρ 3 πr = Q a
D · dS =
Q
r>a
But
H
D · dS = 4πr2 Dr
Dr =
(
Qr
4πa3
Q
4πr2
University of California, Berkeley
r<a
r>a
EECS 117 Lecture 8 – p. 10/11
Charge Sheet
Consider an infinite plane that is charged uniformly with
surface charge density ρs
By symmetry, the flux through the sides of a centered
cylinder intersecting with the plane is zero and equal at
the top and bottom
The flux crossing the top, for instance, is simply DdS ,
where D only can have a ŷ component by symmetry
The total flux is thus 2DdS . Applying Gauss’ Law
2DdS = ρs dS
The electric field is therefore
(
ρs
y>0
2
ǫ
0
E = ŷ
−ρs
2ǫ0 y < 0
University of California, Berkeley
EECS 117 Lecture 8 – p. 11/11