Defined electric field by F = q E

Can be expressed by field lines
Defined flux of electric field

Φ =
E ⋅ d a
Note the sign convention: positive if coming out
Gauss’s Law

E
d a = 4 π q
Useful for solving E fields with symmetries

Spherical charge distribution
E
E
=
⎧
⎨
⎪
Q r 2
Qr
ˆ for r
≥
R
ˆ for r
<
R
R 3
Q
R 2
R
R
S E r r
1

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
Problem: Calculate the electric field at a distance z from a positively charged infinite plane charge

Surface charge density: σ =
Use Gauss again area z

Which surface to use? E

What symmetry do we have?

Consider a cylinder 

Area A and height 2 z
E field must be vertical

How do we know that?
2

Total flux Φ total
= Φ top
+ Φ side
+ Φ bottom


Side is parallel to E  E
⋅ d a
=
0  No flux
Top and bottom are symmetric  Same flux
Φ total
=
2
Φ top
=
2 AE ( z )
Charge inside the cylinder is q inside
=
A
σ
Using Gauss 

E ( z )
=
2
πσ
Don’t forget the direction! z
The result is worth remembering:
E
=
⎩⎪
+
2
ˆ z for z
>
0
−
2
ˆ z for z
<
0
E
Infinite sheet of charge produces uniform E field of 2 πσ above and below
Place two oppositely-charged large sheets in parallel

Consider an area A of them z
E top
E fields from the two sheets overlap and add up + σ

Between the sheets: E = 4 πσ

Cancel each other outside
−σ
Two sheets also attract each other (obviously)

Top sheet feels
E bottom
= −
2
πσ ˆ

The force on area A of the top sheet is
F
= σ
A E bottom
= −
2
πσ 2 A ˆ = −
E 2
8
π
A ˆ
E bottom
3

Imagine we move the top sheet upward by a distance d


We must do work W = Fd =
E
8
π
2
Ad
The energy of the system increases + σ by W z
E top
E bottom
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
Space with E holds energy with a volume density u
=
E
8
π
2
Total electrostatic energy of a system is
U
=
E 2
8
π dV
Will come back to this…
Electrostatic force is conservative

I said this in Lecture 1 without proof
Given F = q E , the above statement is equivalent to
Line integral
P
2
P
1
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
r
E = q r 2 r ˆ d s
r
2
Line integral
r
1 r
2 q r 2 r ˆ ⋅ d s is path independent r
1
q
4

r movement, i.e. dr
ˆ ⋅ d s

r
1 r
2 r q
2 r ˆ ⋅ d s =
r
1 r
2 r q
2 dr = q
⎛
⎝⎜
1 r
1
−
1 r
2
⎞
⎠⎟
The integral depends only on r
1 i.e., is path-independent
and r
2
,
Generalize using Superposition:
Line integral ∫ P
2
P
1
E ⋅ d s for any electrostatic field E has the same value for all paths from P
1
to P
2
q
r
1
dr
r d s
E = q r 2 r ˆ
r 2
For the special case of P
1
= P
2
, 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 d s
A

Consider two paths (A and B) from P
1 to P
2

Path independence means

P
2
P
1
E ⋅ d s
A
=
Corollary above means
P
2
P
1
E ⋅ d s
B
P
1 d s
B

P
2
P
1
E ⋅ d s
A
+
P
1
P
2
E ⋅ ( − d s
B
) = 0
Will use this later when we do the “curl”
P
2
5

φ
21
P
2
P
1
≡ − E ⋅ d s
∫
P
1
P
2
Electric potential difference between P
1
and P
2
Note the negative sign!


To move a charge q from P
1
to P
2
, you must do work
W =
P
2
P
1
− q E ⋅ d s = q φ
21
As a result, the energy of the system increases by q ϕ
21
We can fix P
1
to a reference point and re-interpret this as a scalar function of the position r
r
( r ) ≡ − E ⋅ d s
0
Electric potential at position r
Reference point for the electric potential is arbitrary

If you choose e.g., point B instead of point A as the reference

φ ref = B
( r ) = − ∫
B r
E ⋅ d s r
= − E ⋅ d s
A
∫ A ∫ −
B
E ⋅ d s = φ ref = A
( r )
Electric potential is defined up to an arbitrary constant
+ const .

Just like an indefinite integral
The potential difference is physical (linked to work and energy) and must be free from arbitrary constant

φ
21
= − ∫ P
2
P
1
E ⋅ d s = − ∫
0
P
2 E ⋅ d s + ∫
0
P
1
The constant cancels in the difference
E ⋅ d s = φ ( P
2
) − φ ( P
1
)
6

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 × 10 9 esu

Since the rest of the world uses SI, we will convert to SI when we deal with real-world (esp. EE) problems
Recall in calculus: F ( x ) =
f ( x ) dx f ( x ) =
We can “reverse” the line integral as well d dx
F ( x )

φ = −
E ⋅ d s
E
= −∇ φ gradient
Gradient is “ how quickly the function ϕ varies in space ”

In x-y-z coordinate system:
∇ φ
NB: gradient is a vector field
= ˆ
∂ φ
∂ x
+ ˆ
∂ φ
∂ y
+ ˆ
∂ φ
∂ z

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 ) = ˆ cos( x )sin( y ) + ˆ sin x cos y
∇ φ ( x , y )
φ ( x , y )
© Prof. G. Sciolla, MIT
φ ( x , y ) = sin x with lines of constant values of ϕ
⋅ sin y

In 3-d, we find surfaces of constant potential
= equipotential surfaces

For a single point charge, equipotential surfaces are concentric spheres
Electric field is perpendicular to the equipotential surfaces

This is generally true for gradient of any function
∇ φ ( x , y )
© Prof. G. Sciolla, MIT
8

Electric potential due to a charge distribution is

φ =
N
j = 1 q r j j discrete
φ = ∫
Continuous case maybe 1-d, 2-d, or 3-d dq r continuous
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
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 σ sdsd θ
The potential due to the red bit is
Integrate: d φ =
σ sdsd θ z 2 + s 2
φ = σ
0
2 π d θ
= 2 πσ
⎡
⎣⎢ z
= 2 πσ
( z 2
2
0 a sds
+ s 2
+ a 2 z 2 + s 2
⎤
⎦⎥ s = a s = 0
− z
)
NB: absolute value in case z < 0
ds s
a
d
θ
9

Can we calculate E
)

(
We need to take gradient, which means we need to know the x, y, and zdependence

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 = − ˆ
= − 2 πσ
⎝⎜
∂
∂ z
= 2 πσ
⎛
−
(
∂ φ
∂ z z 2 z 2 z
+ a 2
+ a 2
± 1
⎞
⎠⎟
− z
) z ˆ
ˆ for z > 0 z < 0
ds s
z
a
d
θ
φ = 2 πσ
( z 2 + a 2 − z
)
E = 2 πσ
⎛
⎝⎜
− z 2 z
+ a 2
± 1
⎞
⎠⎟
ˆ for z > 0 z < 0
Dimension: [ ϕ ] = (charge)/(length), [ E ] = (charge)/(length) 2

NB: [ σ ] = (charge)/(length) 2
φ
Far away from the disk (| z | >> a )
= 2 πσ z
(
1 + ( a z ) 2

− 1
)
⎯ z  a → 2 πσ
Same as a point charge with Q = π a 2 σ z
(
1 + 1
2
( a z ) 2 − 1
)
=
π a 2 σ z
Close to the disk (| z | << a )
E
⎯ ⎯⎯ ± 2 πσ ˆ

Same as the infinite sheet of charge
10

We know a system of N charges has a total energy of
U =
1
2
N
∑ j = 1
∑ k ≠ j q j r jk q k =
1
2
N
∑ j = 1 q j
∑ k ≠ j q k r jk
=
1
2
N
∑ j = 1 q j
∑ k ≠ j
Generalizing to continuous distributions
φ jk
U =
1
2
ρφ dv
Potential at the jth charge due to the
other charges

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
The potential is proportional to ρ , i.e.,
Work to go from s to s + ds is
Integrate W = ∫
0
1 s ds ∫ ρφ dv dW
=
1
2
∫
=
ρφ
φ
(
′ d
= dv
ρ ′ s
)
φ
ρ ′ = s ρ s = 0 → 1
φ ′ dv =
( ds ρ )( s φ ) dv
Will come back to this…
Charge is distributed with a volume density ρ ( r )
Draw a surface S enclosing a volume V
Guass’s Law:
∫
S
E ⋅ d a = 4 ∫
V dv Total charge in V
Now, make V so small that ρ is constant inside V

∫
S
E ⋅ d a = 4 πρ V for very small V
As we make V smaller, the total flux out of S scales with V
Therefore: lim
V → 0
S
E ⋅ d a
V
= 4 πρ

LHS is “ how much E is flowing out per unit volume ”

Let’s call it the divergence of E
11

div E ≡ lim
V → 0
S
E ⋅ d a
V
= 4 πρ
In the smallV limit, the integral depend on volume, but not on the shape  We can use a rectangular box


Consider the left ( S
1
) and right ( S
2
) walls
S
1
∫
S
2
E ⋅ d a
E ⋅ d a
Sum =
(
E
= E ( x , y , z ) ⋅ ( − ˆ ) dydz
= E ( x + dx , y , z ) ⋅ ˆ dydz x
( x + dx ) − E x
( x )
) dydz
=
∂ E x
∂ x dxdydz
Add up all walls:
S
E ⋅ d a =
⎛
⎝
∂ E x
∂ x
+
∂ E y
∂ y dy
+ dz
S
1
E ( x , y , z )
S
2 dx
E ( x + dx , y , z )
∂ E z
∂ z
⎞
⎠
V =
( )
V div E
We now have Gauss’s Law for a very small volume/surface
   div E = 4 πρ
where div E ≡ ∇ ⋅ E =
⎛
⎝
⎜
∂ E
∂ x x +
∂
∂
E y y +
∂
∂
E z z
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 ⋅ d a
Applying this theorem to Gauss’s Law (of electromagnetism) gives us the integral and differential versions we now know
12

Let’s calculate div E for


E = r q
2 r ˆ
We can do this by expressing E in x-y-z : E = x
Or we can express div in spherical coordinates
2 q
+ y 2 + z 2

∇ ⋅ F =
Since only E r r
1
2
∂ ( r 2 F r
∂ r
)
+ r
1 sin θ
is non-zero, we get
∂ ( F
θ sin
∂ θ
θ )
+
1 r sin θ
∂ F
φ
∂ φ x ˆ x + ˆ y + ˆ z
∇ ⋅ E =
1 r 2
∂ ( r 2 E r
∂ r
)
=
1 r 2
∂ q
∂ r
= 0
This is correct — we have no charge except at r = 0 x 2 + y 2 + z 2

At r = 0, 1/ r 2 gives us an infinity

That’s OK because a “point” charge has an infinite density
Let’s give the “point” charge a small radius R

We did this in Lecture 2, and the solution was
E
=
⎧
⎨
⎪
Q r 2 r ˆ for r
≥
R
Qr r ˆ for r
<
R
R 3
This part is same as a point charge.
We know div E = 0.
Let’s work on this part

For r < R , ∇ ⋅ E =
1 r 2
∂ ( r 2 E r
∂ r
)
=
1 r 2
∂
∂ r ⎝⎜
⎛ Qr 3 ⎞
R 3 ⎠⎟
=
3 Q
R 3

The charge density of the sphere is ρ =
4 π
3
Q
R 3
It works everywhere (as long as ρ is finite)
∇ ⋅ E = 4 πρ
13

2
We found earlier and

Are they same?
U =
8 π dV U =
1
2
ρφ dv
Consider the divergence of the product E ϕ
∇ ⋅ ( E φ ) =
∂
∂ x
( E x
φ ) +
∂
∂ y
( E y
φ ) +
∂
∂ z
( E z
φ )
=
∂ E x
∂ x
φ + E x
∂ φ
∂ x
+
∂ E y
∂ y
φ + E y
∂ φ
∂ y
+
∂ E z
∂ z
φ + E z
∂ φ
∂ z
= ( ∇ ⋅ E ) φ + E ⋅ ( ∇ φ ) = 4 πρφ − E 2

Integrate LHS over very large volume and use Divergence Theorem

V
∇ ⋅ ( E φ ) dv =
S
( E φ ) ⋅ d a
Integral of RHS must be 0, too
= 0 assuming E φ → 0 at far away
4
dv −
E 2 dv = 0
1
2
ρφ dv =
E
8 π
2 dv
Used Gauss’s Law on infinite sheet of charge


Defined electric potential by line integral


Uniform electric field E = 2 πσ above and below the sheet
Electric field has energy with volume density given by
φ
21
= −
P
2
P
1
E ⋅ d s = φ ( P
2
) − φ (
Electric field is negative gradient of electric potential
Differential form of Guass’s Law:

P
1
)
φ =
N
j = 1
E u
= unit: erg/esu = statvolt q r
= j
4
E
φ =
Equivalent to the integral form via Divergence Theorem
∫
= dq r
E
8
π
2
−∇ φ
V div F dv =
A
F ⋅ d a
14