Download Physics 416G : Solutions for Problem set 7

Physics 416G : Solutions for Problem set 7
Due : October 16, 2015
Please read Griffiths sections 2.4, 2.5 and 3.2, pages 91-112, 124-130.
We covered the material from slightly different point of view in the class.
Please show all the details of your computations including intermediate steps.
1
Problem 1
We derived in the class that the Electrostatic energy between the two dipole p~1 and p~2 is given by
~2 =
U = −~
p1 · E
1 p~1 · p~2 − 3(~
p1 · r̂)(~
p2 · r̂)
.
4π0
r3
~ 2 is the electric field from the dipole p~2 acting on dipole p~1 . We are going to use this to solve
Here E
the following problem.
a) Two coplanar dipoles are oriented as shown in the left figure. Find the equilibrium value of the
angle θ0 for the dipole p0 if the angle θ is fixed.
b) A dipole p is situated a distance z above an infinite grounded conducting plane as in the right figure.
The dipole makes an angle θ with the perpendicular to the plane. Find the energy. If the dipole is free
to rotate, in what orientation will it come to rest?
Hint : To solve the problem more easily, imagine the opposite dipole below the conducting ground by
making a reflected image at z = 0. Then you can get the electric potential or electric field from the
image dipole acting on the dipole above the conducting plane.
Sol: a) Using the formula we can evaluate the Electrostatic potential energy. Let us choose the z
coordinate as the line go through both the dipoles and use the spherical coordinate. Then
U=
pp0 cos(θ + θ0 ) − 3 cos θ cos θ0
pp0 sin θ sin θ0 + 2 cos θ cos θ0
=−
.
3
4π0
r
4π0
r3
The same result can be obtained by using the electric field from on dipole action to another. To find
the Electrostatic energy minimum, which will be the equilibrium, we can take a derivative of U with
the angle θ0 and set it to 0. Then
∂U
∝ sin θ cos θ0 − 2 cos θ sin θ0 = 0 ,
∂θ0
−→
tan θ = 2 tan θ0 .
b) One can solve this problem with the method of image that we have developed in class.
Once you solve the first part of the problem, we can use the result given there in part a). Imagine the
left dipole as the image of the right dipole in the left figure by setting p0 = p, θ0 = θ. Then
U =−
p2 sin2 θ + 2 cos2 θ
.
4π0
r3
Again, we want to find the equilibrium. Thus by taking a derivative with respect to θ, we get
2 sin θ cos θ − 4 cos θ sin θ = 0. Thus θ = 0, π, which gives minimum energy and stable. Thus the
dipole want to be perpendicular to the conducting surface. For θ = π/2, the potential energy becomes
maximum.
2
Problem 2
A metal sphere (metal is a conductor) of radius R, carrying charge Q, is surrounded by a thick
concentric metal shell (inner radius a, outer radius b, as in the figure). The metal shell carries no net
charge.
a) Find the surface charge densities at each surface, σR at R, σa at a, and σb at b.
b) Find the potential at the center, using infinity as the reference point.
Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as
at infinity).
c) How do your answers to a) and b) change?
d) Find the potential difference V between the sphere and the shell in terms of charge Q.
e) What is the capacitance of the system with the sphere and the shell?
Sol: a) Due to the properties of the conductor, especially the fact that the electric field inside the
conductor vanishes and the charges will arrange itself to the surfaces of the conductor, we can read off
the charge densities.
σR =
Q
,
4πR2
σa = −
Q
,
4πR2
σb =
Q
.
4πR2
b) Potential inside the conductor is constant and that of a charge located at the origin with charge Q
1 Q
is given by 4π
. Thus we can get the potential as
0 r
φ=













Potential
1 Q
φb<r = 4π
+ c4 ,
0 r
φa<r<b = c3 ,
1 Q
+ c2 ,
φR<r<a = 4π
0 r
φr<R = c1 ,
Constant
c4 = 0
1 Q
c3 = 4π
0 b
1
c2 = 4π0 ( Qb − Q
a)
Q
Q
1
c1 = 4π0 ( b − a + Q
R)
B.C.
φ(r = ∞) = 0
φb<r (r = b) = φa<r<b (r = b)
φa<r<b (r = a) = φR<r<a (r = a)
φR<r<a (r = R) = φr<R (r = R)
Q
1
Thus the potential at the center is φr=0 = 4π
( Qb − Q
a + R ).
0
If we connect the outer surface to a grounding wire, we drain the charge of that surface.
c) The charge σb = 0, while the charge densities at r = R, a are not modified. The potential at the
center will be modified in such a way that the potential φr=b = 0 at r = b and φb<r = 0. Thus we get
Q
1
φr=0 = 4π
(− Q
a + R ).
0
d) The potential difference V is
1
1
1
V = φr<R (r = R) − φR<r<a (r = a) =
−
Q.
4π0 R a
e) This the capacitance is given by
1
1
1
Q
V =
−
Q=
,
4π0 R a
C
−→
C = 4π0
aR
.
a−R
3
Problem 3
This problem is an extension of the problem 3 of the Homework set 6.
During the lecture, we learned that the potential, in the presence of Azimuthal symmetry (potential
is independent of the angle φ), can be expressed as
φ(r, θ) =
∞ X
Al r l +
l=0
Bl
rl+1
Pl (cos θ) ,
in the spherical polar coordinate, where Pl represent Legendre Polynomial.
As before, a charge density σ0 (θ) = k cos θ with constant k is glued over the surface of a spherical shell
of radius R. Until now the setup is the same as before.
Now we add a conducting sphere of radius a < R, at potential V0 , inside the concentric spherical shell
of radius R.
a) Using the same technique, compute the potential in each region: (i) r > R, (ii) a < r < R, and
r < a.
b) Find the induced surface charge σa (θ) on the conductor.
c) Compute the total charge of this system.
d) Check that your answer is consistent with the behavior of potential at large r.
Sol: a) Due to the Azimuthal symmetry, we have the general form of the potential. For the three
different regions, we can have the following forms.

P∞
l
φR<r = l=0 rB

l+1 Pl (cos θ) ,
P∞
l
l
φ=
Pl (cos θ) ,
φ
= l=0 Cl r + rD
l+1
 a<r<R
φr<a = V0 ,
where we use the fact that potential at infinity vanishes and that inside the conducting sphere is
constant.
We need to determine Bl , Cl , Dl and V0 . To do this, we use the following boundary conditions:
• (i) φ is continuous at R. This gives
V0 =
∞ X
Cl al +
l=0
Dl
al+1
Pl (cos θ) −→
l
Cl al + aD
l+1 = 0 ,
−→
Dl
Cl + a = V0 ,
Dl = −a2l+1 Cl
Dl = aV0 − aC0
l 6= 0 ,
l=0,
Thus there are two independent conditions.
• (ii) φ is continuous at a. This gives
∞ ∞
X
X
Dl
Bl
l
Cl R + l+1 Pl (cos θ) =
Pl (cos θ) −→ Bl = Cl R2l+1 + Dl .
R
Rl+1
l=0
l=0
• Combining these two conditions together we arrive
Bl = Cl (R2l+1 − a2l+1 ) ,
l 6= 0 ,
B0 = (R − a)C0 + aV0 ,
l=0.
• (iii) The difference of radial derivative at r = R is given by the charge density [∂r φR<r − ∂r φa<r<R ]r=R =
−σ0 (θ)/0 . This gives
−
∞
X
l=0
(l + 1)
∞ X
Bl
Dl
k
l−1
P
(cos
θ)
−
lC
R
−
(l
+
1)
Pl (cos θ) = − P1 (cos θ) .
l
l
Rl+2
Rl+2
0
l=0
Thus we have another two conditions.
(l + 1)(Bl − Dl ) = −lR2l+1 Cl ,
2
k
,
C1 + 3 (B1 − D1 ) =
R
0
l 6= 1 ,
l=1.
• Combining all together we get
Bl = Cl = Dl = 0 ,
3
3
R −a
,
30
B0 = D0 = aV0 ,
B1 = k
l>1,
1
,
30
C0 = 0 ,
C1 = k
a3
,
30
l=0.
D1 = −k
l=1
Thus we have the following potentials.

3
3
φR<r = aVr 0 + k R3−a

2 cos θ ,

0r
3
φ=
φa<r<R = aVr 0 + k 310 1 − ar2 cos θ ,


φr<a = V0 ,
b) The charge density induced on the surface of the inner conducting sphere can be computed as
∂φ
0 V0
σa (θ) = −0
= −k cos θ +
.
∂r r=a
a
c) Total charge of the system is the integral of the surface charges induced.
Z
0 V 0
4πa2 = 4πa0 V0 ,
Qa = σa (θ)dS =
a
which is a total charge.
d) At very large r, we can ignore the 1/r2 contribution in φR<r . Thus
aV0
R3 − a3
aV0
+k
cos θ ≈
,
r
30 r2
r
1
1
aV0
=
Q=
4πa0 V0 =
.
4π0
4π0
r
φR<r =
φR<r
Thus we can check that they agree with each other.
4
Problem 4 : A model of metal surface (Bonus problem, optional)
A surprisingly realistic microscopic model for the charge density of a semi-infinite metal (with z = 0
as its macroscopic surface) consists of a positive charge distribution
n̄
z<0
n+ (z) =
0
z>0
and a negative charge distribution
n− (z) =
n̄ 1 − 12 eκz
1 −κz
2e
z<0
z>0
a) Sketch n+ (z) and n(z) on the same graph and give a physical explanation of why they might be
reasonable description of a metal surface.
b) Calculate the dipole moment p~ per unit area of surface for this system.
c) Calculate the electrostatic potential φ(z) and sketch it in the interval −∞ < z < ∞.
d) Calculate and explain the value of φ(∞) − φ(−∞).
e) Calculate the total electrostatic energy per unit area of this system.
Sol: a) The form of n+ (z) “smears out” the charge due to the positive nuclei but recognizes that
there is a well-defined “last layer” of nuclei at z = 0. The form of n− (x) models the fact that the
electron wave functions “spill out” into the vacuum beyond the last row of nuclei. See the figure.
b) There is only a z−component to the dipole moment p~ by symmetry. So, the dipole moment per
unit area is
Z 0
Z ∞
Z ∞
n̄
n̄
pz =
dzz(n+ − n− ) =
dzeκz +
dae−κz = − 2 .
2
κ
−∞
−∞
0
c) The total charge density is ρ(z) = n+ (z) − n− (z) = −sgn(z) 21 n̄e−κ|z| . It must be that E = E(z)ẑ
by symmetry. Therefore, because sgn(z) = d|z|/dz, Gauss’ law gives
dE(z)
ρ(z)
n̄ d|z| −κ|z|
=
=−
e
.
dz
0
20 dz
This can be integrated to
n̄ −κ|z|
e
.
20 κ
E(z) =
The integration constant is zero because only a charged sheet produces an electric field at infinity. The
electric field is finite at z = 0 so the potential must be continuous there. Hence, if we let φ(0) = 0,
z
Z
n̄
dz E(z ) = −
20 κ
0
φ(z) = −
0
0
Z
z
0
dz 0 e−κ|z | =
0
d) This potential gives φ(∞) − φ(−∞) = − on̄κ2 =
pz
0 .
n̄
sgn(z)(e−κ|z| − 1) .
20 κ2
The right side is the change in potential which
occurs across a double layer. Rather than a sudden jump, the change is spread out over the entire
length of the system.
e) The total energy per unit area is
U=
0
2
Z
∞
−∞
dzE 2 (z) =
n̄2
80 κ2
Z
0
−∞
dze2κz +
Z
0
∞
dae−2κz
=
n̄2
.
80 κ3
5
Problem 5 : Field lines (Bonus problem, optional)
The z-axis coincides with the symmetry axis of a flat disk of radius a in the x−y plane. The disk carries
a uniform charge per unit area σ < 0. The rim of the disk carries an additional uniform charge per
unit length λ > 0. Use a side (edge) view and sketch the electric field lines everywhere assuming that
the total charge of the disk is positive. Your sketch must have enough detail to reveal any interesting
topological features of the field line pattern.
Sol: The field lines above are drawn so that
(i) the lines far from the disk resemble those of a positive point charge;
(ii) field lines very near the disk (but away from the rim) resemble the field lines near an infinite sheet
of negative charge; and
(iii) the field lines very near the rim resemble the field lines near an infinitely long positive charge line.
When these things are done, there are inevitably points in space where the field lines must cross. This
is allowed if E = 0 at those isolated points.
Refer to C.L. Pekeris and K. Frankowski, Physical Review A 36, 5118 (1987).