Problem Set 2 Solution
Phys 182 - Fall 2010
Assigned: Friday, Sept. 10 Due: Friday, Sept. 17
1
Griffiths 2.15
(i) Qenc = 0 , so E = 0
(ii)
I
E · da =
E · (4πr2 ) =
=
=
=
Ẽ =
Qenc
ǫ0
Z
1
ρdτ
ǫ0
Z
k 2
1
r sin θdrdθdφ
ǫ0
r2
Z
4πk r
dr
ǫ0 a
4πk
(r − a)
ǫ0
k r−a
r̂
ǫ0
r2
Z
4πk b
(iii) E · (4πr ) =
dr
ǫ0 a
4πk
(b − a)
=
ǫ0
k b−a
Ẽ =
r̂
ǫ0
r2
2
1
2
I
Griffiths 2.18
Qenc
ǫ0
1 4 3
2
E · (4πr ) =
πr ρ
ǫ0 3
ρ
r̃
Ẽ =
3ǫ0
E · da =
ρ
(r̃+ − r̃− )
3ǫ0
ρ
=
d̃
3ǫ0
So Ẽtotal =
3
Griffiths 2.20
(1) ∇ × E1 = k x̂
∂
∂x
xy
(2) ∇ × E2 = k x̂
∂
∂x
y2
ŷ
∂
∂y
2yz
ẑ
∂
∂z
3zx
= k[x̂(0 − 2y) + ŷ(0 − 3z) + ẑ(0 − x)] 6= 0
So E1 is an impossible electrostatic field.
ŷ
∂
∂y
2xy + z 2
ẑ
∂
∂z
2yz
So E2 is a possible electrostatic field.
= k[x̂(2z − 2z) + ŷ(0 − 0) + ẑ(2y − 2y)] = 0
Going along the indicated path:R
Part I: y = z = 0, dy = dz = 0, I E · dl = ky 2 dx = 0.
Ry
R
Part II: x = x0 , z = 0, dx = dz = 0, II E · dl = 2kx0 0 0 ydy = kx0 y02 .
Rz
R
Part III: x = x0 , y = y0 , dx = dy = 0, III E · dl = 2ky0 0 0 zdz = ky0 z02 .
V (x0 , y0 , z0 ) = −
Z
0
(x0 ,y0 ,z0 )
E · dl
= −k(x0 y02 + y0 z02 )
V (x, y, z) = −k(xy 2 + yz 2 )
2
4
Griffiths 2.21
1 q
r̂
4πǫ0 r2
V (r) = −
E · dl
1 qr

∞

r < R: E =
r̂
4πǫ0 R3
r
Z r
1 q
q
1
So for r > R : V (r) = −
dr
=
2
4πǫ
r
4πǫ
r ∞
0
0
∞
q 1
=
4πǫ0 r
Z r
Z R
1 qr
1 q
dr
−
dr
and for r < R : V (r) = −
4πǫ0 r2
4πǫ0 R3
R
∞
q
1
1 r 2 − R2
=
− 3
4πǫ0 R R
2
2
q 1
r
=
3− 2
4πǫ0 2R
R
q 1
q ∂ 1
r̂ = −
r̂
When r > R , ∇V =
4πǫ0 ∂r r
4πǫ0 r2
q 1
so, E = −∇V =
r̂
4πǫ0 r2
r2
q 1
q r
−2r
q 1 ∂
3 − 2 r̂ =
r̂ = −
r̂
When r < R , ∇V =
4πǫ0 2R ∂r
R
4πǫ0 2R R2
4πǫ0 R3
q r
so, E = −∇V =
r̂
4πǫ0 R3
Z
5
r



r > R: E =
Griffiths 2.23
V (0) = −
Z
0
∞
E · dl
Z a Z 0
k (b − a)
k (r − a)
y
0dr
dr −
dr −
=−
ǫ0 r 2
ǫ0 r 2
a
∞
b 1 1
k
a
k (b − a)
+a
ln
−
−
=
ǫ0
b
ǫ0
b
a b
k
b
=
ln
ǫ0
a
Z b
6
Griffiths 2.25
a) V =
2q
1
q
4πǫ0 z 2 +
d
2
3
1
4πǫ0
Z
L
b) V =
1
4πǫ0
Z
R
c) V =
h
iL
p
λ
λdx
√
=
ln x + z 2 + x2
4πǫ0
−L
z 2 + x2
−L
#
"
√
2
2
L+ z +L
λ
√
ln
=
4πǫ0
−L + z 2 + L2
0
R
σ p 2
2πσ p 2
σ2πrdr
√
r + z 2 0 =
R + z2 − z
=
4πǫ0
2ǫ0
r2 + z 2
In each case the symmetries give
a) Ẽ = −
∂V
∂V
∂V
=
= 0, so Ẽ = −
ẑ.
∂x
∂y
∂z
2z
1
2qz
1
1
2q −
ẑ
23 ẑ = 4πǫ 3
4πǫ0
2 2
2
0
d
d 2 2
2
z + 2
z + 2
z
1
1
1
√
√
√
−
ẑ
4πǫ0 L + z 2 + L2
−L + z 2 + L2
z 2 + L2
1
z
−2L
√
=−
ẑ
4πǫ0
z2
z 2 + L2
2Lλ
1
√
ẑ
=
2
4πǫ0 z z + L2
b) Ẽ = −
σ
z
√
− 1 ẑ
2ǫ0
R2 + z 2
σ
z
ẑ
=
1− √
2ǫ0
R2 + z 2
c) Ẽ = −
If the right-hand charge in (a) is −q, then V = 0, which suggests that Ẽ = −∇V = 0. The point is that
∂V
∂V
we only know V on the z axis, and from this we cannot compute Ex = −
or Ey = −
. In part (a)
∂x
∂y
we knew from symmetry that Ex = Ey = 0, but this would no longer be the case. Now Ẽ points in the x
direction, so knowing V on the z axis is insufficient to determine Ẽ.
7
Griffiths 2.32
a) W =
1
2
R
ρV dτ . From Griffiths 2.21: V =
Z R
ρ 1 q
r2
3 − 2 4πr2 dr
2 4πǫ0 2R 0
R
3
5 R
r
qρ
1 r
3 − 2
=
4ǫ0 R
3
R 5 0
3
qρ
R
=
R3 −
4ǫ0 R
5
2
1
3q
qρ 2
R =
=
5ǫ0
4πǫ0 5 R
ρ
2ǫ0
R2 −
W =
4
r2
3
=
q
1
4πǫ0 2R
3−
r2
R2
b) W =
ǫ0
2
R
1 q
4πǫ0 r 2 r̂;
E 2 dτ . For r > R, Ẽ =
For r < R, Ẽ =
1 qr
4πǫ0 R3 r̂.
"Z
#
Z R 2
∞
r
1
ǫ0 q 2
(4πr2 dr) +
(4πr2 dr)
W =
2 (4πǫ0 )2 R r4
R3
0
∞
R !
1 q2
1
1 r5
=
−
+ 6
4πǫ0 2
r R
R
5 0
2
1 q
1 3 q2
1
1
=
=
+
4πǫ0 2 R 5R
4πǫ0 5 R
R
H
c) W = ǫ20 S V E · da + V E 2 dτ , where V is large enough to enclose all the charge but otherwise arbitrary.
1 q
.
Use a sphere of radius a > R, V = 4πǫ
0 r
ǫ0
W =
2
"Z
r=a
1 q
4πǫ0 r
1 q
4πǫ0 r2
r2 sin θdθdφ +
Z
R
0
a 1
4πq 2 1
4πq 2 1
4πq 2
−
+
+
(4πǫ0 )2 a (4πǫ0 )2 5R (4πǫ0 )2
r R
2
2
1 3q
1 q
1
1
1
1
=
=
+
− +
4πǫ0 2 a 5R a R
4πǫ0 5 R
=
ǫ0
2
1 qr
4πǫ0 R3
2
(4πr2 dr) +
Z
a
R
1 q
4πǫ0 r2
2
#
(4πr2 dr)
As a → ∞, the contribution from the surface integral goes to zero, while the volume integral picks up the
slack.
8
Griffiths 2.35
a) σR =
q
−q
q
; σa =
; σb =
2
2
4πR
4πa
4πb2
b) V (0) = −
=
Z
0
∞
E · dl = −
b
∞
1 q
q
q
+ −
4πǫ0 b R a
c) σb → 0 ; V (0) = −
=
9
Z
Z
1 q
4πǫ0 r2
a
∞
(0)dr −
Z
a
1 q
q
−
4πǫ0 R a
R
dr −
Z
1 q
4πǫ0 r2
b
a
(0)dr −
dr −
Griffiths 2.36
a) σa =
−qb
qa + qb
−qa
; σb =
; σR =
4πa2
4πb2
4πR2
5
Z
Z
a
R
0
R
(0)dr
1 q
4πǫ0 r2
dr −
Z
0
R
(0)dr
b) Eout =
c) Ea =
1 qa + qb
r̂ , where r = vector from center of large sphere.
4πǫ0 r2
1 qb
1 qa
rˆa , Eb =
rˆb
2
4πǫ0 ra
4πǫ0 rb2
d) Zero.
e) σR changes (but not σa or σb ); Eoutside changes (but not Ea or Eb ); force on qa and qb still zero.
10
Griffiths 2.39
Say the charge on the inner cylinder be Q, for a length L. By Gauss’ Law:
Z
Q
1
E · da = E · 2πs · L = Qenc =
ǫ0
ǫ0
Q 1
E=
ŝ
2πǫ0 L s
Potential difference between the cylinders is:
V (b) − V (a) = −
Z
a
b
Q
E · dl = −
2πǫ0 L
Z
a
b
Q
1
ds = −
ln
s
2πǫ0 L
b
a
As set up here, a is at the higher potential, so V = V (a) − V (b) =
C=
11
Q
2πǫ0 L
ln
2πǫ0 L
2πǫ0
Q
C
, so capacitance per unit length is:
=
=
V
L
ln ab
ln ab
b
a
.
Griffiths 2.40
a) W = (force) × (distance) = (pressure) × (area) × (distance) =
b) W = (energy per unit volume) × (decrease in volume) =
ǫ0 2
E Aǫ
2
E2
(Aǫ)
ǫ0
2
Both a) and b) are equal, confirming that the energy lost is equal to the work done.
6