–1–
Solution to HW Problems
30. Griffiths 2.21. We calculate the electric potential by the path integral V (~r) = −
this problem, the electric field generated by a uniformly charged solid sphere is
~ r) =
E(~
qr
r̂
4πǫ0 R3
~ r) =
E(~
q
r̂
4πǫ0 r 2
when r < R, and
R ~r
~ In
~ · dl.
∞E
when r > R, supposing that the center of the charge ball is at the origin. The electric field only
has a radial component, so the path integral of the electric field becomes only the integral of the
radial component.
For r > R, outside the sphere, we get
V (~r) = −
Z
r
∞
q
q
dr =
2
4πǫ0 r
4πǫ0 r
For r < R, inside the sphere, the integral is conducted in two parts, from infinity to the radius
of the sphere, and from the radius of the sphere to inside the sphere.
Z r
Z R
q
1 2
q 1
q
q
r2
q
2
dr
−
rdr
=
+
(R
−
r
)
=
(3
−
)
V (~r) = −
2
3
4πǫ0 R 4πǫ0 R3 2
8πǫ0 R
R2
R 4πǫ0 R
∞ 4πǫ0 r
The potential is only a function of r, as sketched in Figure 1. At r < R, V (r) ∝ −r 2 , at r > R,
V (r) ∝ 1/r.
The negative gradient of this potential should be the given electric field. For r < R, inside the
sphere, we get
~ (~r) = − ∂V (r) r̂ = − q (− 2r )r̂ = qr r̂
−∇V
∂r
8πǫ0 R R2
4πǫ0 R3
For r > R, outside the sphere, we get
~ (~r) = −
−∇V
The same as the given electric field.
∂V (r)
q
r̂ =
r̂
∂r
4πǫ0 r 2
–2–
Fig. 1.— Sketch of V (r) as function of r for Problem 30.
–3–
31. Griffiths 2.24. From the cylindrical symmetry, it can be seen that the electric potential is only
a function of the distance to the axis V (~r) = V (s). The potential difference between a point on
R s=b
~ Since the
~ · dl.
the axis and a point on the outer shell is given by ∆V = V (b) − V (0) = − s=0 E
electric field only has the ŝ component, given by
E(s) =
ρa2
2ǫ0 s
E(s) =
ρs
2ǫ0
when a < s < b and
when s < a, the integral can be done in two parts as
−
Z
s=b
s=0
~ =−
~ · dl
E
Z
s=a
s=0
ρs
ds −
2ǫ0
Z
s=b
s=a
ρa2
ρa2 ρa2
b
ds = −
−
ln( )
2ǫ0 s
4ǫ0
2ǫ0
a
32. Griffiths 2.28 In this problem, we use the same configuration as in the class example (or book
example 2.7) by choosing a special ~r = (r, θ, φ) = (r, 0, −), i.e., θ = 0, since the spherical symmetry
of the configuration indicates that the potential at distance r is the same regardless of θ or φ.
Therefore, the distance between the field position ~r = (r, 0, 0) and a source position ~r′ = (r ′ , θ ′ , φ′ )
√
is given by ι = r 2 + r ′2 − 2rr ′ cosθ ′ . The electric potential can be computed by
Z
Z 2π Z π Z R
Z
Z R
ρ
2πρ π
1
r ′2 sinθ ′ dr ′ dθ ′
ρdτ ′
r ′2 sinθ ′ dr ′ dθ ′ dφ′
√
√
=
=
V (~r) =
4πǫ0
ι
4πǫ0 φ′ =0 θ′ =0 r′ =0 r 2 + r ′2 − 2rr ′ cosθ ′
4πǫ0 θ′ =0 r′ =0 r 2 + r ′2 − 2rr ′ cosθ ′
We first integrate over θ ′ . Take u = r 2 + r ′2 − 2rr ′ cosθ ′ , we get du = 2rr ′ sinθ ′ dθ ′ , or sinθ ′ dθ ′ =
du/(2rr ′ ). So the integral wrt θ ′ becomes
Z u2 ′
Z π
r −1/2
r′
′ =π
r′
r ′2 sinθ ′ dθ ′
√
u
du = u1/2 |θθ′ =0
= [(r + r ′ ) − |r − r ′ |]
=
r
r
r 2 + r ′2 − 2rr ′ cosθ ′
u1 2r
θ ′ =0
We then conduct integral with respect to r ′ . Inside the sphere, the integral is divided into two
parts of r ′ < r and r ′ > r:
Z
Z
Z
2πρ R r ′
2πρ r 2r ′2 ′
2πρ R
′
′
′
V (r) =
[(r + r ) − |r − r |]dr =
dr +
2r ′ dr ′
4πǫ0 r′ =0 r
4πǫ0 r′ =0 r
4πǫ0 r′ =r
=
Since ρ =
3q
,
4πR3
ρ 2 2
ρ
ρ
1
r +
(R2 − r 2 ) =
(R2 − r 2 )
2ǫ0 3
2ǫ0
2ǫ0
3
we get
V (r) =
q
(3R2 − r 2 )
8πǫ0 R3
–4–
R
33. Griffiths 2.32. (a) To find the electric energy using equation W = 21 ρV dτ , we need to know
the electric potential by a uniformly charged sphere of radius R inside the sphere, since outside the
sphere, ρ = 0 and the integral is zero. V (r) inside the sphere is given in previous assignments (see
Problem 32.) as
q
(3R2 − r 2 )
V (r) =
8πǫ0 R3
which is only a function of r. Therefore, the energy is
Z R
4πρq
3q 2
3q 2
1 5
2
2 2
5
W =
(3R
−
r
)r
dr
=
R
)
=
(R
−
16πǫ0 R3 r=0
16πǫ0 R6
5
20πǫ0 R
R
(b) To find the energy by W = ǫ20 E 2 dτ , the integral is conducted over all space, and we need to
know the electric field in all the space. From previous assignements, we find that the electric field
only has the radial component and is only a function of r, distance from the center of the sphere.
Inside the sphere (r < R),
qr
ρr
=
E(r) =
3ǫ0
4πǫ0 R3
and outside the sphere (r > R),
q
E(r) =
4πǫ0 r 2
The integral is thus divived into two parts:
Z
Z
Z
ǫ0
4πǫ0 ∞
qr
q
4πǫ0 R
2 2
2
W =
) r dr +
)2 r 2 dr
(
(
E dτ =
3
2
2
2
4πǫ
R
2
4πǫ
r
0
0
r=0
r=R
=
5
q
4πǫ0 q 2 1
3q 2
4πǫ0
2R
(
+
(
=
)
)
2 4πǫ0 R3 5
2 4πǫ0 R
20πǫ0 R
same as (a).
(c) Using the expression
Z
I
ǫ0
2
~
~ · da)
W = ( E dτ + V E
2
we may integrate over a spherical volume of radius a (a > R). Then the volume integral becomes
Z
Z
Z
ǫ0
4πǫ0 a
3q 2
q2 1
qr
q
4πǫ0 R
2 2
2 2
2
)
r
dr
+
)
r
dr
=
−
(
(
E dτ =
3
2
2
2
2
20πǫ0 R 8πǫ0 a
r=0 4πǫ0 R
r=R 4πǫ0 r
The surface integral is conducted only at the spherical surface of radius a, where V (a) =
~ = a2 sinθdθdφr̂. So the surface integral becomes
~
E(a)
= q 2 r̂. The surface da
4πǫ0 a
ǫ0
2
I
q
~ = ǫ0 q
~ · da
VE
2 4πǫ0 a 4πǫ0 a2
I
a2 sinθdθdφ =
q
4πǫ0 a ,
and
q
q2
ǫ0 q
2
a
4π
=
2 4πǫ0 a 4πǫ0 a2
8πǫ0 a
Adding the volume integral and the surface integral, we get
W =
3q 2
20πǫ0 R
same as derived in (a) and (b). The solution also shows that as a −→ ∞, the contribution from
the surface integral, which is proportional to 1/a, goes to zero.