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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 Note the sign convention: positive if coming out
Gauss’s Law
Useful for solving E fields with symmetries Spherical charge distribution
Φ = E ⋅ da∫
E ⋅ da∫ = 4πq
E =
Qr2 r for r ≥ R
QrR3 r for r < R
⎧
⎨⎪⎪
⎩⎪⎪
E
r R
QR2
R
r
S E
2
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
Surface charge density:
Use Gauss again Which surface to use?
What symmetry do we have? Consider a cylinder
Area A and height 2z
E field must be vertical How do we know that?
z E
σ =
chargearea
3
Infinite Sheet of Charge Total flux Φtotal = Φtop + Φside + Φbottom
Side is parallel to E No flux Top and bottom are symmetric Same flux
Charge inside the cylinder is
Using Gauss Don’t forget the direction!
z E
The result is worth remembering:
Infinite sheet of charge produces uniform E field of 2πσ above and below
E ⋅da = 0
Φtotal = 2Φtop = 2AE(z)
qinside = Aσ
E(z) = 2πσ
E = +2πσ z for z > 0−2πσ z for z < 0
⎧⎨⎪
⎩⎪
Place two oppositely-charged large sheets in parallel Consider an area A of them
E fields from the two sheets overlap and add up Between the sheets: E = 4πσ Cancel each other outside
Two sheets also attract each other (obviously) Top sheet feels
The force on area A of the top sheet is
−σ
Pair of Charged Sheets
+σ
Etop Ebottom z
Ebottom = −2πσ z
F = σAEbottom = −2πσ 2Az = −E2
8πAz
4
Imagine we move the top sheet upward by a distance d
We must do work
The energy of the system increases by W
Q: Where exactly is this energy? 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 Total electrostatic energy of a system is
−σ
Pair of Charged Sheets
+σ
Etop Ebottom z W = Fd =E2
8πAd
u =E2
8π
U =E2
8πdV∫ Will come back to this…
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
Thanks to the Superposition Principle, we have only to prove this for the electric field generated by a single point charge
Line integral E ⋅ds
P1
P2∫ is path independent
Line integral q
r 2 r ⋅dsr1
r2∫ is path independent
q
r1
r2 r
ds
E =
qr 2 r
5
Line Integral of Electric Field The dot product is the radial component of the movement, i.e. dr
The integral depends only on r1 and r2, i.e., is path-independent
Generalize using Superposition:
qr 2 r ⋅ds
r1
r2∫ =qr 2 dr
r1
r2∫ = q1r1−
1r2
⎛
⎝⎜⎞
⎠⎟
q
r1
r2 r
ds
E =
qr 2 r
r ⋅ds
dr
Line integral E ⋅dsP1
P2∫ for any electrostatic
field E has the same value for all pathsfrom P1 to P2
Corollary For the special case of P1 = P2, the path becomes a loop
This is equivalent to the path-independence Consider two paths (A and B) from P1 to P2
Path independence means
Corollary above means
Will use this later when we do the “curl”
Line integral E ⋅ds∫ of an electrostatic field around
any closed path is zero
dsB P1
P2
E ⋅dsAP1
P2∫ = E ⋅dsBP1
P2∫
E ⋅dsAP1
P2∫ + E ⋅ (−dsB )P2
P1∫ = 0
dsA
6
Electric Potential The line integral is useful enough to have a name
To move a charge q from P1 to P2, you must do work
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
E ⋅ds
P1
P2∫
φ21 ≡ − E ⋅ds
P1
P2∫
Electric potential at position r
Note the negative sign!
W = −qE ⋅ds
P1
P2∫ = qφ21
φ(r) ≡ − E ⋅ds
0
r
∫
Electric potential difference between P1 and P2
Reference Point Reference point for the electric potential is arbitrary If you choose e.g., point B instead of point A as the reference
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
The constant cancels in the difference
φref=B(r) = − E ⋅ds
B
r
∫ = − E ⋅dsA
r
∫ − E ⋅dsB
A
∫ = φref=A(r) + const.
φ21 = − E ⋅ds
P1
P2∫ = − E ⋅ds0
P2∫ + E ⋅ds0
P1∫ = φ(P2) −φ(P1)
7
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:
We can “reverse” the line integral as well
Gradient is “how quickly the function ϕ varies in space”
In x-y-z coordinate system:
NB: gradient is a vector field It points the direction of maximum rate of increase (= uphill)
Hence E points downhill of ϕ
F(x) = f (x)dx∫
f (x) = d
dxF(x)
φ = − E ⋅ds∫ E = −∇φ gradient
∇φ = x ∂φ
∂x+ y ∂φ
∂y+ z ∂φ
∂z
8
Visualizing Gradient in 2-d For a potential
∇φ(x,y) = x cos(x)sin(y) + y sin x cos y φ(x,y) = sin x ⋅sin y
φ(x,y)
∇φ(x,y)
© Prof. G. Sciolla, MIT
Equipotential Surfaces The same potential 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
Electric field is perpendicular to the equipotential surfaces This is generally true for gradient
of any function
φ(x,y) = sin x ⋅sin y
∇φ(x,y)
© Prof. G. Sciolla, MIT
9
Charge Distribution Electric potential due to a charge distribution is
Continuous case maybe 1-d, 2-d, or 3-d
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
φ =
qj
rjj =1
N
∑ φ =
dqr∫discrete continuous
φ =
λr
d∫ φ =
σr
da∫ φ =
ρr
dv∫
Charged Disk Problem: Calculate the electric potential produced by a thin, uniformly charged disk on its axis Disk radius = a, surface charge density σ
Slice the disk into rings, then into the red bits Charge on the red bit is σsdsdθ The potential due to the red bit is Integrate:
z
a
s ds dθ
dφ =
σsdsdθ
z2 + s2
φ = σ dθ sds
z2 + s20
a
∫0
2π
∫
= 2πσ z2 + s2⎡⎣⎢
⎤⎦⎥s=0
s=a
= 2πσ z2 + a2 − z( ) NB: absolute value in case z < 0
10
Electric Field Can we calculate E from ? We need to take gradient, which means we need
to know the x-, y-, and z-dependence
We calculated ϕ only on the z axis
We are in luck — we know E is parallel to the z axis by symmetry
z
a
s ds dθ
φ = 2πσ z2 + a2 − z( )
E = −∇φ = −z ∂φ∂z
= −2πσ ∂∂z
z2 + a2 − z( ) z= 2πσ −
z
z2 + a2±1
⎛
⎝⎜⎞
⎠⎟z for z > 0
z < 0
Check the Solution
Dimension: [ϕ] = (charge)/(length), [E] = (charge)/(length)2
NB: [σ] = (charge)/(length)2
Far away from the disk (|z| >> a)
Same as a point charge with Q = πa2σ
Close to the disk (|z| << a)
Same as the infinite sheet of charge
φ = 2πσ z2 + a2 − z( )
E = 2πσ −
z
z2 + a2±1
⎛
⎝⎜⎞
⎠⎟z for z > 0
z < 0
φ = 2πσ z 1+ (a z)2 −1( ) z a⎯ →⎯⎯ 2πσ z 1+ 1
2 (a z)2 −1( ) = πa2σz
Ez a⎯ →⎯⎯ ±2πσ z
11
Energy and Potential
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
U =
12
qjqk
rjkk ≠ j∑
j =1
N
∑ =12
qj
qk
rjkk ≠ j∑
j =1
N
∑ =12
qj φ jkk ≠ j∑
j =1
N
∑ Potential at the j-th charge due to the
other charges
U =
12
ρφ dv∫
′ρ = sρ s = 0→1
′φ = sφ
dW = (d ′ρ ) ′φ dv∫ = (dsρ)(sφ)dv∫
W = s ds ρφ dv∫0
1
∫ =12
ρφ dv∫ Will come back to this…
We know a system of N charges has a total energy of
Generalizing to continuous distributions Integrate entire space, or where ρ ≠ 0 No need to avoid j = k because “individual”
charge is infinitesimally small
Shrinking Gauss’s Law Charge is distributed with a volume density ρ(r) Draw a surface S enclosing a volume V
Guass’s Law:
Now, make V so small that ρ is constant inside V
As we make V smaller, the total flux out of S scales with V
Therefore:
LHS is “how much E is flowing out per unit volume” Let’s call it the divergence of E
E ⋅da
S∫ = 4π ρdvV∫ Total charge in V
E ⋅da
S∫ = 4πρV for very small V
limV→0
E ⋅daS∫
V= 4πρ
12
Divergence 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
Add up all walls:
divE ≡ lim
V→0
E ⋅daS∫
V= 4πρ
dx
dy
dz
E(x + dx,y,z) E(x,y,z)S1 S2
E ⋅da
S1∫ = E(x,y,z) ⋅ (−x)dydz
E ⋅da
S2∫ = E(x + dx,y,z) ⋅ xdydz
Sum = Ex(x + dx) − Ex(x)( )dydz
=∂Ex
∂xdxdydz
E ⋅da
S∫ =∂Ex
∂x+∂Ey
∂y+∂Ez
∂z
⎛
⎝⎜
⎞
⎠⎟V = ∇ ⋅E( )V
div E
Gauss’s Law, Local Version We now have Gauss’s Law for a very small volume/surface
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,
Applying this theorem to Gauss’s Law (of electromagnetism) gives us the integral and differential versions we now know
divE = 4πρ where divE ≡ ∇ ⋅E =
∂Ex
∂x+∂Ey
∂y+∂Ez
∂z
⎛
⎝⎜
⎞
⎠⎟
divFdv
V∫ = F ⋅daA∫
13
Coulomb Field Let’s calculate div E for
We can do this by expressing E in x-y-z :
Or we can express div in spherical coordinates
Since only Er is non-zero, we get
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
E =
qr 2 r
E =
qx2 + y 2 + z2
xx + yy + zz
x2 + y 2 + z2
∇ ⋅F =
1r 2
∂(r 2Fr )∂r
+1
r sinθ∂(Fθ sinθ)
∂θ+
1r sinθ
∂Fφ
∂φ
∇ ⋅E =
1r 2
∂(r 2Er )∂r
=1r 2
∂q∂r
= 0
Spherical Charge Let’s give the “point” charge a small radius R We did this in Lecture 2, and the solution was
For r < R,
The charge density of the sphere is
It works everywhere (as long as ρ is finite)
E =
Qr2 r for r ≥ R
QrR3 r for r < R
⎧
⎨⎪⎪
⎩⎪⎪
This part is same as a point charge. We know div E = 0.
Let’s work on this part
∇ ⋅E =
1r 2
∂(r 2Er )∂r
=1r 2
∂∂r
Qr 3
R3
⎛
⎝⎜⎞
⎠⎟=
3QR3
ρ =
Q4π3 R3 ∇ ⋅E = 4πρ
14
Energy Again We found earlier and Are they same?
Consider the divergence of the product Eϕ
Integrate LHS over very large volume and use Divergence Theorem
Integral of RHS must be 0, too
U =
12
ρφ dv∫ U =
E 2
8πdV∫
∇ ⋅ (Eφ) = ∂∂x
(Exφ) + ∂∂y
(Eyφ) + ∂∂z
(Ezφ)
=∂Ex
∂xφ + Ex
∂φ∂x
+∂Ey
∂yφ + Ey
∂φ∂y
+∂Ez
∂zφ + Ez
∂φ∂z
= (∇ ⋅E)φ +E ⋅ (∇φ) = 4πρφ − E 2
∇ ⋅ (Eφ)dv
V∫ = (Eφ) ⋅daS∫ = 0 assuming Eφ → 0 at far away
4π ρφ dv∫ − E 2 dv∫ = 0
12
ρφ dv∫ =E 2
8πdv∫
Summary Used Gauss’s Law on infinite sheet of charge Uniform electric field E = 2πσ above and below the sheet Electric field has energy with volume density given by
Defined electric potential by line integral
Electric field is negative gradient of electric potential
Potential due to charge distribution: or
Differential form of Guass’s Law: Equivalent to the integral form via Divergence Theorem
u =E2
8π
φ21 = − E ⋅ds
P1
P2∫ = φ(P2) −φ(P1) unit: erg/esu = statvolt
E = −∇φ
φ =
qj
rjj =1
N
∑ φ =
dqr∫
∇ ⋅E = 4πρ
divFdv
V∫ = F ⋅daA∫