1 Solutions to Problem Set 3, Physics 370, Spring 2014

Transcription

Solutions to Problem Set 3, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 70 points.
~
1. In English, what is an electric field E?
(5 points possible) In a nutshell, if we go by the definition of an elec~ in Griffiths, it is simply the electric force, as determined by
tric field E
Coulomb’s law, per unit charge being acted upon. It describes the effect (in terms of force per unit charge) a particular charge configuration
will have on any new charge brought into the region. Fundamentally,
what is it? Honestly, I am not sure. For now, I will accept this “operational” definition of the force per unit charge and state like Griffiths
does on page 62 that “I can’t tell you what a field is – only how to
calculate it.”
2. Recall Gauss’ Law is stated in two forms by Griffiths: The integral
form of Gauss’ law as stated in equation (2.13)
I
~ · d~a = Qenc
E
(1)
ǫ0
S
which is a variant of the form you may recall seeing in Physics II. There
is also the differential form of Gauss’ law as stated in equation (2.14)
~ ·E
~ = ρ.
∇
ǫ0
(2)
(a) Explain how these two expressions of Gauss’ law can be considered equivalent. For this problem, clearly cite any mathematical
theorems you use to make your argument.
(b) When is Gauss’ law valid? Does Gauss’ law hold even if we work
~
with a surface that is not perpendicular to the electric field E?
Why would we typically choose such surfaces if Gauss’ law works
equally well without such a restriction? Explain your reasoning.
(c) What is “Qenc ”? Of course, it is the “enclosed charge,” but what
is it enclosed by? Does the enclosing surface have to be a real
surface? What makes a good choice of surface?
2
(a) (4 points possible) Gauss’ law is basically a restatement of
Coulomb’s law, expressed in terms of electric flux instead of electric force (if this doesn’t make sense, read Section 2.2 of Griffiths). It specifically states that the “electric flux through any
surface enclosing a charge is Qenc /ǫ0 .” This is the form described
by equation (2.13), where the left
H hand side sums up the electric
~ · d~a over the entire surface. I
flux through the surface ΦE ≡ S E
can re-write this into equation (2.14) using the divergence theorem
(equation 1.56 of the textbook)
Z I
~
~
∇ · V dτ =
V~ · d~a
(3)
V
S
and use it to re-write the integral form of Gauss’ law as the differential form
I
~ · d~a = Qenc
E
(4a)
ǫ0
S
Z ~ ·E
~ dτ = Qenc
(4b)
∇
ǫ0
V
Z Z ρ
~
~
dτ
(4c)
∇ · E dτ =
ǫ0
V
V
~ ·E
~ = ρ
∴∇
(4d)
ǫ0
where in step 4c we used the definition of the charge density.
Those of you reading the textbook should realize this is basically
outlined in pages 68 and 69 of Griffiths.
(b) (3 points possible) Gauss’ Law is fundamentally just a restatement of Coulomb’s law. So whenever Coulomb’s law is valid,
Gauss’ law should also be valid. W haven’t been too specific
about this point yet, but Coulomb’s law is completely valid as the
only description of the force in all electrostatic situations, that
is where the charges are not moving. As such Gauss’ law works
with any surface enclosing any static charge or charges (or even
no charge)... regardless of the orientation of the surface to the
electric field. Of course, for Gauss’ law to be useful, we typically
choose a “Gaussian surface” that exploits the symmetry in the
3
particular charge configuration so that the electric field is perpendicular (or parallel) to the Gaussian surface because it makes the
mathematics much simpler.
(c) (3 points possible) As noted in the answer to the last part
of this problem, the Gaussian surface is an arbitrary choice we
make. So we typically try to exploit the symmetries in the charge
distribution (or rather, in the electric field we expect from such
a charge distribution) by selecting a Gaussian surface that lies
perpendicular (or parallel) to the electric field at that surface,
which makes computation of the total electric flux through that
surface easy. So the enclosed charge is sometimes up to us to
choose, especially in the case of “infinite” distributions of charge.
3. Griffiths Problem 2.7 (tweaked): Show that the electric field a
distance z from the center of a spherical surface of radius R (see Figure
2.11), which carries a uniform charge density σ is zero for the case where
z < R (inside) and Ez = 4πǫq0 z 2 for the case where z > R (outside) and
q is the total charge on the sphere. [Hint: Use the law of cosines to
write in terms of R and θ. Also, when you are working with your
integral, you may have to rework your bounds of integration, thanks
to a u-substitution you will need to apply to solve the problem. When
reworking those limits, you may end up with a square root like the one
shown below. Be sure to take the positive square root:
√
(R
−
z),
if
R
>
z
R2 + z 2 − 2Rz =
(z − R), if R < z
(10 points possible) Since the problem describes a surface charge
4
density σ, we can read the problem as essentially stating “find the elec~ of a thin spherical shell with uniform charge density σ at a
tric field E
point on the z-axis.”
The net electric field will be in the z-direction because of the symmetry
of the charge distribution. The electric field due to the small patch of
the surface indicated in Figure 2.11 is
~ =
dE
dq ˆ
.
4πǫ0 2
(5)
The magnitude of the separation vector, , is given by the law of
cosines,
2
= R2 + z 2 − 2Rz cos θ.
(6)
Therefore:
1
σda
ˆ.
2
2
4πǫ0 (R + z − 2Rz cos θ)
(7)
~ · zˆ = dE ˆ · zˆ = dE cos ψ
dEz = dE
(8)
~ =
dE
Now, thanks to symmetry, we know that any component of the electric field in the xy-pane due to some piece of charge on the sphere is
cancelled by another piece of charge on the sphere. As such, the total
~ can be found by just integrating over the z-component
electric field E
contributions tot he electric field by all the charge elements. The z~ is just
component of dE
where angle ψ between ˆ and zˆ. We can solve for cos ψ using the law
of cosines
R2 =
2
+ z 2 − 2 z cos ψ
2
+ z 2 − R2
cos ψ =
2 z
(R2 + z 2 − 2Rz cos θ) + z 2 − R2
=
2 z
z − R cos θ
.
=
(9a)
(9b)
(9c)
(9d)
Combining equations 6, 7, 8, and 9, we can write the z-component of
5
the electric field as
dEz = dE ˆ · zˆ = dE cos ψ
1
z − R cos θ
σda
=
2
2
4πǫ0 (R + z − 2Rz cos θ)
σda
z − R cos θ
=
.
2
4πǫ0 (R + z 2 − 2Rz cos θ)3/2
(10a)
(10b)
(10c)
The differential area element for a sphere is da = R2 sin θdθdφ, so the
z-component of the electric field is
Z
Z
z − R cos θ
σR2 2π π
Ez =
sin θdθdφ
(11a)
2
4πǫ0 φ=0 θ=0 (R + z 2 − 2Rz cos θ)3/2
Z
z − R cos θ
σR2 π
sin θdθ.
(11b)
=
2ǫ0 θ=0 (R2 + z 2 − 2Rz cos θ)3/2
There are several√ways to do the integral over θ. One is to make the
substitution u = R2 + z 2 − 2zR cos θ, such that
zR sin θ
dθ
R2 + z 2 − 2zR cos θ
sin θ
du
=√
dθ.
2
zR
R + z 2 − 2zR cos θ
du = √
(12a)
(12b)
Making
become
√
√ the corresponding change to the limits of integration
2
2
2
2
u = R + z − 2Rz = |z − R| for θ = 0 and u = R + z + 2Rz =
z + R for θ = π. Using this, we re-write equation 11
Z
σR2 z+R z − R cos θ du
Ez =
(13a)
2ǫ0 u=|z−R|
u2
zR
Z
σR z+R z − R cos θ
du
(13b)
=
2ǫ0 z u=|z−R|
u2
I can eliminate the R cos θ by re-writing it as R cos θ =
R2 +z 2 −u2
.
2z
The
6
integral for the electric field is
σR
Ez =
2ǫ0 z
Z
z+R
σR
=
2ǫ0 z
Z
z+R
u=|z−R|
u=|z−R|
Z z+R
R2 +z 2 −u2
2z
u2
du
(14a)
2z 2 −R2 −z 2 +u2
2z
u2
du
(14b)
z−
z 2 − R 2 + u2
du
u2
u=|z−R|
Z z+R
σR
z 2 − R2
=
+ 1 du
4ǫ0 z 2 u=|z−R| u2
=
σR
4ǫ0 z 2
(14c)
(14d)
Now doing the actual integrals
2
z+R
z − R2
σR
−
+u
Ez =
4ǫ0 z 2
u
|z−R|
2
2
2
σR
z −R
z − R2
=
−
+
+ (z + R − |z − R|)
4ǫ0 z 2
z+R
|z − R|
z 2 − R2 1
σR
−(z − R) +
+ (z + R − |z − R|)
=
4ǫ0 z 2
|z − R|
2
(15a)
(15b)
(15c)
At this point it is convenient to consider separately the two cases z > R
and z < R.
For points outside the sphere z > R and |z − R| = z − R, and Ez is
σR
[−(z − R) + (z + R) + (z + R − (z − R))]
4ǫ0 z 2
σR
=
(2R + 2R)
4ǫ0 z 2
σR2
.
=
ǫ0 z 2
Ez =
(16a)
(16b)
(16c)
The total charge of the sphere is q = 4πR2 σ so that we can write the
field as
q
Ez =
,
(17)
4πǫ0 z 2
which is just the field of a point charge at the origin!
7
For points inside the sphere z < R and |z − R| = R − z so that
σR
[−(z − R) − (z + R) + (z + R − (R − z))]
4ǫ0 z 2
σR
=
(−2z + 2z)
4ǫ0 z 2
= 0,
Ez =
(18a)
(18b)
(18c)
so the field is zero everywhere inside the spherical shell. (Thanks to
Dr. Craig for providing a pre-LATEXed version of an early draft of this
solution.)
4. Griffiths Problem 2.8 (tweaked): Use the electric field inside and
outside a uniformly charged spherical shell as determined in Problem
2.7 to find the field inside and outside a sphere of radius R, which
carries a uniform volume charge density ρ. Express your answers in
terms of the total charge of the sphere, q. What does the field outside
~ as a function of the
the sphere look similar to? Draw a graph of |E|
distance from the center. NOTE: If you were not able to do Problem
2.7, you can still tackle this problem starting with the stated solutions
for the electric field inside and outside the uniformly charged spherical
shell.
(10 points possible) The key here is to think of the sphere as being
a collection of spherical shells. From the previous problem we know
that the electric field of a spherical shell with charge dq is given in my
rephrasing of Problem 2.7 as
dE =
dq
4πǫ0 z 2
(19)
for points outside the sphere and 0 for points inside. Furthermore, since
there is spherical symmetry, the z axis is somewhat arbitrary, so I’ll
just called it the radial coordinate, r ′ . The sphere in this problem has
charge density ρ. For points outside the sphere, a distance r ′ from the
center, the electric field is
Z q
Z q
1
q
dq ′
=
,
(20)
dq ′ =
Er =
′2
′2
4πǫ0 r q′ =0
4πǫ0 r ′2
q ′ =0 4πǫ0 r
8
where q is the total charge of the sphere. Outside the sphere the
field is just that of a point charge at the origin.
Inside the sphere we need to be a bit more careful. In preparation for
doing the integral note that the charge in a spherical shell of thickness
dr ′ should be dq = ρAdr ′ where A = 4πr ′2 is the surface area of a shell,
such that dq = 4πρr ′2 dr ′ . Here the integral for the electric field goes
from r ′ = 0 to r ′ = r. The electric field inside the sphere is
Z r
Z q
ρ
ρr
ρ r ′3
dq ′
′2
′
.
(21)
=
=
r
dr
=
Er =
′2
′2
′2
ǫ0 r r′ =0
ǫ0 r 3
3ǫ0
q=0 4πǫ0 r
Note that the boundary between the solutions inside and outside the
sphere are equal at r ′ = R (the surface of the sphere) implies
q
ρR
=
2
4πǫ0 R
3ǫ0
4
q = πR3 ρ
3
3q
.
ρ=
4πR3
(22a)
(22b)
(22c)
So the complete solution is:
Er =
or
~ =
E
q
,
4πǫ0 r 2
q
r
4πǫ0 R3
if r ≥ R
if r ≤ R
q
rˆ,
4πǫ0 r 2
q
rˆ
r
4πǫ0 R3
if r ≥ R
if r ≤ R
(23)
(24)
A graph of the electric field (done using Maple) is below.
9
(Thanks to Dr. Craig for providing me with a draft pre-LATEXed solution.)
10
5. Griffiths Problem 2.10: A charge q sits at the back corner of a cube,
~ through the shaded area?
as shown in Figure 2.17. What is the flux of E
(HINT: For simplicity, treat the cube drawn in Figure 2.17 as one
octant of a cube with a charge at the center. If you start doing complicated calculus, you are on the wrong path to E&M enlightenment.)
(5 points possible) This problem is computationally very nasty. This
is because across the shaded face of the cube shown, the distance from
that surface to the surface varies, which means the electric field strength
varies, which means the electric flux per unit area varies across the surface. As such, solvingH for the electric flux through that surface by direct
~ · d~a would be very non-trivial.
integration of ΦE = E
Instead, we use the hint provided. Envision the charge at the “center”
of the surface of a larger cube consisting of eight cubes of the size of
those in in Figure 2.17, but with the charge at the center, as shown
here.
Since the charge is centered now, by symmetry it should be clear each of
the six faces of the cube should have 1/6 of the total flux. Furthermore,
11
if we view one of these six faces face-on, the charge is centered, which
means that while the flux per unit area varies across the face, it does so
symmetrically across each quadrant of the face. In other words, each
of the 24 squares (each a quadrant of a face of the cube) that make
up the closed surface get the same electric field flux as every other
one. By Gauss’ law, the total electric flux through any closed surface
H
~ · d~a = Qenc , therefore the flux through the shaded surface
is ΦE = E
H ǫ0
1
~ · d~a = 1 Qenc = 1 q . If you are concerned about
is ΦE,shaded = 24 E
24 ǫ0
24 ǫ0
the fact that this is a cube and not a sphere, note that the symmetry of
the situation means that even though the flux through a given ’square’
varies as you move across the square, all 24 squares will see the same
1
total flux, so the flux through any one square must be 24
the total flux
through the closed surface.
12
6. Griffiths Problem 2.11: Use Gauss’s law to find the electric field
inside and outside a spherical shell of radius R, which carries a uniform
surface charge density of σ. Compare your answer to Problem 2.7.
(10 points possible) Clearly the Gaussian surface here should exploit
the spherical symmetry in the problem, so let’s use a spherical surfaces
illustrated below.
Outs
ide Guassian S
urfa
ce
Inside Guassian
Su
rf
e
ac
r
For a spherical surface, the area element
d~a = (rdθ)(r sin θdφ)ˆ
r = r 2 sinθdθdφˆ
r
(25)
But we don’t even need to concern ourselves with this, since there is
~ is radial and has the same magnispherical symmetry, we know the E
H
~ · d~a = E · A = E(4πr 2 ).
tude over a co-centric spherical surface, so E
For the Inside Gaussian surface,
I
~ · d~a = E(4πr 2 ) = Qenc = 0
E
ǫ0
(26)
~ inside = 0.
therefore E
For the Outside Gaussian surface,
I
2
~ · d~a = E(4πr 2 ) = Qenc = 4πR σ
E
ǫ0
ǫ0
therefore
σR2
~
rˆ.
Eoutside =
ǫ0 r 2
(27)
(28)
These answers are the same as what we arrived at through much more
painful means in Problem 2.7 as shown in equations 18a and 17.
13
7. Griffiths Problem 2.12: Use Gauss’s law to find the electric field
inside a uniformly charged sphere (charge density ρ). Compare your
answer to Problem 2.8.
(10 points possible) Just as Problem 2.8 relied on the results of Problem 2.7, we can use our experience from Problem 2.11 and use the same
Guassian surfaces as in that problem to solve the problem. Its just that
in this case, we will have to consider a charge volume density ρ for radius r < R and from symmetry we again know that for a co-centric
spherical surface the E field is radial and of equal magnitude everywhere
on that surface.
For the Inside Gaussian surface (r < R), the enclosed charge is 43 πr 3 ρ,
thus
I
~ · d~a = E(4πr 2 ) = Qenc = 1 4 πr 3 ρ
(29)
E
ǫ0
ǫ0 3
therefore
~ inside = 1 ρrˆ
r.
(30)
E
3ǫ0
. You can also express this in terms of the total charge on the sphere
since Qtot = 34 πR3 ρ, thus
~ inside =
E
or
Qtot
rˆ
r
4πǫ0 R3
(31)
Qtot
~r
(32)
4πǫ0 R3
This is the same as equation 23 for r < R in our solution of Problem
2.8.
~ inside =
E
14
8. Griffiths Problem 2.18: Two spheres, each of radius R and carrying
uniform charge densities +ρ and −ρ, respectively, are placed so that
they partially overlap (Fig. 2.28). Call the vector from the positive cen~ Show that the field in the region of overlap
ter to the negative center d.
is constant, and find its value. [Hint: Use the answer to Problem 2.12.]
(10 points possible) From Problem 2.12, the field inside the positive
sphere is
~ + = ρ ~r+
(33)
E
3ǫ0
where r+ is the vector from center of the positive charge distribution
to the point in question (as shown in the figure below). The negative
sphere should have a field
~ − = − ρ ~r−
E
3ǫ0
(34)
where r− is the vector from center of the negative charge distribution
to the point in question (as shown below).
rr+
d
+
-rr+
r+
– r-
=d
15
This means the total field should be the sum of equations 33 and 34:
~ =E
~+ + E
~−
E
ρ
=
[~r+ −~r− ] .
3ǫ0
(35a)
(35b)
And by definition the dipole vector, d~ = ~r+ −~r− (see lower right hand
corner of diagram above), thus:
~
~ = ρ d.
E
3ǫ0
(36)
Since d~ is just the seperation between the charges and is constant, the
~ field must be constant in the overlap region (which is inside both
E
~ fields).
spheres, such that equation 32 holds for both sphere’s E

1 Solutions to Problem Set 3, Physics 370, Spring 2014

Transcription

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