Physics 712 – Electricity and Magnetism
Solutions to Midterm Exam, Spring 2016
Please note that some possibly helpful formulas and integrals appear on the second page.
Each question is worth 20 points.
1. A charge distribution in vacuum produces a potential given, in cylindrical coordinates,
  a2 
by    ,  , z   ln 1  2  ; that is, it is independent of both z and . Find the electric
0   
field E everywhere and the charge density   x  everywhere away from the z-axis.
Don’t confuse the charge density with the cylindrical coordinate  . Demonstrate that
there is also a linear charge density along the z-axis, and determine its magnitude.
We find the electric field from the equation
E    ρˆ

1 


2a 2
   a2 
 ρˆ
ln  1  2   
  φˆ
  zˆ    ρˆ
 
 0    

z
 0 1  a 2  2   3
2a 2 ρˆ
.
 0    2  a2 
We then find the charge density using
1 
 2a 2   1 
1 

4a 2
  x     D   0  E   0 
  E     E  z Ez       2  a 2   2 2 2 .
  


   a 
However, these formulas are a little tricky to apply at   0 , which corresponds to the zaxis. To find the charge along the z-axis, draw a small cylinder of radius  and length L centered
on the z-axis. The total charge inside this cylinder will be q V    0  E  nˆ da . Because the
S
electric field is in the radial direction, only the lateral surface will contribute. This surface has an
area of 2 L , so we have
q V    0  E  nˆ da  2 L
S
2a 2
4 La 2
.

   2  a2   2  a2
But in the limit that   0 , this doesn’t vanish, but takes the limiting value q  4 L . This
implies a linear charge density

q
 4 .
L
2. The potential on a sphere of radius a is given by   a,  ,    V cos  . The potential on a
sphere of radius 2a is given by   2a,  ,    V . Find the potential in the region
a  r  2a assuming there are no charges between the two spheres.
We need to write each of these potentials in terms of spherical harmonics. Using the
explicit form of the spherical harmonics, it is not hard to see that
  a,  ,    V
4
3
 Y10  ,   ,   2a, ,    V 4 Y00  ,   .
Now, the general solution of  2   0 in spherical coordinates is given by

  r , ,       Alm r l  Br  l 1  Ylm  ,   .
l
l  0 m  l
By comparison with the given expressions, it seems to make sense to conjecture that only the
values  l , m    0, 0  and  l , m   1, 0  will contribute. We therefore have
  r , ,     A00  B00 r 1  Y00  ,     A10 r  B10 r 2  Y00  ,   .
We want the first pair of terms to vanish at r = a and take the value V 4 Y00 at r = 2a.
This yields the two equations
A00  B00  2a   V 4
1
A00  B00 a 1  0,
The first of these equations implies B00  aA00 , and substituting this into the other, we have
A00  aA00  2a   V 4 ,
1
A00  V 4 ,
1
2
A00  2V 4
and B00  2aV 4 .
We also want the expression A10 r  B10 r 2 to vanish at r = 2a, and take the value V
at r = a. This yields the two equations
A10 a  B10 a 2  V
4
3
 , A10  2a   B10  2a   0 .
2
The second of these equations implies B10  8a 3 A10 . Substituting into the first, we have
A10 a  8a 3 A10 a 2  V
7aA10  V
A10   17 Va 1
4
3
4
3
4
3
,
 ,
 , B10  87 Va 2
4
3
.
If we substitute these back into our expression for the potential, we have
 a
  r ,  ,    2V 4 1   Y00  ,    17 V
 r
4
3
 r
 a
  
8a 2 
 Y10  ,   .
r2 
4
3

We can then substitute in the explicit forms for the spherical harmonics to write our answer as
 8a 2 r 
 a
  r ,  ,    2V 1    17 V  2   cos  .
a
 r
 r
3. Two matching point charges q are placed on opposite
sides at a distance r from the center of a conducting
sphere of radius a. Find the total force on one of the
charges if the sphere is (a) grounded (b) neutral.
a
r
r
This calculation must be done by the method of images.
For a grounded conducting sphere,there will be two point image charges of magnitude
q   qa r at a distance of r   a 2 r from the center. The charges themselves do not provide
any force on themselves, but they DO feel the force from the other charge. The distance between
the two charges is 2r , and the distance from the charge to the two image charges is r  a 2 r .
The total force on the charge on the right will then be


2
xˆ  q 2
qq
qq
  q xˆ


F
4 0  4r 2  r  a 2 r 2  r  a 2 r 2  4 0





a
a
 1 


 4r 2 r  r  a 2 r  2 r  r  a 2 r  2 




q 2 xˆ  1
ar
ar



4 0  4r 2  r 2  a 2 2  r 2  a 2 2 


If you insist that the sphere be neutral, you must add an additional charge 2q right at
the center. We will therefore have


xˆ  q 2
qq
qq
2qq 



4 0  4r 2  r  a 2 r 2  r  a 2 r 2
r2 




q 2 xˆ  1
a
a
2a  q 2 xˆ





4 0  4r 2 r  r  a 2 r 2 r  r  a 2 r 2 rr 2  4 0


F


ar
ar
2a 
 1 


 4r 2  r 2  a 2  2  r 2  a 2  2 r 3 


4. A solid conducting sphere of radius a is surrounded by
a hollow conducting sphere of radius c. A charge Q is
placed on the inner sphere and –Q on the outer sphere.
The region from r = a to r = b is filled with an insulator
with dielectric constant , and the rest is left in vacuum.
Find D and E everywhere between the two spheres,
and the potential difference    c   a . Also find
the bound surface charge density at r = b, and the total
energy.
–Q
2b
+Q
2a 2c
The problem clearly has spherical symmetry, so it makes a lot of sense to assume that all
fields will be strictly radial and depend only on the distance from the center. We therefore have,
for example, D  rˆD  r  . We then use Gauss’s on a sphere of radius r anywhere in the region
a  r  c to see that
Q   D  nˆ da  D  r   da  4 r 2 D  r  ,
S
Dr  
S
Q
Q
, so D  x  
rˆ .
2
4 r
4 r 2
If we divide this by  in the insulating medium and by  0 in vacuum, we find the electric field as
well.
1
1
Qrˆ 
E x 
Dx 

4 r 2  01
  x
ar b,
b  r  c.
The potential difference is the integral of the electric field, so
b
c dr 
1
Q  b dr
Q  1

    a     c         dl   Er dr 


 a 2  b
2 
a
a
4   r
 0 r  4  r a r 0

1
1
1 
Q  1


  
.
4  a b b 0 c 0 
c



b
c
c
I unintentionally asked for the negative of this, which can be fixed by adding a minus sign.
The charge density on the surface is given by
 b  P  nˆ   D   0 E   rˆ 
Q  0 
1   .
4 b 2 
 
The energy can then be computed using
W   w d 3x 
1

8
1
2
3
EDd x 
 1 b 1


  r a  0r

1
2
 1

 8
b
c
c 1
 b1 2 2

2
2 2
d
r
dr
r
dr
D
D




2
  a 
b  0
  4 
 1
1
1
1 

  
.
  a  b  0b  0 c 
 1 b dr 1
 a 2 
0
 r

c
b
dr 

r2 
5. A tokamak is in the shape of a rectangular
cross-section donut centered on the z-axis,
with height h, inner radius a and outer
radius b, as sketched in the cutaway view at
right. A total current I is then sent around
the rectangular direction of the tokamak, so
it goes up through the hole in the center,
a
b
across the top, down on the outside, and
then back to the center, equally at all angles,
so as to generate a magnetic flux density B  x   B   , z  φˆ in cylindrical coordinates.
Find B at all points inside or outside the tokamak, and find the total magnetic energy
stored inside the tokamak.
We are in vacuum, for which B  0 H , or B  0 H . We will use the integral version of
Ampere’s Law, which says that I S   H  x   dl . We will use loops that go around in the
φ̂ direction, such as the two dashed loops sketched in above, while remaining at constant  and z.
Such a loop has a length of 2 , and therefore we have
0 I S  0  H  x   dl   B  x   dl  2 B   , z  ,
B  , z  
0 I S φˆ
.
2
where IS is the current passing through an imaginary surface bounded by these loops. For the
smaller Ampere loop, which lies within the “hole” of the donut, there is not current passing
through it, and therefore we simply find B  0 . Indeed, the same is true of any loop that is not
within the “cake” of the donut, whether it be above, below, around the exterior, or in the hole. In
contrast for any loop that is in the “cake” of the donut, the entire current I passes through the
Ampere loop, so we have Is = I. We therefore have, in summary
Bin 
0 Iφˆ
, B out  0.
2
The energy is then easily computed using
W   w d 3x 

0 I 2
8 2

2
0
1
2
3
BHd x 
d 
b
a
0 I 2
1
2 3

B
x
d
2 0 
8 2

in
d 3x
2
b
20 I 2 h
1
d  h
b

ln    a 
dz
0 I 2 h ln   .
2 0
2
8
4

a
h