Gabriel Barello - Classical Electrodynamics
1.21
a.)
For this problem, we compute
A
Lz ,
the action per unit length. We use g = 1 and will use the function
Ψ(x, y) = Ax(1 − x) · y(1 − y)
as the form of our approximate solution. First we compute
5 Ψ = A ((1 − x) · y(1 − y) − x · y(1 − y), (1 − y) · x(1 − x) − y · x(1 − x), 0)
| 5Ψ |2 = A2 y 2 (1 − y)2 (1 − 2x)2 + x2 (1 − x)2 (1 − 2y)2
And note that gΨ = Ψ. The integrals compute to be
Z
gΨd2 x =
v
Z
A
36
| 5Ψ |2 d2 x =
v
A2
45
Then the minimum of the difference of these is given when A = 54 . Indeed, the second derivative is positive at this value
of A, so it is indeed a minimum (if that wasn’t obvious from the form).
b.)
Figure 1: This is a comparison of the true potential (in blue) and the approximate potential (dashed) The left plot shows
y = .25 and the right shows y = .5. The vertical axis is in units of eV·4π0
1
1.24
Since this setup is π/2 radians rotationally symmetric each of the corner points and each of the edge points are equivalent,
so we really only need to compute one corner, one edge and the middle. Let us label these points {a, b, e} respectively. Recall
the best-average formula for the poisson equation
hhΦiip = hhΦii +
h2
h2
g + hgic
5
10
where
4
1
hΦic + hΦis
5
4
And the c and s stand for cross and square average. (the cross average is the one that makes a ”plus” sign on the grid).
Note, too that h2 = 1/16. Now we are ready to get things lined up. Note the following table:
hhΦii =
ij
a
b
e
i+1,j
b
a
b
i,j+1
0
0
b
i-1,j
0
a
b
i,j-1
b
e
b
i+1,j+1
0
0
a
i-1,j-1
0
b
a
i+1,j-1
e
b
a
i-1,j+1
0
0
a
After some fussing, the iterative equations we get are:
ei
4bi
3
+
+
16
10
160
2ai
ei
bi
1
=
+ + +
5
5
8
40
4bi
a
3
=
+ +
5
4 80
ai+1 =
bi+1
ei+1
Using the jacobian method with 0 · Φ = 1 on all sites initially gives
0.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1.
0.48125
0.361719
0.226863
0.164902
0.119332
0.0940143
0.077463
0.0676442
0.0614758
0.0577307
1.
0.75
0.44875
0.317344
0.216899
0.162304
0.126118
0.104821
0.091378
0.0832388
0.0781876
1.
0.6875
0.457813
0.30743
0.221153
0.165485
0.132254
0.111451
0.0987943
0.0909623
0.0861645
Note that my answers differ by a factor of 4π from the book’s exact solution, because I computed Φ · 0 whereas Jackson
gave 4π0 · Φ.
2
b.)
After ten iterations of the gauss-seidel method using the same initial values we get
Iteration
0
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
a
1
0.48125
0.259176
0.140508
0.0901451
0.0683307
0.0589005
0.0548232
0.0530603
0.0522981
0.0519686
b
1
0.5425
0.271445
0.157308
0.107832
0.086445
0.077198
0.0731999
0.0714713
0.0707239
0.0704007
e
1
0.374813
0.210872
0.13555
0.103169
0.0891607
0.0831043
0.0804858
0.0793536
0.0788641
0.0786524
c.)
We factored out a 1/0 at the begining, so we are actually reporting the value 0 Φ. The following plot compares our
approximate solution to the exact one
Figure 2: This is a log plot of the aproximate solution using the jacobian method (blue) and the gauss seidel method (red)
compared to the exact solution (lines). Unfortunately the lowest exact value is obscured by the log plot, but it is there, I
promise. The vertical axis is in units of Φ · 0 . As you can see the gauss-seidel method converges faster and is more accurate
after 10 iterations.
3
2.1
a.)
The apropriate image charge is −q at position (−d, 0). Then the electric field at a point (0, y) is
q
1
(−dx̂ + y ŷ − dx̂ − y ŷ)
4π0 (d2 + y 2 ) 32
q
d
x̂
=−
2π0 (d2 + y 2 ) 32
E(0, y) =
This is related to the surface charge by the equation σ = 0 E. So,
σ=−
− 3
qd 2
d + y2 2
2π
Here is a plot
Figure 3: This shows the (nondimentionalized) charge density with d = 1
b.)
The force on the charge is the same as the force that would result from the presence of the image charge.
F =−
q2
q2 1
1
=
−
4π0 (2d)2
16π0 d2
c.)
The total force acting on the plane can be computed by computing the integral
Z
Z
(qd)2 2π ∞ 2
(d + r2 )−3 rdrdθ
8π 2 0 0
0
Z
(qd)2 ∞ −3
=
u du
8π0 d2
1 q2
=−
16π0 d2
Fp =
4
u ≡ d2 + r 2
d.)
The work is given by the integral
Z
F · dl = −
W =
=
q2
16π0
∞
Z
x−2 dx
d
1 q2
16π0 d
e.)
The potential energy is given by the integral
Z
2d
F · dl = −
PE =
∞
q2
4π0
Z
∞
x−2 dx
2d
1 q2
=−
8π0 d
This is a factor of two larger, of course this is because as the charge moves away from the plane the charge on the plane
disperses, as if the image charge were moving away at the same rate. We usually define potential energy by holding one
charge fixed while moving the other charge in from infinity. In this case, it is as if we are moving both charges in from infinity
at the same time, resulting in a factor of two difference. One could also think of it in terms of the fields. Potential energy
is stored in the field, but in the current situation, only half the field actually exists (the other half is cancelled out by the
conductor) so there is only half as much energy stored in the field as in the corresponding situation with a real charge instead
of a conducting plane.
f.)
q = e = −1.602 · 10−19 C
d = 1A = 1 · 10−10 m
0 = 8.854 · 10−12 F/m →
W = 3.59959 eV.
2.9
The reaction of the sphere (radius a) to a uniform electric field E = E ẑ can be modeled using image charges of charge ± Qa
R
2
at positions ± aR with 2πQ0 R2 = E and taking the limit R → ∞. In the limit, this becomes an electric dipole, with dipole
3
3
moment p = − 2Qa
R2 ẑ = −4π0 a E ẑ. Which gives a surface charge density
σ0 = −30 E cos(θ)
q
σq = σ0 +
4πa2
Uncharged
(1)
Charged with charge q
(2)
Where θ is measured from the positive z axis. Finally, recall that the force per unit area on a differential element is:
F⊥ /A =
σ2
20
(3)
To find the force on each hemisphere we must only integrate the above formula over each hemisphere. Note that the force
on the hemisperes will be equal and opposite, since the charge densities are equal and opposite. Furthermore, note that the
component of force on a surface element not in the ẑ direction will cancel due to the azimuthal symmetry so we can integrate
instead only the z components of the force.
Applying these observations, the force required to keep the hemispheres from separating is given by
2
Z
F = 2πa
0
π
2
σ2
cos(θ) sin(θ)dθ
2
5
(4)
With 4 and 2, we can calculate the force for the two scenarios given:
a.)
F0 = 9π(aE)2 0
Z
π
2
cos(θ)3 sin(θ)dθ
0
=
9
π(aE)2 0
4
b.)
For a charged sphere we get
π
2
q
2
(−30 E cos(θ) + 4πa
2)
cos(θ) sin(θ)dθ
2
0
Z π2
q
q2
(−60 E cos(θ) 4πa
2 + 16π 2 a4 )
= F0 + 2πa2
cos(θ) sin(θ)dθ
2
0
Z π2
q
q
) cos(θ) sin(θ)dθ
(−60 E cos(θ) +
= F0 +
40 0
4πa2
q
q
= F0 +
(
− 20 E)
40 8πa2
2
Z
Fq = 2πa
And the total force is given by
Fq = F0 +
q
q
(
− 20 E)
40 8πa2
(5)
2.17
a.)
The free space greens function for three dimentions (apart from an arbitrary multiplicative constant) is
G(x, y, z; x0 , y 0 , z 0 ) =
By defining a =
p
1
1
=p
0
2
kx − x0 k
(x − x ) + (y − y 0 )2 + (z − z 0 )2
(x − x0 )2 + (y − x0 )2 we can write the 2-dimentional free-space greens function as
0
0
Z
G(x, y; x , y ) = 2
0
∞
√
a2
1
dx
+ x2
x = kz − z 0 k
Which has the well-known solution (apart from an inessential, though infinite constant. This constant is hand-waved
away by instead integrating to a large value Z instead of to infinity, but I think this will be good enough.)
G(x, y; x0 , y 0 ) = − log((x − x0 )2 + (y − y 0 )2 )
= − log(ρ2 + ρ02 − 2ρρ0 cos(φ − φ0 ))
Where the last line follows from the ”law of cosines”.
b.)
The general solution to laplace’s equation is given by free-space solutions for ρ > ρ0 and ρ < ρ0 which are then patched
together at ρ = ρ0 . To this end, we define two new functions, R and Φ, by the assumption G = R(ρ, ρ0 )Φ(φ, φ0 ). Let us
compute the fourier coefficients of the Φ function for the laplace equation sourced by delta functions.
Z
Z
0
1 ∂
∂R(ρ, ρ0 )
1
R(ρ, ρ0 ) 1
4π
imφ
0
√
√
ρ
e
Φ(φ, φ )dφ −
m2 eimφ Φ(φ, φ0 )dφ = − √ δ(ρ − ρ0 )eimφ
2
ρ ∂ρ
∂ρ
ρ
2π
2π
2π
6
The surface terms in the second term of the l.h.s. canceled because of the cyclic boundary conditions on Φ. We can divide
0
by eimφ and pull out an overall fourier integral to arrive at
Z 2π
0
4πδ(ρ − ρ0 )
1 ∂ ∂R(ρ, ρ0 )
im(φ−φ0 )
0
2 R(ρ, ρ )
e
Φ(φ,
φ
)dφ
=
−
ρ
−m
ρ ∂ρ
∂ρ
ρ2
ρ
0
The fourier-transform factor is constant w.r.t. ρ and likewise the rest is a constant w.r.t. φ so w.l.o.g. we can set the fourier
transform factor equal to a constant, we choose 1, so that
Z
2π
0
1
1 im(φ0 −φ)
eim(φ−φ ) Φ(φ, φ0 )dφ =
→ Φ(φ, φ0 ) =
e
2π
2π
0
∂
4π
∂R(ρ, ρ0 )
m2
ρ
− 2 R(ρ, ρ0 ) = − δ(ρ − ρ0 )
∂ρ
∂ρ
ρ
ρ
Thus we arrive at the general solution in the form
∞
1 X im(φ−φ0 )
e
Rm (ρ, ρ0 )
2π −∞
G=
where Rm satisfies the above equation.
c.)
To complete the solution we solve the above equation in the homogenious case and then patch together the solution at ρ = ρ0 .
The general solution of the homogenious equation is given in Jackson as
R(ρ, ρ0 ) = am (ρ0 )ρm
0
0
m ∈ Z\{0}
0
R(ρ, ρ ) = a0 (ρ ) + b0 (ρ ) log(ρ)
m=0
We need two solutions one for ρ > ρ0 and one for ρ < ρ0 , we will distinguish them with a subscript > and < respectively
on the coefficients. Then, the two full solutions are give by
X
R< (ρ, ρ0 ) = a0< (ρ0 ) + b0< (ρ0 ) log(ρ) +
am< (ρ0 )ρm
m∈Z\{0}
0
0
X
0
R> (ρ, ρ ) = a0> (ρ ) + b0> (ρ ) log(ρ) +
am> (ρ0 )ρm
m∈Z\{0}
There are a few conditions we can impose right off the bat. Namely, the solution needs to be well-behaved at zero, and
be logarithmic as ρ → ∞. With this we can determine that b0< (ρ0 ) = 0, and that R< has no negative powers of ρ and R>
has no positive powers ofρ. This leaves us with
X
R< (ρ, ρ0 ) = a0< (ρ0 ) +
am< (ρ0 )ρm
m>0
0
0
R> (ρ, ρ ) = a0> (ρ ) + b0> (ρ0 ) log(ρ) +
X
am> (ρ0 )ρm
m<0
Next, the solution must be symmetric. That is to say that if ρ < ρ0 then R< (ρ, ρ0 ) = R> (ρ0 , ρ) and likewise, for ρ0 > ρ,
R> (ρ, ρ0 ) = R< (ρ0 , ρ). These conditions state that
a0< (ρ0 ) +
X
am< (ρ0 )ρm = a0> (ρ) + b0> (ρ) log(ρ0 ) +
m>0
0
X
am> (ρ)ρ0m
m<0
0
a0> (ρ ) + b0> (ρ ) log(ρ) +
X
0
am> (ρ )ρ
m<0
m
= a0< (ρ) +
X
m>0
7
am< (ρ)ρ0m
Note that the solutions for various m must independently obey this symmetry, since they are all linearly independent.
Then this tells us that a<0 = A log(ρ0 ), a>0 = 0 and b>0 = A, with A a yet to be determined constant. In addition it must be
that am> (ρ0 ) = Am (ρ0 )m and am< (ρ0 ) = Am (ρ0 )−m with Am constants that will be determined by considering the derivative
discontinuity. So that we finally have the solutions, us to an overall constant in each term
m
ρ
R< (ρ, ρ ) = A log(ρ ) +
Am
ρ0
m>0
m
X
ρ0
0
R> (ρ, ρ ) = A log(ρ) +
Am
ρ
m>0
0
0
X
The last consideration to make is that the difference in the derivatives at ρ = ρ0 should be −4π/ρ0 . The derivative
evaluated (with respect to ρ) near ρ = ρ0 are
m
ρ0
m>0
X
m
A
R>,ρ (ρ → ρ0 , ρ0 ) = 0 −
Am 0
ρ
ρ
m>0
R<,ρ (ρ → ρ0 , ρ0 ) =
X
Am
Each term satisfies the discontinuity condition individually so we get that A = −4π and that Am =
solutions can be written in the form
2π
m.
So that the two
X 1 ρ m
R< (ρ, ρ ) = −2π log(ρ ) + 4π
m ρ0
m>0
X 1 ρ0 m
0
2
R> (ρ, ρ ) = −2π log(ρ ) + 4π
m ρ
m>0
0
02
Which can be conveniently expressed as
R< (ρ, ρ0 ) = −2π log(ρ2> ) + 4π
X 1 ρ< m
m ρ>
m>0
With ρ< (ρ> ) the lesser (greater) of the two radii. Lastly we can plug this back into our original solution. Note that the
full solution (for each m) of the radial equation has a positive and a negative power of m. We threw out the poorly-behaved
powers of m, but for the well behaved ones we get 2 contributions, one from the positive m in the series, and one from the
negative m in the above series. Thus this ammounts to adding the positive-m term to the negative m term, causing the
imaginary parts of the exponential to cancel (as needs to be the case to itnerprit this as a potential), leaving a factor of 2 cos,
so in the end we get
G = − log(ρ2> ) + 2
m
∞
X
1 ρ<
cos(m(φ − φ0 ))
m
ρ
>
1
2.20
a.)
Using the greens function of problem 2.17 the potential is given as
σ(ρ0 , φ0 )
G(ρ, φ; ρ0 , φ0 )ρdρdφ
4π0
!
m
3
∞
X
λ X
1 ρ<
n
2
=
(−1)
− log(ρ> ) + 2
cos(m(φ − nπ/2))
4π0 n=0
m ρ>
m=1
Z
Φ(ρ, φ) =
8
Where ρ> (ρ< ) is the greater (lesser) of ρ and a, but since the log term has no φ dependence it cancels when the sum over
n is performed, thus we can write the solution as
Φ(ρ, φ) =
m X
∞
3
λ X 1 ρ<
(−1)n cos(m(φ − nπ/2))
2π0 m=1 m ρ>
n=0
for a moment lets focus on the sum of cosines. Writing this out in full and manipulating we find that
cos(mφ) − cos(mφ − mπ/2) + cos(mφ − 2mπ/2) − cos(mφ − 3mπ/2) = (cos(mφ) − cos(mφ − mπ/2))(1 + (−1)m )
={
2 cos(mφ)
0
if m = 4k + 2, k ∈ Z
else
So that we can write the solution as
Φ(ρ, φ) =
4k+2
∞
ρ<
λ X 1
cos((4k + 2)φ)
π0
2k + 1 ρ>
k=0
b.)
By manipulating our solution for Φ we quickly arrive at
4k+2
∞
X
ρ> iφ
λ
1
Φ(ρ, φ) =
Re
e
π0
2k + 1 ρ<
!
k=0
Note the identity (page 75 of jackson)
X Zn
1
1+Z
= log
n
2
1−Z
nodd
Where for our case we have Z =
iφ
ρ> e
ρ<
2
so that
λ
Φ=
Re log
2π0
λ
Re log
=
2π0
eiφ )2
1 + ( ρρ>
<
!!
1 − ( ρρ>
eiφ )2
<
> iφ
(1 − i ρρ<
e )(1 + i ρρ>
eiφ )
<
(1 −
ρ> iφ
ρ< e )(1
+
!!
ρ> iφ
ρ< e )
In the interior we have that ρ< = ρ and ρ> = a and the solution takes the form
λ
(z − ia)(z + ia)
Φ=
Re log
= Re(w(z))
2π0
(z − a)(z + a)
In the exterior ρ< = a and ρ> = ρ and we have
λ
(a − iz)(a + iz)
Φ=
Re log
2π0
(a − z)(a + z)
But we can multiply the interior by i2 = −1 without changing the real part (since the real part of log depends only on
the magnitude of z).
9
λ
(ia + z)(ia − z)
Re log
2π0
(a − z)(a + z)
λ
(z − ia)(ia + z)
=
Re log
2π0
(z − a)(z + a)
Φ=
= Re(w(z))
Thus, we can identify
Φ = Re(w(z))
So that the potential is exactly the real part of the funtion w(z) defined in Jackson. Now, as is stated in problem 2.3 of
Jackson, the potential of an infinite line charge is the logarithm of the distance. This solution reflects that beautifully by
converting the cartesian components of position into a complex variable and then taking the real part. The four factors in
the argument of teh logarithm correspond exactly to two positive line charges at ±ia, or ±aŷ, and two negative line charges
at ±a, or ±ax̂, where x and y have been identified with the real and imaginary axes respectively.
c.)
The cartesian components of the field are given by
∂Φ sin θ ∂Φ
−
)
∂ρ
ρ ∂φ
∂Φ cos θ ∂Φ
Ey = −(sin(θ)
+
)
∂ρ
ρ ∂φ
Ex = −(cos(θ)
Note that for ρ < a
∞
∂Φ
2λ X ρ 4k+1
=
cos((4k + 2)φ)
∂ρ
aπ0
a
k=0
∞
X
2λ
1 ∂Φ
=−
ρ ∂ρ
aπ0
k=0
ρ 4k+1
sin((4k + 2)φ)
a
So the full expressions are
2λ
Ex = −
aπ0
2λ
=
aπ0
2λ
Ey = −
aπ0
=
2λ
aπ0
∞
X
ρ 4k+1
k=0
∞
X
a
!
(cos((4k + 2)φ) cos(θ) + sin(θ) sin((4k + 2)φ))
!
ρ 4k+1
cos((4k + 1)φ)
a
k=0
∞
X
!
ρ 4k+1
(cos((4k + 2)φ) sin(θ) − cos(θ) sin((4k + 2)φ))
a
k=0
!
∞
X
ρ 4k+1
sin((4k + 3)φ)
a
k=0
Keeping only the first two terms, we get
2λ
ρ
ρ5
Ex = −
cos(φ) +
cos(5φ)
aπ0 a
a
2λ
ρ
ρ5
Ey = −
sin(3φ) +
sin(7φ)
aπ0 a
a
10
When y = 0, φ = 0 or π and the x component of the electric field is given by
Ex = ±
2λ
aπ0
ρ ρ5
±
a a
with the ± corresponding to the negative or positive axis respectively. The relative strength of the k = 1 contribution
and k = 2 contribution is
k=1
1
=± 4
k=2
ρ
2.26
Consider the 2-dimentional region ρ ≥ α, 0 ≤ φ ≤ β bounded by conducting surfaces held at zero potential.The large ρ
value of the potential Φ is determined by some unknown charge distribution at ρ >> α and thus this may be considered as
a laplace problem with dirichlet boundary conditions.
a.)
The general solution to laplace’s equation in polar coordinaes is given, after separation of variables, by
R(p) = aρν + bρ−ν , Ψ(φ) = A cos(νφ) + B sin(νφ)
| ν>0
R(p) = a0 + b0 log(ρ), Ψ(φ) = A0 + B0 φ
| ν=0
Given the boundary conditions we can conclude that A0 = B0 = A = 0 and that ν = n πβ . So, only the sin term of the
ν > 0 solution will appear. Lastly we have the condition that Φ(α, φ) = 0 which gives us the condition
a(α)ν + b(α)−ν = 0 → b = −a(α)2ν
Which gives us the general form of the solution
Φ(ρ, φ) =
X
Cn (α)
nπ
β
ρ nπ
β
α
n∈Z
!
nπ
α β
nπ
−
sin( φ)
ρ
β
b.)
Keeping only the lowest nonvanishing term we get
Φ0 (ρ, φ) = C1 (α)
π
β
ρ πβ
α
πβ !
α
π
−
sin( φ)
ρ
β
By which we can calculate the electric fields at the surfaces via the equation En̂ = − ∂Φ
∂ n̂ to be
πβ +1 !
α
π
+
sin( φ)
α
ρ
β
π !
∂Φ
1 ∂Φ
π πβ ρ πβ
α β
π
Eφ (ρ, φ) = −
=−
= C1 α
−
cos( φ)
∂ n̂
ρ ∂φ
βρ
α
ρ
β
π π
Eρ (ρ, φ) = C1 α β −1
β
ρ πβ −1
11
With Eρ zero at φ = 0,β and Eφ = 0 at ρ = α. Then, the surface charge density is given by σ = 0 E⊥ , or
π
π0 πβ −1
sin( φ)
α
β
β
πβ !
π0 πβ ρ πβ
α
α
= σ |φ=β
= C1
−
βρ
α
ρ
σ |ρ=α = C1
σ |φ=0
c.)
Consider β = π. The linear charge density looks like this
Figure 4: This plot shows the linear charge density. Note that the charge density on the cylinder goes as sin(φ) but since it
is a half-circle, it works out that the linear charge density is constant! Of course, this is a plot of the non-dimentionalized
function with α = 1
Now lets convert the polar vectors into cartesian vectors! I will use x̂ parallel to the plane and ŷ perpendicular to the
plane. Then the electric field in x and y coordinates, for large ρ as is required to be far away from the cylinder, is given by
Ex
Ey
=
Eρ cos(φ) − Eφ sin(φ)
Eρ sin(φ) + Eφ cos(φ)
sin(φ) cos(φ) − cos(φ) sin(φ)
cos(φ)2 + sin(φ)2
0
1
= C1
= E0
So, indeed the field is constant and perpendicular to the plane far from the cylinder.
Now, we have the charge density on the cylinder, we just need to integrate over φ. This works out to be
Z
Z
σρ=α = 2αE0 0
sin(φ) = 4E0 0 α.
φ
Now, the surface charge density on an infinite conducting plane is σ = E0 0 so indeed the surface charge on the cylinder
is twice that contained on a strip on width 2α on an infinite conducting plane. Of course, all the surface charge densities are
actually charge density per unit length in the z direction. Next we need to show that the extra charge density is taken from
the surrounding area. To show this we will take the difference between the surface charge density of this system and that of
an infinite conducting plane, and integrate this from ρ = α to ∞, and multiply by 2, of course. This works out to be:
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Z
∞
Z
∞
(σφ − σ0 )dρ = 2E0 0
2
α
1−
α
α2
−
1
dρ
ρ2
= 2E0 0 α
Exactly what we needed! Now, just for completeness’ sake, note that the difference between these two surface charges
goes to zero for large ρ so indeed the statment that the charge is taken from nearby points on the plane is accurate. This
also answers the last question regarding the surface charge on a wide strip being the same wether or not the bump is present.
Since the difference in surface charge densities between the two cases goes to zero as ρ becomes large, the total surface charge
within that large ρ strip will be the same between the two cases, since the total charge density on the two surfaces is the
same, and the surface charge density at large ρ is the same.
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