Problem Set 1 Solution
Phys 182 - Fall 2010
Assigned: Wednesday, Sept. 1 Due: Friday, Sept. 10
1
Griffiths 1.13
r̃ = (x − x0 )x̂ + (y − y 0 )ŷ + (z − z 0 )ẑ; r =
p
(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 .
∂ 2
∂ 2
∂ 2
(r )x̂ +
(r )ŷ +
(r )ẑ
∂x
∂y
∂z
= 2(x − x0 )x̂ + 2(y − y 0 )ŷ + 2(z − z 0 )ẑ
a) ∇(r2 ) =
= 2r
1
∂ 1
∂ 1
∂ 1
b) ∇( ) =
( )x̂ +
( )ŷ +
( )ẑ
r
∂x r
∂y r
∂z r
(x − x0 )
(y − y 0 )
(z − z 0 )
=−
x̂
−
ŷ
−
ẑ
r3
r3
r3
1
= − 2 r̂
r
c)
2
∂ n
∂r
rx
(r ) = nrn−1
= nrn−1
= nrn−1 rˆx
∂x
∂x
r
so ∇(rn ) = nrn−1 r̂
Griffiths 1.16
We have r =
p
x2 + y 2 + z 2 .
∂ x
∂ y
∂ z
( )+
( )+
( )
∂x r3
∂y r3
∂z r3
3
3 ∂r
∂r
∂r
= 3 − 4 (x
+y
+z )
r
r
∂x
∂y
∂z
3
3 x2
y2
z2
= 3 − 4( +
+ )
r
r r
r
r
3
3
= 3− 3
r
r
=0
∇·v =
So ∇ · v = 0 everywhere except at the origin (r=0), where it is infinite.
1
3
Griffiths 1.35
a) Using ∇ × (f A) = f (∇ × A) − A × (∇f ):
Z
Z
Z
f (∇ × A) · da =
∇ × (f A) · da + [A × (∇f )] · da
S
S
IS
Z
=
f A · dl + [A × (∇f )] · da
P
S
b) Using ∇ · (A × B) = B · (∇ × A) − A · (∇ × B):
Z
Z
Z
B · (∇ × A)dτ =
∇ · (A × B)dτ +
A · (∇ × B)dτ
V
IV
Z V
= (A × B) · da +
A · (∇ × B)dτ
S
4
V
Griffiths 1.38
1
1 ∂
a) ∇ · v1 = 2 (r2 r2 ) = 2 4r3 = 4r
r ∂r
r
Z
Z
Z
(∇ · v1 )dτ = (4r)(r2 sinθdrdθdφ) = 4
R
r3 dr
Z
0
Z
Z
v1 · da =
(r2 r̂) · (r2 sinθdrdθdφr̂) = r4
π
Z
sinθdθ
0
Z
π
dφ = 4πR4
0
Z
2π
sinθdθ
0
2π
dφ = 4πR4
0
Z
1
1 ∂
b) ∇ · v2 = 2 (r2 2 ) = 0 , so (∇ · v2 )dτ = 0
r ∂r
r
Z
Z
Z
1
2
v2 · da = ( 2 r̂) · (r sinθdrdθdφr̂) = sinθdθdφ = 4π
r
R
The divergence is zero everywhere except at the origin, where it blows up, so our calculation of (∇ · v2 )dτ
is incorrect.
5
Griffiths 1.48
R
First Method: use Eq. 1.99 to write J = e−r (4πδ 3 (r))dτ = 4πe−0 = 4π.
Second Method: integrating by parts (use Eq. 1.59).
Z
I
r̂
r̂
∂
−r
J =−
·
∇(e
)dτ
+
e−r 2 · da . But∇(e−r ) = ( e−r )r̂ = −e−r r̂
2
r
r
∂r
S
Z V
Z
1 −r
r̂
=
e 4πr2 dr + e−r 2 · r2 sinθdrdθdφr̂
r2
r
Z R
Z
= 4π
e−r dr + e−R sinθdθdφ , with R → ∞.
0
= 4π
2
δ 3 (~r)
6
Z
δ 3 (~r)f (~r)d3~r = f (0)
Z
= δ(r)f (r)dr
Z
δ(r)
= 4π
f (r)dr
4π
Z
Z
δ(r)
= sinθdθdφ
f (r)r2 dr
4πr2
Z
δ(r)
=
f (r, θ, φ)r2 sinθdrdθdφ
4πr2
Z
δ(r)
f (~r)d3~r
=
4πr2
δ(r)
δ 3 (~r) =
4πr2
7
Levi-Civita Symbol
~ × ~r)i = εijk ∂j rk , but ∂j rk = 0 for j 6= k
a) (∇
~ × ~r)i = 0 for all i
so (∇
~ × ~r = 0
and ∇
~ × (∇
~ × A))
~ i = εijk ∂j εklm ∂l Am
b) (∇
= εkij εklm ∂j ∂l Am
= (δil δjm − δim δjl )∂j ∂l Am
= ∂j ∂i Aj − ∂j ∂j Ai , and interchanging order of the partials
= ∂i ∂j Aj − ∂j ∂j Ai
~ i (∇
~ · A)
~ − ∇2 A
~i
=∇
~ × (∇
~ × A)
~ = ∇(
~ ∇
~ · A)
~ − ∇2 A
~
∇
8
Griffiths 2.5
R
1
( λdl
The horizontal components cancel, leaving E = 4π
2 cosθ)ẑ.
0 R d
Here d2 = r2 + z 2 and cosθ = dz are constants, while dl = 2πr. So:
E=
1
λ(2πr)z
ẑ
4π0 (r2 + z 2 )3/2
3
9
Griffiths 2.6
Break the disk into rings of radius r and thickness dr. Total charge of a ring is σ · 2πr · dr = λ · 2πr, so
λ = σdr is the ”line charge” of each ring. Now we can use our results from Griffiths 2.5 for the field of each
ring.
1 σdr(2πr)z
ẑ
4π0 (r2 + z 2 )3/2
Z
r
1
2πσzẑ R
σz 1
]ẑ
=
dr =
[ −√ 2
R + z2
4π0 0 (r2 + z 2 )3/2
20 z
Ering =
Edisk
σ
20 ẑ.
1 Q
4π0 z 2 with
For R z the second term → 0, so Eplane =
For z R,
10
√
1
R2 +z 2
≈ z1 (1 −
2
R
2z 2 ),
so E =
Q = πR2 σ.
Griffiths 2.9
a) ρ = 0 ∇ · E = 0
1 ∂ 2
1
(r · kr3 ) = 0 2 k(5r4 ) = 50 kr2 .
r2 ∂r
r
I
b) By Gauss’s Law: Qenc = 0
By direct integration: Qenc
E · da = 0 (kR3 )(4πR2 ) = 4π0 kR5 .
Z
Z R
Z
= ρdτ =
(50 kr2 )(4πr2 dr) = 20π0 k
0
11
R
r4 dr = 4π0 kR5 .
0
Griffiths 2.10
Think of this cube as 1 of 8 surrounding the charge. Each of the 24 squares which make up the surface of
this larger cube gets the same flux as every other one. so:
Z
Z
1
E · da
E · da =
24
whole large cube
one face
Z
q
q
By Gauss’s Law, the latter is
. Therefore
E · da =
0
240
one face
4