Download Solution to HW Problems

–1–
Solution to HW Problems
25. Griffiths 2.15. Given the spherical symmetry of the configuration, we can see that the Gaussian
surface is concentric sphere, on which the electric field is radial and its magnitude is a constant. Here
we take such coordinates that the center of the sphere is at the origin, i.e., ~r′ = ~0, so ~ι = ~r − ~r′ = ~r.
~ = ~0.
(i) In the region r < a, a sperical Gaussian surface encloses zero charge, so E
(ii) In the region a < r < b, a concentric Gaussian sphere of radius r encloses some fraction of the
total charge, which is
Z
Z r Z π Z 2π
k 2
Qenc = ρdτ =
r sinθdrdθdφ = 4πk(r − a)
2
r=a θ=0 φ=0 r
H
~ = E(r)4πr 2 . Using Gauss’s law,
~ · da
Meanwhile the total flux through the spherical surface is E
H
~ = Qenc /ǫ0 , we get
~ · da
E
k(r − a)
E(r) =
ǫ0 r 2
(iii) In the region r > b, the total charge enclosed in a sperical Gaussian surface is
Z
Z b Z π Z 2π
k 2
′
r sinθdrdθdφ = 4πk(b − a)
Qenc == ρdτ =
2
r=a θ=0 φ=0 r
H
~ = E(r)4πr 2 . Using Gauss’s law,
~ · da
And the total flux through the spherical surface is E
H
~ = Qenc /ǫ0 , we get
~ · da
E
k(b − a)
E(r) =
ǫ0 r 2
~ = E(r)r̂. The sketch of E(r)
In all three regions, the vector electric field points radially, so E
is given in Figure 1. Note that the profile between a and b may vary depending on the values of a
and b.
26. Griffiths 2.18. In this problem, we can directly use the conclusion of the class example. The
electric field generated by one sphere of charge ρ centered at r~1 ′ inside the sphere is given by
ρ~
ι1
ρ
ρι1
ιˆ1 =
=
(~r − r~1 ′ )
E~1 =
3ǫ0
3ǫ0
3ǫ0
The electric field generated by the other sphere of charge −ρ centered at r~2 ′ inside the sphere is
given by
−ρι2
−ρ~
ι2
−ρ
E~2 =
ιˆ2 =
=
(~r − r~2 ′ )
3ǫ0
3ǫ0
3ǫ0
So the net electric field generated by the two charge spheres in the overlapping region (i.e., inside
both spheres) is the superposition of E~1 and E~2 , which is
~ = E~1 + E~2 = ρ (r~2 ′ − r~1 ′ ) = ρ d~
E
3ǫ0
3ǫ0
It is seen that the net electric field in this region is constant with the same magnitude and direction
~
along d.
–2–
Fig. 1.— Sketch of E(r) as a function of r.
–3–
27. Griffiths 2.16. The charge distribution in this problem has a cylindrical symmetry, so the
Gaussian surface is a concentric cylindrical shell with radius s. To make a closed surface, we also
choose the height of the cylinder as H, and the cylindrical surface and the top and bottom surfaces
of the cylinder make the closed surface on which we use Gauss’s law. On the cylindrical surface, the
~ = E(s)ŝ//n̂, and at the same distance
electric field is parallel to the surface normal anywhere E
from the central axis of the cylinder, E(s) is constant. At the top and bottom surfaces, the electric
field is perpendicular to the surface normal n̂ = ẑ, so the electric flux through the top and bottom
surfaces are zero. Therefore, the left-hand side of the Gauss’s law, the electric flux through the
~ only on the cylindrical surface, which is
surface, becomes the surface integral of E
Z
Z
Z
~
~
E · da = E(s)ŝ · ŝda = E(s) da = E(s)A = E(s)2πsH
at any distance s.
R
(i) When s < a, the charge enclosed is given by Qenc = ρdτ . Since ρ is a constant, it can be taken
R
out of the integral, then Qenc = ρ dτ = ρV , where V is the volume of the cylinder with radius s,
H
~ = Qenc /ǫ0 , we get E(s)2πsH = ρπs2 H/ǫ0 , leading to
~ · da
and V = πs2 H. Using Gauss’s law E
E(s) =
ρs
2ǫ0
(ii) When a < s < b, the charge enclosed in the Gaussian surface is the total volume charge in the
solid cylinder of radius s and height H, which is Qenc = ρV = ρπa2 H. So using Gauss’s law, we
get
ρa2
E(s) =
2ǫ0 s
(iii) When s > b, the Gaussian surface will enclose the charge in the inner cylinder as well as the
charge on the outer cylindrical shell. Since these two parts of charges cancel each other to keep the
cable neutral, so Qenc = 0, thus E(s) = 0.
The sketch of E(s) as a function of s is given in Figure 2. Note that the vector of the electric field
~ = E(s)ŝ.
is E
28. Griffiths Problem 2.17. From the symmetric geometry, we know that the electric field in this
configuration must be in the direction of ŷ, and at the same distance |y| to the xz plane, the
magnitude of the field is a constant, while the fields at −y and at y point to opposite directions.
~ = |E(y)|ŷ, and at −y, E
~ = −|E(y)|ŷ. Therefore, we may choose a Gaussian pillbox with
At y, E
two surfaces parallel to the xz plane and the area being A, and its length along the y-axis from −y
to y. The electric flux is non-zero only through the two surfaces parallel to the xz plane at ±y, so
H
~ = 2|E(y)|A.
~ · da
E
R
Inside the slab |y| < d, the total charge enclosed in the Gaussian pillbox is Qenc = ρdτ = ρV =
ρ2|y|A. Using Gauss’s law, we get 2|E(y)|A = ρ2|y|A/ǫ0 , |E(y)| = ρ|y|/ǫ0 . Taking into account
the direction, E(y) = ρy/ǫ0 .
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Outside the slab |y| > d, the total charge enclosed is Qenc = ρ2dA, so |E(y)| = ρd/ǫ0 or E(y) =
ρd y
ǫ0 |y| .
~ = E(y)ŷ.
The sketch of E(y) is given in Figure 3. Note that the electric field is a vector E
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Fig. 2.— Sketch of E(s) as a function of s.
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Fig. 3.— Sketch of E(y) as a function of y.
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~ ×E
~ = ~0. For the following
29. Griffiths Problem 2.20. The electrostatic field is curl free, or ∇
example fields, we shall compute the curl of the field first.
~ ×E
~ = (0 − 2kz)x̂ + (0 − 3kz)ŷ + (0 − kx)hatz 6= 0. So this field cannot be an electrostatic
(a) ∇
field.
~ ×E
~ = (2kz − 2kz)x̂ + (0 − 0)ŷ + (2ky − 2ky)ẑ = ~0. So this field is an electrostatic field. The
(b) ∇
R ~r
R
~ = − ~~r (Ex dx + Ey dy + Ez dz). Here the reference
~ · dl
electric potential is given by V (~r) = − O~ E
O
~ = (0, 0, 0). It is noted that this potential is path independent. However, to
is at the origin O
determine it, we need to design a path for the integral. We choose a simple path in three parts as
such: from (0, 0, 0) to (x, 0, 0), then to (x, y, 0) and then to (x, y, z), as shown in Figure 4. Along
such a path, we find
Z z
Z y
Z z
Z y
Z x
2kyzdz = −k(xy 2 +yz 2 )
2kxydy−
Ez (x, y, z)dz = 0−
Ey (x, y, 0)dy−
Ex (x, 0, 0)dx−
V (~r) = −
0
0
0
0
~ gives the electric field as given.
Now we can check whether -∇V
~ (~r) = −
−∇V
the same as the given field.
∂V
∂V
∂V
x̂ −
ŷ −
ẑ = ky 2 x̂ + k(2xy + z 2 )ŷ + 2kyzẑ
∂x
∂y
∂z
0
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z
(x, y, z)
(0, 0, 0)
(x, 0, 0)
x
Fig. 4.— Sketch of the integral path for Prob. 29
y
(x, y, 0)