Problem Set 3 Solution
Phys 182 - Fall 2010
Assigned: Friday, Sept. 17 Due: Friday, Sept. 24
1
Griffiths 3.1
√
The argument is exactly the same as in Griffiths section 3.1.4, except that since z < R, z 2 + R2 − 2zR =
q
1 q
1
(R − z) instead of (z − R). Hence Vavg = 4πǫ
[(z + R) − (R − z)] = 4πǫ
. If there is more than one
0 2zR
0 R
1 Qenc
charge inside the sphere, the average potential due to interior charges is 4πǫ0 R , and the average due to
Qenc
the exterior charges is Vcenter , so Vavg = Vcenter + 4πǫ
.
0R
2
Griffiths 3.2
A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV .
But we know that Laplace’s equation allows no local minima for V. What looks like a minimum in the figure
must in fact be a saddle point, and the box ’leaks’ through the center of each face.
3
Griffiths 3.6
Place image charges +2q at z = −d and −q at z = −3d. Total force on +q is:
1
q2
1 1
q
2q
−q
1
−2q
29q 2
ẑ = −
ẑ =
− + −
ẑ
F=
+
+
4πǫ0 (2d)2
(4d)2
(6d)2
4πǫ0 d2
2 8 36
4πǫ0 72d2
4
Griffiths 3.7
(a) From Fig. 3.13: ı =
√
√
r2 + a2 − 2ra cos θ; ı′ = r2 + b2 − 2rb cos θ.Therefore:
R2
q
R
q′
√
,
while
b
=
=
−
ı′
a r2 + b2 − 2rb cos θ
a
q
q
= − q
= −q
4
R2
a
ar 2
r2 + R
+ R2 − 2ra cos θ
R
a2 − 2r a cos θ
R
Clearly, when r = R, V → 0.
1
(b) σ = −ǫ0 ∂V
∂n . In this case,
∂V
∂n
=
∂V
∂r
at the point r = R. Therefore:
2
1 2
ra 2
a
−1 2
−ǫ0 q
2
−3/2
−3/2
2r − 2a cos θ
(r + a − 2ra cos θ)
(2r − 2a cos θ) + (R + ( ) − 2ra cos θ)
σ(θ) =
4πǫ0
2
2
R
R2
r=R
2
a
q
−(R2 + a2 − 2Ra cos θ)−3/2 (R − a cos θ) + (R2 + a2 − 2Ra cos θ)−3/2
=−
− a cos θ
4π
R
a2
q
(R2 + a2 − 2Ra cos θ)−3/2 R − a cos θ −
+ a cos θ
=
4π
R
q
2
2
2
2
−3/2
=
(R − a )(R + a − 2Ra cos θ)
Z4πR
Z
q
Induced = σda =
(R2 − a2 ) (R2 + a2 − 2Ra cos θ)−3/2 R2 sin θdθdφ
4πR
π
−1 2
q
(R2 − a2 )2πR2
(R + a2 − 2Ra cos θ)−1/2
=
4πR
Ra
0
1
1
q 2
2
(a − R ) √
−√
=
2a
R2 + a2 + 2Ra
R2 + a2 − 2Ra
p
But a > R (else q would be inside), so R2 + a2 − 2Ra = a − R
q 2
q
1
q
1
2
=
=
(a − R )
−
[(a − R) − (a + R)] =
(−2R)
2a
a+R a−R
2a
2a
qR
= q′
=−
a
(c) The force on q due to the sphere is the same as the force of the image charge q ′ :
R 2
qq ′
1
1
q 2 Ra
1
1
−
=
=
−
q
F =
4πǫ0 (a − b)2
4πǫ0
a
(a − R2 /a)2
4πǫ0 (a2 − R2 )2
To bring q in from infinity to a, we do work:
a
Z
q2 R a
1
1
q2 R
q2 R
1
a′
′
W =
−
da
=
=
−
4πǫ0 ∞ (a′2 − R2 )2
4πǫ0
2 (a′2 − R2 ) ∞
4πǫ0 2(a2 − R2 )
5
Griffiths 3.9
(a) Image problem: λ above, −λ below. Taking an arbitrary point of reference at s = a.
λ
The potential of λ is V+ = − 2πǫ
ln
0
λ
The potential of −λ is V− = 2πǫ0 ln
s+ a ,
s− a ,
where s+ is distance from λ.
where s− is distance from −λ.
s−
λ
ln
Total V = 2πǫ
s+ .
p 0
p
2
2 , and s =
y 2 + (z + d)2 , so
and s+ = y + (z − d)
−
2
2
+(z+d)
λ
ln yy2 +(z−d)
V = 4πǫ
2
0
2
∂V
∂V
∂V
. Here
=
, evaluated at z = 0.
∂n
∂n
∂z
2(z − d)
2(z + d)
λ
− 2
σ(y) = −ǫ0
4πǫ0 y 2 + (z + d)2
y + (z − d)2 z=0
2λ
λd
d
−d
=−
=−
− 2
4π y 2 + d2
y + d2
π(y 2 + d2 )
(b) σ = −ǫ0
6
Griffiths 3.10
Image charges: −q at (a, −b) and (−a, b); q at (−a, −b).
"
1
q
1
p
V (x, y) =
+p
2
2
2
2
4πǫ0
(x − a) + (y − b) + z
(x + a) + (y + b)2 + z 2
#
1
1
−p
−p
(x + a)2 + (y − b)2 + z 2
(x − a)2 + (y + b)2 + z 2
2
1
1
1
q
−
x̂ −
ŷ + √
[cos θx̂ + sin θŷ]
F=
4πǫ0
(2a)2
(2b)2
(2 a2 + b2 )2
p
p
where cos θ = a/ a2 + b2 , sin θ = b/ a2 + b2
b
a
q2
1
1
=
− 2 x̂ +
− 2 ŷ
16πǫ0
a
b
(a2 + b2 )3/2
(a2 + b2 )3/2
1 1
−q 2
−q 2
q2
W =
2
+2
+2 √
4 4πǫ0 (2a)
(2b)
(2 a2 + b2 )
q2
1 1
1
√
=
− −
16πǫ0
a b
a2 + b 2
For this to work, θ must be an integer divisor of 180◦ . Thus 180◦ , 90◦ , 60◦ , 45◦ , etc work, but no others.
7
Griffiths 3.15
The boundary conditions are:

(i)
V = 0 when x = 0,





(ii) V = 0 when x = a,



(iii) V = 0 when y = 0,

(iv)





(v)



(vi)
V = 0 when y = a,
V = 0 when z = 0,
V = V0 when z = a.
We want sinusoidal functions in x and y, exponential in z:
X(x) = A sin (kx) + B cos (kx) , Y (y) = C sin (ly) + D cos (ly) , Z(z) = Ee
√
k2 +l2 z
+ Ge−
√
k2 +l2 z
(i) ⇒ B = 0; (ii) ⇒ k = nπ/a; (iii) ⇒ D = 0; (iv) ⇒ l = mπ/a; (v) ⇒ E + G = 0. Therefore
p
Z(z) = 2E sinh (π n2 + m2 z/a)
3
Putting this all together, and combining the constants, we have:
V (x, y, z) =
∞ X
∞
X
Cn,m sin (nπx/a) sin (mπy/a) sinh (π
n=1 m=1
p
n2 + m2 z/a)
Evaluate Cn,m by imposing boundary condition (vi):
i
p
XXh
Cn,m sinh (π n2 + m2 ) sin (nπx/a) sin (mπy/a)
V0 =
From Eqns. 3.50 and 3.51:
Cn,m sinh (π
Therefore
V (x, y, z) =
p

2 Z a Z a
if n or m is even,
0,
2
n 2 + m2 ) =
V0
sin (nπx/a) sin (mπy/a)dxdy 16V0

a
, if both are odd.
0
0
π 2 nm
16V0
π2
√
1
sinh (π n2 + m2 z/a)
√
sin (nπx/a) sin (mπy/a)
nm
sinh (π n2 + m2 )
n=1,3,5... m=1,3,5...
X
X
3.71-¿3.72
8
Griffiths 3.18
The polynomial is odd, so we only need P1 and P3 .
V0 (θ) = k cos (3θ) = k[4 cos3 θ − 3 cos θ] = k[αP3 (cos θ) + βP1 (cos θ)]
1
5α
3
3
3
3
4 cos (θ) − 3 cos θ = α (5 cos (θ) − 3 cos θ) + β cos θ =
cos θ + β − α cos θ
2
2
2
3
8
so: α = ; β = −
5
5
k
and V0 (θ) = [8 cos3 θ − 3 cos θ]
5
∞
X

Al rl Pl (cos θ) sin θ), for r < R



l=0
Now V (r, θ) = ∞
X

Bl



Pl (cos θ),
for r > R

rl+1
l=0
where
Z
2l + 1 π
V0 (θ)Pl (cos θ) sin θdθ
2Rl 0
Z π
Z π
2l + 1 k
8
P
(cos
θ)P
(cos
θ)
sin
θdθ
−
3
P
(cos
θ)P
(cos
θ)
sin
θdθ
=
3
l
1
l
2Rl 5
0
0
2
2
k 2l + 1
8
δl3 − 3
δl1
=
5 2Rl
2l + 1
2l + 1
k 1
=
[8δl3 − 3δl1 ]
5 Rl
Al =
4
From continuity of V at R, we have Bl = Al R2l+1 ,so
Bl =
k 1
[8δl3 − 3δl1 ]
5 Rl+1
Therefore
 3
r
r
k


8
P
(cos
θ)
,
r≤R
P
(cos
θ)
−
3
1
3

5
R
R
" #
V (r, θ) =
2
4

k
R
R


8
P3 (cos θ) − 3
P1 (cos θ) , r ≥ R

5
r
r
Or

k r
r 2
2


[5
cos
θ
−
3]
−
3
,
r≤R
cos
θ
4

5 R
R
" #
2
V (r, θ) =
2

R
k R

2

cos θ 4
[5 cos θ − 3] − 3 , r ≤ R

5 r
r
Finally
σ(θ) = ǫ0
∞
X
(2l + 1)Al Rl−1 Pl (cos θ)
t=0
epsilon0 k
[−9P1 (cos θ) + 56P3 (cos θ)]
5R
epsilon0 k
cos θ[140 cos2 θ − 93]
=
5R
=
9
Griffiths 3.19
From Eq. 3.83 and 3.69:
σ(θ) = ǫ0
∞
X
(2l + 1)Al Rl−1 Pl (cos θ) , Al =
t=0
2l + 1
2Rl
Z
π
V0 (θ)Pl (cos θ) sin θdθ
0
So we have:
σ(θ) =
10
∞
ǫ0 X
(2l + 1)2 Cl Pl (cos θ) , with Cl =
2R t=0
Z
π
V0 (θ)Pl (cos θ) sin θdθ
0
Griffiths 3.20
Set V=0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the
potential of a uniformly charged spherical shell:
R2
1 Q
V (r, θ) = −E0 r − 2 cos θ +
r
4πǫ0 r
5