Download Physics 417G : Solutions for Problem set 2

Physics 417G : Solutions for Problem set 2
Due : January 29, 2016
Let us refresh our memory!
Here are some problems with electric dipole moment.
Please show all the details of your computations including intermediate steps.
1
Problem 1
Three point charges are located at a distance d from the origin, −q at (0, −d, 0), −q at (0, d, 0), and q
at (0, 0, d).
a) Draw the rough picture of the charge configuration. Write down the volume charge distribution
ρ(~r0 ) in terms of delta functions.
b) We have the following multipole expansion expressions
r̂i r̂j Qij
Q r̂i pi
1
+ ··· ,
+ 2 +
φ(~r) =
4π0 r
r
r3
Z
Z
Z
1
3 0
0
3 0 0
0
Q = d x ρ(~r ) , p~ = d x ~r ρ(~r ) , Qij =
d3 x0 [3ri0 rj0 − δij r02 ]ρ(~r0 ) ,
2
for the three lowest orders. Now compute the potential upto the two lowest orders at the location |~r|
which is much bigger than d, |~r| d. Express your answer in the spherical polar coordinate.
c) Compute the electric field for the answer given in b).
d) Using |~
p| = qd which is a dipole moment, write down the electric field for the dipole moment p~ in
the coordinate free from.
Sol: a) The charge density is
ρ(~r0 ) = −qδ(x)δ(y + d)δ(z) − qδ(x)δ(y − d)δ(z) + qδ(x)δ(y)δ(z − d) .
The charges are depicted in the figure.
b) Multipole expansion of the potential is given by
Z
Z
3 0
0
Q = d x ρ(~r ) = −q ,
p~ = d3 x0~r0 ρ(~r0 ) = −qd(x̂ − x̂) + qdẑ = qdẑ ,
Thus
−q qd cos θ
1
+
+ ··· ,
φ(~r) =
4π0 r
r2
where we use the spherical polar coordinate
x̂ = r̂ sin θ cos φ + θ̂ cos θ cos φ − φ̂ sin φ ,
ŷ = r̂ sin θ sin φ + θ̂ cos θ sin φ + φ̂ cos φ ,
~ we get
c) By taking a derivative E = −∇φ,
q
r̂
d E(r, θ, φ) =
− 2 + 3 2 cos θr̂ + sin θθ̂ + · · · .
4π0
r
r
d) The electric field for the dipole moment p~ in coordinate free from is given by
~ dipole =
E
1 3(~
p · r̂)r̂ − p~
.
4π0
r3
ẑ = r̂ cos θ − θ̂ sin θ .
2
Problem 2
Here we consider a dipole and its interaction with the electric field, torque, force and energy.
a) In a uniform electric field, a charge neutral object does not have a net force. Yet, it can have a net
torque. Derive the expression of a torque for the dipole depicted in the left figure with respect to the
point O.
b) In a non-uniform electric field, F~+ does not exactly cancel F~− . Thus there will be a net force on
the dipole in addition to the torque. Derive the expression of the net force by summing over the force
and by evaluating the difference of the electric field of the two charge +q and −q in the same figure.
~
c) Derive the energy of ideal dipole p~ in an electric field E.
d) Derived the electrostatic energy between the two dipole p~1 and p~2 by using an electric field due to
one of the two dipoles.
e) Two coplanar dipoles are oriented as shown in the right figure. Find the equilibrium value of the
~ = ~r × F~ . Thus we have
angle θ0 for the dipole p0 if the angle θ is fixed. Sol: a) A torque is given by N
~ × (−q E)
~ net = ~r+ × F~+ + ~r− × F~− = 1 d~ × (q E)
~ + (−d)
~ = q d~ × E
~ = p~ × E
~ .
N
2
b) The net force can be evaluate as
~+ − E
~ − ) = q∆E
~ = q(d~ · ∇)
~ E
~ = (~
~ E
~ ,
F~ = F~+ + F~− = q(E
p · ∇)
~ = (d~ · ∇)
~ E
~ along the d~ direction.
where we choose ∆E
c) The energy can be obtained by integrating the force when the dipole is not a function of position.
~ E
~ = ∇(~
~ p · E),
~ and thus
Then we get F~ = (~
p · ∇)
Z
Z
~
~
~ p · E)
~ = −(~
~ ,
p · E)
U = − F · dl = − d~l · ∇(~
where we use the potential vanishes at infinity.
The same result can be obtained from potential φ(~r) as
~ − qφ(~r) = q(φ(~r + d)
~ − φ(~r)) = −q
U = qφ(~r + d)
Z
~
r +d~
~ = −(~
~ ,
d~l · E
p · E)
~
r
where the integral is done for an ideal dipole.
d) Using the coordinate independent expression for an electric field from a dipole, we get
~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.
e) 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
=
−
.
4π0
r3
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 .
3
Problem 3
In this problem we are going to show that the average field inside a sphere of radius R, due to all the
charge within the sphere, is
~ average = − 1 p~ .
E
4π0 R3
(1)
where p~ is the total dipole moment. There are several ways to prove this delightfully simple result.
Here we take the following approach.
a) Show that the average field due to a single charge q at point ~x inside the sphere is the same as the
field at ~x due to a uniformly charged sphere with ρ = −q/(4πR3 /3), namely
Z
1
q(~x − ~x0 )
~ average = − 1
.
E
d3 x0
3
4π0 4πR /3
|~x − ~x0 |3
where ~x − ~x0 is the vector from ~x, point of source charge, to the integration volume d3 x0 .
b) The right side of the expression (1) can be found from Gauss’s law for the uniformly charged solid
sphere, which can be written in terms of the dipole moment of q.
c) Use the superposition principle to generalize the result of b) to an arbitrary charge distribution.
d) Show that the average field over the sphere due to all the charges outside is the same as the field
they produce at the center.
0
~ = 1 q(~x0 −~x3) . Thus if we
Sol: a) The electric field due to a point charge q at ~r to the point ~x0 is E
4π0 |~
x −~
x|
average the field over a volume, we get the expression
Z
Z
1
1
1
x − ~x0 )
3 ~
3 0 q(~
~
Eaverage =
d
x
E
=
−
d
x
.
4πR3 /3
4π0 4πR3 /3
|~x − ~x0 |3
R (~x−~x0 )
~ρ = 1
ρ |~x−~x0 |3 . This time ~x
The electric field at ~r due to uniform charge ρ over the sphere is E
4π0
0
is the field point and ~x is the source point of the charge density. This gives the same effect of changing
~ average = E
~ ρ.
the sign of the charge. Thus if we put ρ = −q/(4πR3 /3), we get the same expression, E
b) Let us consider the Gauss surface inside a uniform charge sphere with charge density ρ. Then we
get the electric field
~ ρ = 1 ρ~r = − q ~r = − p~
E
.
30
4π0 R3
4π0 R3
~ average is the sum of the individual averages. And
c) If there are many charges inside the sphere, E
p~total is the sum of the individual dipole moments. Thus this gives
~ average = −
E
p~
.
4π0 R3
d) For the charge Q outside the sphere, the same argument follows. Thus we compute the average
~ average inside the sphere due to Q as the electric field due to the charge density ρ =
electric field E
−Q/(4πR3 /3).
3
~ average = E
~ ρ = 4πR /3 ∗ ρ r̂ = − Q ~r .
E
4π0
r2
4π0 r3
This is precisely the field produced by Q (at ~r) at the center of the sphere. So the average field (over
the sphere) due to a point charge outside the sphere is the same as the field that same charge produces
at the center. And by superposition, this holds for any collection of exterior charges.
4
Problem 4
a) The word “dipole” comes from a geometrical construction due to Maxwell. This can be thought
~ Now let s → 0. If
as two equal and opposite charges (“poles”) q/s at opposite ends of a vector sd.
~
p~ = q d, the electrostatic potential of the resulting “point dipole” located at ~r0 is
"
#
q/s
1
1
1
q/s
~
=−
p~ · ∇
.
φ(~r) = lim
−
~
s→0 4π0 |~
|~r − ~r0 |
4π0
|~r − ~r0 |
r − ~r0 − sd|
Show that the result is exact in the limit s → 0.
~ 2 φ(~r), obtain the expression of ρdipole in
b) Using the result of a) and Poisson equation ρdipole = −0 ∇
terms of a derivative of delta function.
c) The electric field of a point electric dipole must be given by the result of problem 1 d) when ~r 6= ~r0 .
What happens precisely at ~r = ~r0 ? Using the results of problem 3, argue that
3(~
p · n̂)n̂ − p~ 4π
~r − ~r0
1
~
−
p~δ(~r − ~r0 ) ,
n̂ =
.
(2)
E(~r) =
4π0
|~r − ~r0 |3
3
|~r − ~r0 |
Sol a) The computation goes as
#
"
q/s
1
q/s
−
φ(~r) = lim
~
s→0 4π0 |~
|~r − ~r0 |
r − ~r0 − sd|
1
q/s
1 ~ ~ 2 q/s
q/s
q/s
~
~
= lim
− sd · ∇
+ (sd · ∇)
+ ··· −
s→0 4π0 |~
r − ~r0 |
|~r − ~r0 | 2
|~r − ~r0 |
|~r − ~r0 |
1
1
~
=−
p~ · ∇
.
4π0
|~r − ~r0 |
The result is exact because all the higher order terms of the Taylor expansion vanishes in the limit
s → 0.
b) Let us take the Laplacian to the expression, then
~2 1
~ 2 φ(~r) = 1 p~ · ∇
~ ∇
~ r − ~r0 ) .
ρdipole = −0 ∇
= −~
p · ∇δ(~
4π
|~r − ~r0 |
~ 2 1 = −4πδ(~r − ~r0 ). As far as we know, point electric dipoles with this singular
Where we use ∇
|~
r −~
r0 |
charge density do not exist in nature. Nevertheless, it is a useful tool for computation when the size
of a charge distribution is very small compared to all other characteristic lengths in an electrostatic
problem.
c) Given that the point dipole was constructed from two delta function point charges, it is reasonable
to guess that a delta function lies at the heart of the electric field of the dipole as well. To check this,
it is simplest to use the previous problem and a spherical integration volume V centered at ~r0 . In that
case,
Z
~ r) = − p~ .
d3 xE(~
(3)
30
V
On the other hand, it is clear from the symmetry of the problem that the first term in (2) integrates
to zero over the volume of any origin-centered sphere. Therefore, if this behavior persists as V → 0,
(3) will be true if the total electric field of a point dipole is given by (2).