Download Report

Solutions to Problem Set 4, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 55 points.
1. In English, explain what is the point behind Griffith’s visualization of
the relationships between various electrostatic quantities in his Figure
2.35, reproduced below? What is he attempting to illustrate?
(5 points possible) Figure 2.35 is a stab at illustrating the relationship between three different physical quantities, the charge density ρ,
~ and the electric potential V . There are six formulas
the electric field E
that we derived that inter-relate these three physical quantities, which
are shown on the two arrows connecting each quantity to each other
quantity. That’s all that Griffiths is trying to illustrate, that given any
one of these three physical quantities, it is (in principle) possible to
derive the other two.
Solutions to Problem Set 4, Physics 370, Spring 2014
2
2. Griffiths Problem 2.22 (Tweaked): Find the potential a distance s
from an infinitely long straight wire that carries a uniform line charge
density λ. Compute the gradient of your potential, and check that it
~ field. HINT: You will need to choose a reference
yields the correct E
point a fixed distance s0 from the wire for the potential. Be sure to
explain why you need to do this. Also, for the sake of brevity, you may
use the solution for the electric field of an infinitely long straight wire
we worked out in lecture by citing it.
(10 points possible) We start with an infinitely long straight wire
with linear charge density λ. Because the charge distribution is infinite,
we can’t set the reference point at infinity (you could cite the undefined
integral at infinity as a reason for doing this). So instead, we choose a
reference point a fixed distance s0 from the wire. When we did problem 2.13 in class, we worked out that the electric field of an infinitely
~ = λ sˆ. Given the symmetry, it is convenient
long straight write is E
2πǫ0 s
to choose a path from s0 to s straight away from the wire such that
d~l = ds′ sˆ (where I am using s′ instead of s to avoid confusion between
the coordinate s and the limit s). In that case, via the definition of the
electric potential (eqn. 2.21):
Z s
Z
λ
λ
s0
1
λ
~
~
ds = −
[ln(s) − ln(s0 )] =
ln .
V = − E · dl = −
2πǫ0 s0 s
2πǫ0
2πǫ0
s
(1)
We can check this by taking the gradient. Using the equations listed
on the inside front cover of Griffiths, we find in cylindrical coordinates
is
ˆ
~ = −∇V
~ = − ∂V sˆ − ∂V zˆ − 1 ∂V φ.
E
(2)
∂s
∂z
s ∂φ
Here the potential only depends on s so
λ
λ s −s0
∂
s0
λ
~
sˆ = −
E=−
ln
sˆ =
sˆ
2
∂s 2πǫ0
s
2πǫ0 s0 s
2πǫ0 s
(3)
like we expected. (Thanks to Dr. Craig for providing a pre-LATEXed
draft of this solution.)
Solutions to Problem Set 4, Physics 370, Spring 2014
3
3. Griffiths Problem 2.25 (Tweaked): In class, we worked out Grif~ on the
fiths Problem 2.6, which asked us to find the electric field E
z-axis above a uniformly charged disk with surface charge density σ
and radius R (as illustrated in Figure 2.34c below) via direct integration of the contributions to the electric field of all the charges in the
disk.
After a bit of work, we found the solution to be
σ
z
~ =
E
1− √
zˆ.
2ǫ0
z 2 + R2
(a) Using Equation 2.30
1
V (~r) =
4πǫ0
Z
σ(r~′ )
da′ ,
find the electric potential V at a distance z above the center of
that same charge distribution.
~ = −∇V
~
(b) Compute the electric field based on this potential using E
and confirm it matches the result we worked out.
(c) Let’s say we made the left half of the disk (where x < 0) have surface charge density −σ. This would imply that for every positive
charge on the right half of the disk (where x > 0) there would be
a corresponding negative charge on the left half (where x < 0).
What would be the electric potential on the z-axis in this case?
~ = −∇V
~ . There is a probCompute the electric field using E
~ field, what is it? HINT: If you
lem with this solution for the E
are doing extensive calculations for this last problem, you are not
achieving E&M elightenment.
Tackling the problem as suggested:
Solutions to Problem Set 4, Physics 370, Spring 2014
4
(a) (4 points possible) As suggested, we start by using equation
2.30
Z
1
σ(r~′ ) ′
da ,
(4)
V (~r) =
4πǫ0
where in this case, for a circular disk, it will be easiest to use a
polar coordinate system where we integrate in radius r ′ from the
center of the disk outward and θ′ around the disk’s axis such that
the area element becomes
da′ = (dr ′ )(r ′dθ′ ) = r ′ dr ′dθ′
(5)
P
r
z
σ
r’
θ
da’
Furthermore, considering the diagram shown here of the relationship between r ′ , z, and ths separation vector magnitude r, we can
see that
√
= r ′2 + z 2 .
(6)
Combining this information with the fact that the surface charge
density σ is constant, equation 4 becomes
Z R Z 2π
σ
r ′ dr ′
√
V (0, 0, z) =
dθ′
(7a)
4πǫ0 r′ =0 θ′ =0 r ′2 + z 2
Z R
r ′ dr ′
σ
√
=
(7b)
2ǫ0 r′ =0 r ′2 + z 2
To solve this integral, we can use u-substitution where u = r ′2 +z 2
Solutions to Problem Set 4, Physics 370, Spring 2014
such that du = 2r ′ dr ′ and therefore the integral becomes:
Z 2 2
σ 1 R +z −1/2
u
du
V (0, 0, z) =
2ǫ0 2 u=z 2
σ 1 1/2 R2 +z 2
2u
=
u=z 2
2ǫ0 2
h
i
√
σ
V (0, 0, z) =
R2 + z 2 − z
2ǫ0
5
(8a)
(8b)
(8c)
where equation 8c is the correct expression for the electric potential on the z-axis.
(b) (3 points possible) Applying equation (2.23)
~ = −∇V
~
E
∂V
∂V
∂V
=−
xˆ −
yˆ −
zˆ
∂x
∂y
∂z
(9a)
(9b)
we can find the electric field in this case, noting that by symmetry,
= ∂V
= 0 on the z-axis, therefore
we know ∂V
∂x
∂y
~ 0, z) = − ∂V zˆ
E(0,
∂z
i
σ ∂ h√ 2
2
R + z − z zˆ
=−
2ǫ0 ∂z
σ 1
1
√
=−
2z − 1 zˆ
2ǫ0 2 R2 + z 2
z
σ
√
− 1 zˆ
=−
2ǫ0
R2 + z 2
σ
z
~
E(0, 0, z) =
1− √
zˆ
2ǫ0
z 2 + R2
(10a)
(10b)
(10c)
(10d)
(10e)
which is the same equation we got via direct integration of the
charge distribution in Griffiths Problem 2.6 in class.
(c) (3 points possible) If one half of the disk is positively charged
and the other side is negatively charged, we end up with an electric
potential of zero on the z-axis, since the potential contributed by
any positive charge is precisely balanced by the corresponding
negative charge equidistant from the point P on the z-axis. This
Solutions to Problem Set 4, Physics 370, Spring 2014
6
can be confirmed computationally by re-working our solution to
part (a)
Z R Z π
Z R Z 2π
r ′ dr ′
r ′ dr ′
σ
−σ
′
√
√
V (0, 0, z) =
dθ +
dθ′
4πǫ0 r′ =0 θ′ =0 r ′2 + z 2
4πǫ0 r′ =0 θ′ =π r ′2 + z 2
(11a)
Z R
Z
R
σ
r ′ dr ′
r ′ dr ′
σ
√
√
=
−
(11b)
4ǫ0 r′ =0 r ′2 + z 2 4ǫ0 r′ =0 r ′2 + z 2
= 0.
(11c)
Using this solution to the electric potential, we can compute the
electric field using equation (2.23) and get
~ = −∇V
~ = 0!
E
(12)
Of course, this makes no sense, why would the electric field be zero
directly between two regions of completely opposite charge? In
fact, the field should be pointing horizontally, in the direction away
from the positively charged side of the disk, toward the negatively
charged side of the disk, that is in the −ˆ
x direction. We can see
this by considering the electric field due the charge on the right
half of the disk, Eright , on some point on the z-axis versus the
electric field due to a corresponding point (with opposite charge)
on the left half of the disk, Elef t , as illustrated in the figure below.
z axis
t
gh
E ri
E
lef
t
Etotal
−σ
+σ
So why did this approach fail? It failed because we only know
the potential on the z-axis and nowhere else, we have no hope of
computing Ex = − ∂V
or Ey = − ∂V
given V only on the z axis.
∂x
∂y
Solutions to Problem Set 4, Physics 370, Spring 2014
7
In part (b) we didn’t have a problem because symmetry implied
= ∂V
= 0”, but this is no longer the case. Put another
“ ∂V
∂x
∂y
way (Griffith’s way) part (b) relied on exploiting the symmetry in
~ points
the problem, however, now with that symmetry broken “E
in the x direction, so knowing V on the z axis is insufficient to
~
determine E.”
4. Griffiths Problem 2.34 (tweaked): In lecture, we worked through
Griffiths Problem 2.21 where we determined the electric potential for
a uniformly charged solid sphere of radius R and charge q to be:
(
3q
− 8πǫq0 R3 r 2 , r ≤ R
8πǫ0 R
V =
q
,
r > R.
4πǫ0 r
Using this, find the energy stored in a uniformly charged solid sphere
of radius R and charge q. Do it two different ways:
(a) Use equation 2.43 from the textbook
Z
1
ρV dτ.
W =
2
(b) Use equation 2.45 from the textbook
Z
ǫ0
W =
E 2 dτ.
2 allspace
Don’t forget to integrate over all space.
We are to find the total energy stored in a sphere of radius R with total
charge q uniformly distributed throughout the sphere.
(a) (5 points possible) The first way we will find the energy is from
the integral
Z
1
ρV dτ.
(13)
W =
2
As noted in the question, we determined the electric potential in
this case was
(
3q
− q 3 r2, r ≤ R
V = 8πǫq 0 R 8πǫ0 R
(14)
,
r
>
R,
4πǫ0 r
Solutions to Problem Set 4, Physics 370, Spring 2014
3q
For a uniform charge density I know ρ = Vq = 4πR
3.
( 2
ρR
− 6ǫρ0 r 2 , r ≤ R
2ǫ0
V = ρR
3
,
r > R,
3ǫ0 r
8
(15)
and the charge density, ρ is
ρ(r) =
(
ρ, r ≤ R
0, r > R.
(16)
[NOTE: You could solve this problem without using charge den3q
sity at all, since charge density within the sphere is ρ = 4πR
3 , you
could make that substitution in equation 13 and solve the integral
in terms of total charge. This might even be easier.] The integral
(13) is zero for r > R because the charge density is zero there.
The volume element in spherical coordinates is dτ = 4πr 2dr,1 so
inside the sphere the integral is
2
Z
ρ 2
ρR
1 R
(17a)
−
r (4πr 2)dr
ρ
W =
2 0
2ǫ0
6ǫ0
Z
πρ2 R 2 2 r 4
=
R r − dr
(17b)
ǫ0 0
3
R
r5
πρ2 R2 r 3
(17c)
−
=
ǫ0
3
15 0
πρ2 R5 4
=
(17d)
3ǫ0 5
4πρ2 R5
(17e)
=
15ǫ0
3q 2
=
.
(17f)
20πǫ0 R
1
You could set this up with a full volume integral with dτ = r2 sin θdθdφdr, but since
there is no angular dependence, when you integrate over the θ and φ components, you
would just get 4π. I therefore skipped those components and just wrote dτ = 4πr2 dr.
You could also take the approach that since you know only the radial component changes,
2
2
since the total volume τ = 43 πr3 and dτ
dr = 4πr , therefore dτ = 4πr dr. Either approach
gets you to the proper volume element.
Solutions to Problem Set 4, Physics 370, Spring 2014
9
(b) (5 points possible) Now we find the same energy again, but in
a different way. We start with the expression
Z
ǫ0
E 2 dτ,
(18)
W =
2 all space
where E is the electric field. I can solve for the electric field E
given the potential using the relation:
~ = −∇V
~
E
(19)
Recall the potential only had a radial dependence, so the gradient
in spherical coordinates is just:

3q
q
2
− ∂
−
r
rˆ, r ≤ R
3
∂r 8πǫ0 R
~ =
8πǫ0 R
(20)
E
q
− ∂
rˆ
r > R,
∂r 4πǫ0 r
Solving this, we find the electric field is
(
q
r, r ≤ R
~ = 4πǫ0 R3 rˆ
E
q
rˆ,
r > R.
4πǫ0 r 2
(21)
The energy breaks up into two pieces because there are two separate forms for the electric field inside and outside the sphere:


Z
Z
ǫ0
E 2 dτ +
E 2 dτ  .
W = 
(22)
2
inside
The first piece is
Z
Z R
2
E dτ =
0
inside
q
r
4πǫ0 R3
2
outside
R
q2
r 5 q2
.
(4πr )dr =
=
4πǫ20 R6 5 0
20πǫ20 R
2
(23)
Outside the sphere we have
∞
2
Z
Z ∞
q2
q
q 2 1 2
2
=
E dτ =
.
(4πr
)dr
=
−
4πǫ0 r 2
4πǫ20 r R
4πǫ20 R
R
outside
(24)
Solutions to Problem Set 4, Physics 370, Spring 2014
Adding these two together gives
1
3q 2
ǫ0 q 2
+
1
=
,
W =
2 4πǫ20 R 5
20πǫ0 R
10
(25)
which agrees with the result from the previous method in (17f).
(Thanks to Dr. Craig for providing a pre-LATEXed draft of this solution.)
5. Griffiths Problem 2.38: A metal sphere of radius R, carrying charge
q, is surrounded by a thick concentric metal shell (inner radius a, outer
radius b, as in Figure 2.48, shown below). The shell carries no net
charge.
(a) Find the surface charge σ at R, at a, and at b.
(b) Find the potential at the center, using infinity as the reference
point.
(c) Now the outer surface is touched to a grounding wire, which lowers
its potential to zero (same as infinity). How do your answers to
(a) and (b) change?
(a) (3 points possible) I use the fact that since the inner sphere is a
solid conductor, all the charge will reside at ths surface, therefore
by the definition of surface charge density:
σR =
q
4πR2
(26)
For the inner surface of the outer sphere (r = a), I can use Gauss’
Law with a spherical Gaussian surface co-centric with the inner
Solutions to Problem Set 4, Physics 370, Spring 2014
11
metal sphere such that a < r < b. Since I know the electric field
within a conductor must be zero, I can state:
Z
~ · d~a = Qenc
(27a)
E
ǫ0
Qenc = 0
(27b)
Therefore, since the total enclosed charge must be zero, the inner
surface of the outer sphere must have −q total charge, therefore
the charge density is:
−q
σa =
(28)
4πa2
Finally, since the outer sphere is electrically neutral as a whole,
the total charge on the outer surface of the outer sphere must be
+q, and so:
q
(29)
σb =
4πb2
(b) (4 points possible) If I use infinity as the reference point (such
that V (∞) = 0), then I can compute the potential at the center
as:
Z 0
~ · d~l
V =−
E
(30)
∞
~ changes with location, so we need to break
The difficulty is the E
~ = 0. Outside the
up the integral. Within the conductors,
E
H
~ · d~a = Qenc ). I note
conductors we can use Gauss’ Law ( E
ǫ0
that since the charge is distributed in a spherically symmetric
pattern, a spherical Gaussian surface co-centric with the charge
distribution makes the most sense. For the volumes not within
the actual conductors (R < r < a and r > b), the enclosed charge
is just +q, therefore:
E(4πr 2 ) =
~ =
E
q
ǫ0
q
rˆ
4πǫ0 r 2
(31a)
(31b)
~ information and a radial line for the line integral (so
Using this E
Solutions to Problem Set 4, Physics 370, Spring 2014
12
d~l = drˆ
r), then equation 30 becomes:
Z b
Z a
Z R
Z 0
~
~
~
~
~
~
~ · d~l (32a)
V (0) = −
E · dl −
E · dl −
E · dl −
E
∞
b
a
R
Z a
Z 0
Z R
Z b
q
q
dr −
dr −
(0)dr −
(0)dr
=−
2
2
b
R
a 4πǫ0 r
∞ 4πǫ0 r
(32b)
Z b
Z R q
dr
dr
=
−
−
(32c)
2
2
4πǫ0
∞ r
a r
q
1
1
1
V (0) =
(32d)
+ −
4πǫ0 b R a
(c) (3 points possible) If we drain the surface charge from the outer
surface, this will only drain out the charge at outer surface. The
induced charge at the inner surface (r = a) remains. Therefore,
the enclosed charge for a Guassian surface r > a will be zero,
~ = 0 for r > a. As such, the appropriate version of the
therefore E
potential line integral shown in Equation 30 is:
Z a
Z R
Z 0
~ · d~l −
~ · d~l −
~ · d~l
V (0) = −
E
E
E
(33a)
∞
a
R
Z 0
Z a
Z R
q
dr −
(0)dr
(33b)
=−
(0)dr −
2
R
∞
a 4πǫ0 r
Z R
q
q
−
dr
(33c)
=
2
4πǫ0
a 4πǫ0 r
q
1
1
V (0) =
(33d)
−
4πǫ0 R a
6. Griffiths Problem 2.41: Two large metal plates (each of area A) are
held a distance d apart. Suppose we put a charge Q on each plate;
what is the electrostatic pressure on the plates? HINT: Read Section
2.5.3.
(10 points possible) Electrostatic pressure between two conductors
is defined in equation 2.52 as:
P =
ǫ0 2
E
2
(34)
Solutions to Problem Set 4, Physics 370, Spring 2014
13
where E here is the field just outside the surface. So clearly we need to
determine the magnitude of the electric field just
√outside the plates. If I
A (where I am assumassume the seperation between
the
plates
d
≪
√
ing squarish plates so A roughly represents the order of magnitude
length of a side of the plates), then let me take the electric field of a
single plate to be equal to the electric field due to an uniformly charged
infinite plate (given by equation 2.17 of the textbook). I’ll probably
be a little off near the edges of the plates (due to edge effects where
the electric field is not perfectly perpendicular to the plates’ surfaces),
but hey, I just want to have a good idea of the electrostatic pressure
involved:
~ = 1 σˆ
E
n
(35)
2ǫ0
Then in this case σ =
Q
A
and
~ =
E
Q
n
ˆ
2ǫ0 A
(36)
If the two plates hold the same charge, then between the plates, the
electric fields of the two plates are in the opposite directions, so the
total field between the plates is zero. Outside of the plates, the electric
fields of the two plates point in the same directions, so they easily sum.
We can summarize this as:
(
between plates,
~ = 0,
(37)
E
Q
n
ˆ
,
outside
plates.
ǫ0 A
Using the electric field outside the plates given by Equation 37, we can
now compute the electrostatic pressure using equation 34 we can find:
2
ǫ0 2 ǫ0
Q
Q2
P = E =
=
(38)
2
2 ǫ0 A
2ǫ0 A2
This is the pressure pushing the plates apart. I know this because there
is no electric field between the plates and the electric field due to the
plates is pointed away from the plates, so it is forcing the plates apart.