### About Global Documents

**Global Documents provides you with documents from around the globe on a variety of topics for your enjoyment. **

**Global Documents utilizes edocr for all its document needs due to edocr's wonderful content features. Thousands of professionals and businesses around the globe publish marketing, sales, operations, customer service and financial documents making it easier for prospects and customers to find content. **

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