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Symmetries & Idealizations Homework 8
Due 10/19/16 @ 4:00 pm
PRACTICE:
1. What do you think will be the flux through the cylindrical surface that is placed as
shown in the constant vector field in the figure on the left? What if the cylinder is
placed upright, as shown in the figure on the right? Explain.
Solution:
The flux is zero through the cylinder on the left, since the vector field is parallel to
the cylinder. The flux is also zero through the cylinder on the right, since the flux in
through the back is equal to the flux out through the front.
2. Calculate the divergence of each of the following vector fields. You may look up the
formulas for divergence in curvilinear coordinates.
~ = z 2 x̂ + x2 ŷ − y 2 ẑ
(a) F
~ = e−x x̂ + e−y ŷ + e−z ẑ
(b) G
~ = yz x̂ + zx ŷ + xy ẑ
(c) H
(d) ~I = x2 x̂ + z 2 ŷ + y 2 ẑ
~ = xy x̂ + xz ŷ + yz ẑ
(e) J
~ = s2 ŝ
(f) K
~ = r3 φ̂
(g) L
Solution:
~ =0
~ ·F
∇
~ = −e−x − e−y − e−z
~ ·G
∇
~ =0
~ ·H
∇
~ · ~I = 2x
∇
1
~ = 2y
~ ·J
∇
~ = 3s
~ ·K
∇
~ =0
~ ·L
∇
REQUIRED:
~ through each side of the
3. A charge q sits at the corner of a cube. Find the flux of E
cube. Do not do a long calculation (either by hand or by computer)!
Solution:
First of all, there is no flux through the three faces which meet at the charge, since
the electric field is parallel to those faces. By symmetry, the flux through each of the
remaining faces must be the same.
Furthermore, the total flux through the cube gives the enclosed charge. But the charge
sits in a corner, where we could have 8 such cubes meet. The charge must be evenly
distributed among these cubes, so only an eighth is “inside” the original cube.
The flux through each of the three opposite faces must therefore be
Q
1 Q
=
3 80
240
4. Use the cross product to find the components of the unit vector n̂ perpendicular to the
plane shown in the figure below, i.e. the plane joining the points {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.
Solution:
Take the cross product of any two vectors in the plane. For example, the vector −ı̂ + ̂
that points from (1, 0, 0) to (0, 1, 0) and the vector −ı̂ + k̂ that points from (1, 0, 0)
to (0, 0, 1). The result is perpendicular to the plane. Make sure that the right hand
rule gives the vector out of the plane in the direction that you want. Normalize the
resulting vector to obtain:
1 n̂ = √ ı̂ + ̂ + k̂
3
2
5. Consider the vector field F~ = (x + 2)x̂ + (z + 2)ẑ.
(a) Calculate the divergence of F~ .
Solution:
∂
∂
∂
Fx +
Fy + Fz
∂x
∂y
∂z
∂
∂
∂
=
(x + 2) +
(0) + (z + 2)
∂x
∂y
∂z
= 2
~ · F~ =
∇
(b) In which direction does the vector field F~ point on the plane z = x? What is the
value of F~ · n̂ on this plane where n̂ is the unit normal to the plane?
Solution:
On the plane z = x, the x and z components of F~ are equal, therefore F~ points
in the ı̂ + k̂ direction, i.e. parallel to the plane. Therefore we see, without any
further calculation, that F~ · n̂, the flux of F~ through this plane is zero.
(c) Verify the divergence theorem for this vector field where the volume involved is
drawn below.
Solution:
The divergence theorem states that the flux of any (sufficiently smooth) vector
field through any closed surface is the same as the integral of the divergence of
that vector field over the interior of the surface.
I
F~ · n̂ dA =
Z
~ · F~ dτ
∇
Calculating the left-hand side: We showed in part (b) that there is no flux through
the angled plane. There is also no flux through the triangular end pieces because F~
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has no ̂ component (perpendicular to those planes). The (outward) flux through
the square boundary in the x, y-plane is:
Z 1Z 1
[(x + 2)x̂ + 2ẑ] · (−ẑ) dx dy = −2
0
0
The flux through the remaining vertical side is (at x = 1):
Z 1Z 1
[3x̂ + (z + 2)ẑ] · (x̂) dy dz = 3
0
0
Therefore, the left-hand side of the divergence theorem is −2 + 3 = 1.
Calculating the right-hand side, we have a constant divergence of 2, integrated
over the prism shape which is half of the unit cube. Therefore, we get just the
divergence, 2, times the volume of the prism, 1/2. So the right-hand side of the
divergence theorem also is 1, and we have verified the divergence theorem for this
special case.
6. You can use Gauss’s law to find the gravitational field due to a spherical shell of matter
(or equivalently, the electric field due to a spherical shell of charge):

0
M
r−
−G
~g =
b3 −a3

r̂
−G M
r2
a3
r2
r̂
for r < a
for a < r < b
(1)
for r > b
(If you want extra practice with Gauss’s law, this is a good question to practice on!)
b
a
O
Q
This problem explores the consequences of the divergence theorem for this shell.
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(a) Using the given value of the gravitational field, find the divergence of the gravitational field everywhere in space. You will need to divide this question up into
three parts: r < a, a < r < b, and r > b.
Solution:
The divergence in spherical coordinates is:
~ · ~g = 1 ∂ (r2 gr ) + 1 ∂ (sin θgθ ) + 1 ∂ (gφ )
∇
r2 ∂r
r sin θ ∂θ
r sin θ ∂φ
Since our vector field is radial, (i.e. has only an r̂-component), only the first of
these terms is relevant. Notice that you have to multiply the r-component of the
vector field by r2 before differentiating. Therefore:

0
1 ∂ 2  −G M
~
∇ · ~g = 2 (r )
b3 −a3

r ∂r
−G M
r2
(
0
=
−G b33M
= const
−a3
0
r−
a3
r2
for r < a
for a < r < b
for r > b
for r < a
for a < r < b
for r > b
(b) Discuss the physical meaning of the divergence in this particular example.
Solution:
Just as the divergence of the electric field at any point is proportional to the charge
density at that point, the divergence of the gravitational field is proportional to
the mass density at that point.
~ · ~g = −4πGρ
∇
Therefore, we are not surprised that the divergence we found in part (a) is zero
both inside and outside the shell and constant in the shell.
(c) For this gravitational field, verify the divergence theorem on a sphere, concentric
with the shell, with radius Q, where a < Q < b.
Solution:
The divergence theorem states that the flux of any (sufficiently smooth) vector
field through any closed surface is the same as the integral of the divergence of
that vector field over the interior of the surface.
I
Z
~ · ~g dτ
~g · n̂ dA = ∇
Calculating the left-hand side: The gravitational field is constant over the surface
of the sphere of radius Q and points radially outward, in the same direction as the
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unit normal. Therefore, the flux reduces to the the magnitude of the gravitational
field times the surface area of the sphere.
I
M
a3
Q3 − a3
2
~g · n̂ dA = −G 3
Q
−
4πQ
=
−4πGM
b − a3
Q2
b 3 − a3
Calculating the right-hand side, we see from part (a) that the divergence is conTherefore we must
stant with value − b3GM
3 −a3 for a < r < b and 0 elsewhere.
integrate a constant over the volume of the spherical shell that lies inside Q.
Z
4π
Q 3 − a3
3GM
3
3
~ · ~g dτ = −
(Q
−
a
)
=
−4πGM
∇
b 3 − a3 3
b 3 − a3
We have verified the divergence theorem for the special case of a spherical shell
of matter with uniform mass density.
(d) Discuss how this example would change if you were discussing the electric field of
a uniformly charged spherical shell.
Solution:
The mathematics of static electric fields is exactly the same as the mathematics
of static gravitational fields. Therefore the only changes for a uniform spherical
shell of charge are:
~
~g → E
1
4π0
ρ mass density → ρ charge density
−G →
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