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Transcript
```Electromagnetism
Physics 15b
Lecture #3
Gauss’s Law
Electric Potential
Purcell 1.13–2.9
What We Did Last Time
Defined electric field by F = qE

Can be expressed by field lines
Defined flux of electric field Φ = ∫ E ⋅ da

Gauss’s Law
∫ E ⋅ da = 4π q
R
Useful for solving E fields with symmetries
  Spherical charge distribution
E

⎧
⎪
⎪
E=⎨
⎪
⎪⎩
Q
r̂ for r ≥ R
r2
Qr
r̂ for r < R
R3
E
S
Note the sign convention: positive if coming out
r
Q
R2
R
r
1
Today’s Goals
Continue with Gauss’s Law

Apply to infinite sheets of charge
Discuss energy in the electric field

Empty space with E has energy?
Define electric potential ϕ
by line-integrating electric field
Closely related to energy
  Vector calculus connects electric potential
to electric field and vice versa


Derive differential form of Gauss’s Law
Connect electric field and charge density
  More vector calculus

J.C.F. Gauss, 1777-1855
Infinite Sheet of Charge
Problem: Calculate the electric field at a distance z from a
positively charged infinite plane
charge
  Surface charge density: σ =
area
Use Gauss again
Which surface to use?
  What symmetry do we have?
  Consider a cylinder 
  Area A and height 2z

z
E
E field must be vertical

How do we know that?
2
Infinite Sheet of Charge
Total flux Φtotal = Φtop + Φside + Φbottom
Side is parallel to E  E ⋅ da = 0  No flux
  Top and bottom are symmetric  Same flux

Φ total = 2Φ top = 2AE(z)
Charge inside the cylinder is
z
qinside = Aσ
E
Using Gauss  E(z) = 2πσ

Don’t forget the direction!
⎧⎪ +2πσ ẑ for z > 0
E=⎨
⎪⎩ −2πσ ẑ for z < 0
The result is worth remembering:
Infinite sheet of charge produces
uniform E field of 2πσ above and below
Pair of Charged Sheets
Place two oppositely-charged large sheets in parallel

Consider an area A of them
z
E fields from the two sheets


Etop
Ebottom
+σ
Between the sheets: E = 4πσ
Cancel each other outside
−σ
Two sheets also attract each
other (obviously)

Top sheet feels

The force on area A of the top sheet is
Ebottom = −2πσ ẑ
F = σ AEbottom
E2
= −2πσ Aẑ = −
Aẑ
8π
2
3
Pair of Charged Sheets
Imagine we move the top sheet upward by a distance d
E2
Etop
Ebottom
  We must do work W = Fd =
8π
  The energy of the system increases
+σ
by W
Q: Where exactly is this energy?
−σ
Note that the volume of the space
between the sheets increased by Ad
  This is also where E field exists

E2
Space with E holds energy with a volume density u =
8π
Total electrostatic energy of a system is
E2
U=∫
dV
Will come back to this…
8π
Line Integral of Electric Field
Electrostatic force is conservative

I said this in Lecture 1 without proof
Given F = qE, the above statement is equivalent to
Line integral

∫
P2
P1
E ⋅ d s is path independent
Thanks to the Superposition Principle,
we have only to prove this for the electric
field generated by a single point charge
Line integral
∫
r2
r1
E=
q
r̂
r2
ds
r
q
r
r̂ ⋅ d s is path independent 1
2
r
r2
q
4
Line Integral of Electric Field
The dot product r̂ ⋅ ds is the radial component of the
movement, i.e. dr
∫
r2
r1

q
r̂ ⋅ d s =
r2
∫
r2
r1
⎛ 1 1⎞
q
dr = q ⎜ − ⎟
2
r
⎝ r1 r2 ⎠
E=
dr
The integral depends only on r1 and r2,
i.e., is path-independent
ds
Generalize using Superposition:
Line integral
∫
P2
P1
q
r̂
r2
r
E ⋅ d s for any electrostatic
field E has the same value for all paths
from P1 to P2
r1
r2
q
Corollary
For the special case of P1 = P2, the path becomes a loop
Line integral ∫ E ⋅ d s of an electrostatic field around
any closed path is zero
This is equivalent to the path-independence
Consider two paths (A and B) from P1 to P2
  Path independence means

∫
P2
P1

∫
P2
P1
E ⋅ d sB
Corollary above means
∫
P2
P1

E ⋅ d sA =
P1
dsA
P2
dsB
P1
E ⋅ d s A + ∫ E ⋅ (− d sB ) = 0
P2
Will use this later when we do the “curl”
5
Electric Potential
The line integral
P2
∫
P2
P1
E ⋅ d s is useful enough to have a name
φ21 ≡ − ∫ E ⋅ d s Electric potential difference between P1 and P2
P1
Note the negative sign!

To move a charge q from P1 to P2, you must do work
W=

∫
P2
P1
−qE ⋅ d s = qφ21
As a result, the energy of the system increases by qϕ21
We can fix P1 to a reference point and re-interpret this as
a scalar function of the position r
r
φ (r) ≡ − ∫ E ⋅ d s Electric potential at position r
0
Reference Point
Reference point for the electric potential is arbitrary

If you choose e.g., point B instead of point A as the reference
r
r
A
B
A
B
φref =B (r) = − ∫ E ⋅ d s = − ∫ E ⋅ d s − ∫ E ⋅ d s = φref = A (r) + const.

Electric potential is defined up to an arbitrary constant
  Just like an indefinite integral
The potential difference is physical (linked to work and
energy) and must be free from arbitrary constant
P2
P2
P1
P1
0
0
φ21 = − ∫ E ⋅ d s = − ∫ E ⋅ d s + ∫ E ⋅ d s = φ (P2 ) − φ (P1)

The constant cancels in the difference
6
Unit
Dimension of electric potential is (electric field)×(length)

Since electric field is (force)/(charge), this equals to (force)×(length)/
(charge) = (energy)/(charge)
Unit of electric potential is erg/esu = statvolt
In SI, the unit of electric potential is volt = joule/coulomb
300 volt = 1 statvolt
1 coulomb = 3×109 esu

Since the rest of the world uses SI, we will convert to SI when we
deal with real-world (esp. EE) problems
Electric Field ↔ Potential
Recall in calculus: F(x) = ∫ f (x) dx
f (x) =
We can “reverse” the line integral as well
φ = −∫ E ⋅ds
E = −∇φ
d
F(x)
dx

Gradient is “how quickly the function ϕ varies in space”

In x-y-z coordinate system:
∇φ = x̂
∂φ
∂φ
∂φ
+ ŷ
+ ẑ
∂x
∂y
∂z
NB: gradient is a vector field

It points the direction of maximum rate of increase (= uphill)

Hence E points downhill of ϕ
7
For a potential φ (x,y) = sin x ⋅ sin y
∇φ (x,y) = x̂ cos(x) sin(y) + ŷ sin x cos y
∇φ (x,y)
φ (x,y)
Equipotential Surfaces
The same potential φ (x,y) = sin x ⋅ sin y can be expressed
with lines of constant values of ϕ
In 3-d, we find surfaces of
constant potential
= equipotential surfaces
  For a single point charge,
equipotential surfaces are
concentric spheres

∇φ (x,y)
Electric field is perpendicular
to the equipotential surfaces

This is generally true for gradient
of any function
8
Charge Distribution
Electric potential due to a charge distribution is
N
φ=∑
j =1

qj
φ=
discrete
rj
∫
dq
r
continuous
Continuous case maybe 1-d, 2-d, or 3-d
φ=
λ
∫ r d
φ=
σ
∫r
da
φ=
ρ
∫ r dv
Since potential is a scalar, it is often easier to calculate
than the electric field
  Once you have ϕ, you can always get E from −∇ϕ
An example is in order
Charged Disk
Problem: Calculate the electric potential produced by a thin,
uniformly charged disk on its axis
z

Disk radius = a, surface charge density σ
Slice the disk into rings, then into the red bits
Charge on the red bit is σsdsdθ
σ sdsdθ
The potential due to the red bit is dφ =
z2 + s 2
  Integrate:


2π
a
sds
0
0
z +s
φ = σ ∫ dθ ∫
2
ds
2
dθ
s
s =a
= 2πσ ⎡⎢ z 2 + s 2 ⎤⎥
⎣
⎦s =0
= 2πσ
(
z +a − z
2
2
)
a
NB: absolute value
in case z < 0
9
Electric Field
Can we calculate E from φ = 2πσ


(
)
z 2 + a2 − z ?
We need to take gradient, which means we need
to know the x-, y-, and z-dependence
z
We calculated ϕ only on the z axis
We are in luck — we know E is parallel
to the z axis by symmetry
E = −∇φ = − ẑ
= −2πσ
∂
∂z
(
∂φ
∂z
ds
)
dθ
s
z 2 + a 2 − z ẑ
⎛
⎞
z
= 2πσ ⎜ −
± 1⎟ ẑ
⎝
⎠
z 2 + a2
a
for z > 0
z<0
Check the Solution
φ = 2πσ
(
z 2 + a2 − z
)
⎛
⎞
z
E = 2πσ ⎜ −
± 1⎟ ẑ
⎝
⎠
z 2 + a2
for z > 0
z<0
Dimension: [ϕ] = (charge)/(length), [E] = (charge)/(length)2

NB: [σ] = (charge)/(length)2
Far away from the disk (|z| >> a)
φ = 2πσ z

( 1+ (a z) − 1) ⎯⎯⎯→ 2πσ z (1+
z a
2
1
2
)
(a z)2 − 1 =
π a 2σ
z
Same as a point charge with Q = πa2σ
Close to the disk (|z| << a)
z a
E ⎯⎯⎯
→ ±2πσ ẑ

Same as the infinite sheet of charge
10
Energy and Potential
We know a system of N charges has a total energy of
q j qk 1 N
q
1 N
1 N
Potential at the j-th
U = ∑∑
= ∑ q j ∑ k = ∑ q j ∑ φ jk
2 j =1 k ≠ j r jk
2 j =1 k ≠ j r jk 2 j =1 k ≠ j
charge due to the
other charges
Generalizing to continuous distributions
U=
1
ρφ dv
2∫
Integrate entire space, or where ρ ≠ 0
  No need to avoid j = k because “individual”
charge is infinitesimally small

Does the factor 1/2 make sense?
Imagine increasing ρ slowly everywhere as ρ ′ = s ρ s = 0 → 1
  The potential is proportional to ρ, i.e., φ ′ = sφ
  Work to go from s to s + ds is dW = (d ρ ′)φ ′ dv = (ds ρ )(sφ ) dv

∫
1
  Integrate W =
∫0 s ds ∫ ρφ dv = 2 ∫ ρφ dv
1
∫
Will come back to this…
Shrinking Gauss’s Law
Charge is distributed with a volume density ρ(r)
Draw a surface S enclosing a volume V
Guass’s Law:
∫
S
E ⋅ da = 4π ∫ ρ dv
Total charge in V
V
Now, make V so small that ρ is constant inside V
∫
S

E ⋅ da = 4πρV for very small V
As we make V smaller, the total flux out of S scales with V
Therefore:
∫
lim
S
V →0
E ⋅ da
V
= 4πρ
LHS is “how much E is flowing out per unit volume”
  Let’s call it the divergence of E

11
Divergence
div E ≡ lim
V →0
∫
S
E ⋅ da
V
= 4πρ
In the small-V limit, the integral depend on volume, but not
on the shape  We can use a rectangular box

Consider the left (S1) and right (S2) walls
∫
∫
S1
S2
dz
E ⋅ da = E(x,y,z) ⋅ (− x̂)dydz
E ⋅ da = E(x + dx,y,z) ⋅ x̂dydz
(
)
dy
Sum = E x (x + dx) − E x (x) dydz
=

∂E x
dxdydz
∂x
S1 E(x,y,z) E(x + dx,y,z)
S2
dx
⎛ ∂E
∂E y ∂E z ⎞
E ⋅ da = ⎜ x +
+
⎟ V = ∇ ⋅E V
S
∂y
∂z ⎠
⎝ ∂x
(
∫
)
div E
Gauss’s Law, Local Version
We now have Gauss’s Law for a very small volume/surface
divE = 4πρ
⎛ ∂E
∂E y ∂E z ⎞
where div E ≡ ∇ ⋅E = ⎜ x +
+
⎟
∂y
∂z ⎠
⎝ ∂x
Connects local properties of E with the local charge density
  Integrate over a volume and you get Gauss’s Law back

Gauss’s Divergence Theorem (this is math!)
For any vector field F,

∫
V
div F dv =
∫
A
F ⋅ da
Applying this theorem to Gauss’s Law (of electromagnetism) gives
us the integral and differential versions we now know
12
Coulomb Field
Let’s calculate div E for E =
q
r̂
r2
q
x̂x + ŷy + ẑz
2
2
x + y + z x2 + y 2 + z2
  Or we can express div in spherical coordinates
2
1 ∂(r Fr )
1 ∂(Fθ sin θ )
1 ∂Fφ
∇ ⋅F = 2
+
+
∂r
r sin θ
∂θ
r sin θ ∂φ
r

We can do this by expressing E in x-y-z : E =

Since only Er is non-zero, we get
∇ ⋅E =
2
2
1 ∂(r Er ) 1 ∂q
= 2
=0
r 2 ∂r
r ∂r
This is correct — we have no charge except at r = 0
At r = 0, 1/r2 gives us an infinity
  That’s OK because a “point” charge has an infinite density

Spherical Charge
Let’s give the “point” charge a small radius R

We did this in Lecture 2, and the solution was
⎧
⎪
⎪
E=⎨
⎪
⎪⎩
Q
r̂ for r ≥ R
r2
Qr
r̂ for r < R
R3
This part is same as a point charge.
We know div E = 0.
Let’s work on this part
2
1 ∂(r Er ) 1 ∂ ⎛ Qr 3 ⎞ 3Q
= 2
=
2
∂r
r
r ∂r ⎜⎝ R 3 ⎟⎠ R 3

For r < R, ∇ ⋅E =

The charge density of the sphere is ρ =
Q
R3
4π
3
∇ ⋅E = 4πρ
It works everywhere (as long as ρ is finite)
13
Energy Again
We found earlier U =

Are they same?
1
E2
∫ 8π dV and U = 2 ∫ ρφ dv
Consider the divergence of the product Eϕ
∂
∂
∂
(E φ ) +
(E φ ) + (E zφ )
∂x x
∂y y
∂z
∂E x
∂φ ∂E y
∂φ ∂E z
∂φ
=
φ + Ex
+
φ + Ey
+
φ + Ez
∂x
∂x ∂y
∂y
∂z
∂z
∇ ⋅ (Eφ ) =
= (∇ ⋅E)φ + E ⋅ (∇φ ) = 4πρφ − E 2

Integrate LHS over very large volume and use Divergence Theorem
∫
V

∇ ⋅ (Eφ ) dv =
∫
S
(Eφ ) ⋅ da = 0
Integral of RHS must be 0, too
4π ∫ ρφ dv − ∫ E 2 dv = 0
assuming Eφ → 0 at far away
1
E2
ρφ
dv
=
∫ 8π dv
2∫
Summary
Used Gauss’s Law on infinite sheet of charge
Uniform electric field E = 2πσ above and below the sheet
E2
  Electric field has energy with volume density given by u =

8π
Defined electric potential by line integral
P2
φ21 = − ∫ E ⋅ d s = φ (P2 ) − φ (P1)
unit: erg/esu = statvolt
P1


Electric field is negative gradient of electric potential
Potential due to charge distribution: φ =
N
∑r
j =1
Differential form of Guass’s Law:

qj
E = −∇φ
or φ =
j
∫
dq
r
∇ ⋅E = 4πρ
Equivalent to the integral form via Divergence Theorem
∫
V
div F dv =
∫
A
F ⋅ da
14
```
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