Massachusetts Institute of Technology
Department of Physics
Physics 8.07
Fall 2005
Problem Set 2 Solutions
Problem 1: Potential and Electric Field with Cylindrical Symmetry
a) The ring of charge lies in the plane z = 0. The potential at x = y = 0, z > 0
is
1
1 Z
dq ,
Vring (z) =
4π0 (s2 + z 2 )1/2
where x0 = s cos φ, y 0 = s sin φ gives the Cartesian coordinates of a point
on the ring, and dq = λ(sdφ) is the element of charge. Carrying out the
integration over φ, we get
Vring (z) =
λs
.
20 (s2 + z 2 )1/2
b) The disk of uniform charge density σ can be regarded as a set of rings of
linear charge density σds. Thus,
Vdisk (z) =
σ ZR
s ds
.
20 0 (s2 + z 2 )1/2
Carrying out the integration gives
Vdisk (z) =
σ √ 2
z + R2 − |z| .
20
~ = Ez (z)~ez along the z-axis, where
c) By symmetry, E
dVdisk
σz
Ez (z) = −
=
dz
20
1
1
−√ 2
|z|
z + R2
!
.
For z 2 R2 this yields the electric field of a uniform surface charge, Ez =
(σ/20 ) sign(z). For z 2 R2 , it yields the electric field of a point charge,
Ez = Q sign(z)/(4π0 z 2 ) where Q = σπR2 .
1
Problem 2: Griffiths Problems 2.12 and 2.18 (p. 75)
a) From Gauss’s law,
4πr2 Er =
giving
4πρr3
1
Qenc =
,
0
30
~ = ρr ~er .
E
30
b) Let ~r1 be the displacement vector going from the center of the first sphere
(of charge density ρ) to the field point. Let ~r2 be the displacement vector
going from the center of the second sphere (of charge density −ρ) to the field
point. If the field point is within both spheres, the electric field is given by
linear superposition using the results of part (a):
~ = ρ (~r1 − ~r2 ) = ρ d~ ,
E
30
30
where d~ is the vector connecting the center of the first sphere with the center
of the second sphere.
Problem 3: Griffiths Problem 2.49 (p. 108)
a) Using linear superposition with q → ρ(~x 0 ) d3 x0 ,
~ x) =
E(~
1 Z ρ(~x 0 )(~x − ~x 0 )
1
0
0
1
+
|~
x
−
~
x
|
e−|~x−~x |/λ d3 x0 .
0
3
4π0
|~x − ~x |
λ
~ ×E
~ = 0. To see this, note that for a single
b) A potential must exist because ∇
0
~
~ ×E
~ =0
point charge at ~x = 0, E = Er (r)~er where ~r = ~x. For this case, ∇
~ is constructed to be translationally invariant (i.e., it
everywhere. Because E
~ E
~ = 0 for a point
doesn’t matter where we put the origin), it follows that ∇×
charge regardless where it is located. Using linear superposition, it follows
~ ×E
~ = 0 for any static charge distribution.
that ∇
c) The potential of a point charge at r = 0, such that Er = −dV /dr, is readily
found by guessing (and using the method of trial and error) to be
qe−r/λ
.
V (r) =
4π0 r
This is called the Yukawa potential and it gives the electrostatic potential
that would arise in quantum electrodynamics if the photon had a mass h̄/(λc)
where 2πh̄ is Planck’s constant and c is the speed of light. In this case the
2
photon would be a “massive spin-one field” and electromagnetic forces would
not act much further than a distance λ. Although this modification appears
not to describe real electromagnetism (because the photon is massless), it
does describe (after some elaboration) the weak nuclear force responsible for
radioactive decay. The weak force carrying particles are the W and Z bosons
which are quite massive so that λ for those forces is smaller than the size of
a proton.
d) Using the potential of part (c) and Er = (q/0 )(1 + R/λ)e−R/λ , for a sphere
of radius R one easily obtains
I
−R/λ
R
~ a = qe
E·d~
1+
0
λ
and
q
R −R/λ
1 Z
2
V
(r)4πr
dr
=
1
−
1
+
e
,
λ2
0
λ
which immediately yield the desired result,
I
q
1 Z
~
.
E · d~a + 2 V (r) d3 x =
λ
0
e) To generalize the result of part (d), we can take the divergence of the answer
to part (a):
~ ·E
~ =
∇
0
1
1 Z
0
~ · ~x − ~x
ρ(~x 0 )∇
1 + |~x − ~x 0 | e−|~x−~x |/λ d3 x0
0
3
4π0
|~x − ~x |
λ
Z
0
0
0 ρ(~x )(~x − ~x ) (~x − ~x )
1
1
0
−|~
x−~
x 0 |/λ 3 0
·
|~
x
−
~
x
|
e
dx
+
1
−
1
−
4π0
|~x − ~x 0 |3
λ|~x − ~x 0 |
λ
1
1
= ρ(~x ) − 2 V (~x ) .
0
λ
!
In deriving this we have used
0
0
~ x − ~x 0 | = ~x − ~x and ∇
~ · ~x − ~x
∇|~
|~x − ~x 0 |
|~x − ~x 0 |3
!
= 4πδ 3 (~x − ~x 0 )
as well as the generalization of the answer to part (c) for an arbitrary charge
distribution,
1 Z ρ(~x 0 ) −|~x−~x 0 |/λ 3 0
V (~x ) =
e
dx .
4π0 |~x − ~x 0 |
The desired result follows at once by integrating over an arbitrary volume
and using the divergence theorem,
I
Z
1
1 Z
3
~
E · d~a + 2 V d x = Qenc , Qenc ≡ ρ d3 x .
λ
0
3
f) Figure 2.35 of the text is modified only for the formulas on the left and right
~ x ) for
sides of the triangle. The modified Coulomb integrals for V (~x ) and E(~
0
a given source charge distribution ρ(~x ) have been given above. Similarly, the
modification of Gauss’s law has been given in part (e) in both its differential
and integral forms. The only other modification is the Poisson equation,
which becomes
1
1
2
∇ − 2 V =− ρ.
λ
0
4