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PHY 5346
HW Set 3 Solutions – Kimel
4. 2.6 We are considering two conducting spheres of radii ra and rb respectively. The charges on
the spheres are Qa and Qb.
a) The process is that you start with qa1 and qb1 at the centers of the spheres, and sphere a
then is an equipotential from charge qa1 but not from qb1 and vice versa. To correct this we use
the method of images for spheres as discussed in class. This gives the iterative equations given in the
text.
b) qa1 and qb1 are determined from the two requirements
∑
j=1
∞
qaj = Qa and ∑
j=1
∞
qbj = Qb
As a program equation, we use a do-loop of the form
∑
j=2
n
qaj =
−raqbj − 1
dbj − 1
and similar equations for qbj, xaj, xbj, daj,dbj. The potential outside the spheres is given
by
φx⃗ =
1
4π 0
∑
j=1
n
qaj
x⃗ − xajk̂
+∑
j=1
n
qbj
x⃗ − dbjk̂
This potential is constant on the surface of the spheres by construction.
And the force between the spheres is
F =
1
4π 0
∑
j,k
qajqbk
d − xaj − xbk2
c) Now we take the special case Qa = Qb, ra = rb = R, d = 2R. Then we find, using the iteration
equations
xaj = xbj = xj
x1 = 0, x2 = R/2, x3 = 2R/3, or xj =
j − 1
j
R
qaj = qbj = qj
qj = q, q2 = −q/2, q3 = q/3, or qj =
−1 j+1
j
q
So, as n → ∞
∑
j=1
∞
qj = q∑
j=1
∞
−1 j+1
j
= q ln2 = Q → q = Q
ln2
The force between the spheres is
F =
1
4π 0
q2
R2
∑
j,k
−1 j+k
jk 2 − j−1
j
− k−1
k
2 =
1
4π 0
q2
R2
∑
j,k
−1 j+kjk
j + k2
Evaluating the sum numerically
F =
1
4π 0
q2
R2
0.0739 =
1
4π 0
Q2
R2
1
ln22
0.0739
Comparing this to the force between the charges located at the centers of the spheres
Fp =
1
4π 0
Q2
R24
Comparing the two results, we see
F = 4
1
ln22
0.0739Fp = 0.615Fp
On the surface of the sphere
φ =
1
4π 0
∑
j=1
∞
qj
R − xj
=
q
4π 0R
∑
j=1
∞
−1 j+1
Notice
1
1 + 1
= ∑
j=1
∞
−1 j+1
So φ =
1
4π 0
q
2R
=
1
4π 0
Q
2 ln2R
=
Q
C
→
C
4π 0R
= 2 ln2 = 1. 386
HW Set 3 Solutions – Kimel
4. 2.6 We are considering two conducting spheres of radii ra and rb respectively. The charges on
the spheres are Qa and Qb.
a) The process is that you start with qa1 and qb1 at the centers of the spheres, and sphere a
then is an equipotential from charge qa1 but not from qb1 and vice versa. To correct this we use
the method of images for spheres as discussed in class. This gives the iterative equations given in the
text.
b) qa1 and qb1 are determined from the two requirements
∑
j=1
∞
qaj = Qa and ∑
j=1
∞
qbj = Qb
As a program equation, we use a do-loop of the form
∑
j=2
n
qaj =
−raqbj − 1
dbj − 1
and similar equations for qbj, xaj, xbj, daj,dbj. The potential outside the spheres is given
by
φx⃗ =
1
4π 0
∑
j=1
n
qaj
x⃗ − xajk̂
+∑
j=1
n
qbj
x⃗ − dbjk̂
This potential is constant on the surface of the spheres by construction.
And the force between the spheres is
F =
1
4π 0
∑
j,k
qajqbk
d − xaj − xbk2
c) Now we take the special case Qa = Qb, ra = rb = R, d = 2R. Then we find, using the iteration
equations
xaj = xbj = xj
x1 = 0, x2 = R/2, x3 = 2R/3, or xj =
j − 1
j
R
qaj = qbj = qj
qj = q, q2 = −q/2, q3 = q/3, or qj =
−1 j+1
j
q
So, as n → ∞
∑
j=1
∞
qj = q∑
j=1
∞
−1 j+1
j
= q ln2 = Q → q = Q
ln2
The force between the spheres is
F =
1
4π 0
q2
R2
∑
j,k
−1 j+k
jk 2 − j−1
j
− k−1
k
2 =
1
4π 0
q2
R2
∑
j,k
−1 j+kjk
j + k2
Evaluating the sum numerically
F =
1
4π 0
q2
R2
0.0739 =
1
4π 0
Q2
R2
1
ln22
0.0739
Comparing this to the force between the charges located at the centers of the spheres
Fp =
1
4π 0
Q2
R24
Comparing the two results, we see
F = 4
1
ln22
0.0739Fp = 0.615Fp
On the surface of the sphere
φ =
1
4π 0
∑
j=1
∞
qj
R − xj
=
q
4π 0R
∑
j=1
∞
−1 j+1
Notice
1
1 + 1
= ∑
j=1
∞
−1 j+1
So φ =
1
4π 0
q
2R
=
1
4π 0
Q
2 ln2R
=
Q
C
→
C
4π 0R
= 2 ln2 = 1. 386