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PHYS2012
ELECTROMAGNETIC PROPOERTIES OF MATTER
WORKSHOP 3
Question 1
Sulfur has a density of 2.1103 kg.m-3, is number 16 in the periodic table and has an
atomic mass of 32. Sulfur has a dielectric constant of 4. The internal electric field in a
piece of sulfur is 107 V.m-1. Assuming all electrons to be displaced by the same distance
d with respect to the nucleus, find d. How does d compare to the size of an atom?
Comment on this result.
e = 1.6010-19 C NA = 6.021023 particles.mol-1
0 = 8.8510-12 C2.N.m-2
1
2
3
4
5
6
7
8
9
10
Consider a single sulfur atom. Draw a diagram showing the electron distribution.
The internal electric field in a piece of sulfur is 107 V.m-1. Draw a diagram
showing the electron distribution and the electric field.
On the diagram show the electric dipole moment of a sulfur atom (vector), the
charge that has been separated and the separation distance d and the equation for
the electric dipole moment of a single sulfur atom.
What does the internal electric field do to the piece of sulfur?
Determine the electric susceptibility of the sulfur.
Determine the number density of the sulfur.
Determine the electric dipole moment of a sulfur atom.
Determine the charge separation distance d.
Comment on the magnitude of d.
You can also approach the problem from an algebraic point of view.
Express the separation distance d in terms of the following quantities
r, o, E, M (molar mass), Z (atomic number), , NA
Question 2
Consider two concentric conducting spherical shells with radii R1 and R2. The space
between the shells is taken up by a material with dielectric constant εr. What is the
capacitance C? Show as R2→R1 (d = R2-R1) and for large radii of curvature that you
obtain the result for a parallel plate capacitor.
Determine the capacitance of Earth, assuming it to a conductor.
k
1
4 0
 9  109
4 0 
1
9  109
RE  6.38  106 m
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Sunday, April 30, 2017
1
1
Assume vacuum filled capacitor to start.
Radius of inner conducting sphere (charge +Q), R1
Radius of outer conducting sphere (charge –Q), R2
Draw a labeled diagram of the physical situation.
Show that the electric field between the spheres is
E
Q
4 0 r2
Show the Gaussian surface on your diagram.
2
Derive the following equation for the magnitude of the potential in the region
between the two spheres
|V |
Q
R2
4   0 R1
dr
Q 1 1 
Q R2  R1

  
2
r
4   0  R1 R2  4   0 R1 R2
3
Derive an expression for the capacitance C.
Comment on your result – what does C depend upon?
4
Modify the equation you derived in (2.3) to account for the presence of a
dielectric between the metal spheres.
When a dielectric is inserted between the plates of a capacitor, does the
capacitance C decrease, increase or stay the same? Explain your answer.
5
6
7
Show as R2→R1 (d = R2-R1) and for large radii of curvature you obtain the result
for a parallel plate capacitor
  A
C r 0
d
Show that the capacitance of an isolated sphere (radius R) is given by
C  4  R
Determine the capacitance of Earth, assuming it to be conducting.
CEarth = ? F
Question 3
A metallic ring of cross-sectional area 2.5 cm2 and mean radius 400 cm and relative
permeability 1500 is wound uniformly with 3000 turns of wire. If a current of 1.6 A
passes through the wire, find the mean B-field and the magnetization of the ring.
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Solution Question 1
 = 2.1x103 kg.m-3
Z = 16
M = 32x10-3 kg
E = 107 V.m-1
r = 4
density of sulfur
atomic number
molar mass
internal electric field
dielectric constant
(1 to 3)
Induced dipole moment – sulfur atom
E
-8e
+16e
-8e
-8e
Zero electric field –
helium atom
symmetric  zero
dipole moment
+16e
-8e
d
A
B
effectively charge +16e at A and -16e at B
dipole moment p = 16 e d
p
(4)
(5)
The electric field polarizes the sample
sulfur
P  e  0 E  n p
r  4
e   r  1  3
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
+
(6)

msample
V
n
 NA
p
 0 e E
(7)
M
n
p2/em/wks_3_09.doc
+
+
+
+
+
+
 b
Nm
M

n
V
NA
 2.1  10  6.02  10  atoms.m
3

of
E
 b
23
32  10
3
-3
 3.95  1028 atoms.m-3
8.85 10  3 1 10  C.m  6.722  10

12
3.95  10
Sunday, April 30, 2017
7
28
33
C.m
3
(8)
p  Z ed
(9)
6.722  1033 

p
d

2.6  1015 m
Z e (16)(1.602  1019 )
The separation distance d is much smaller than the diameter of the atom
d ~ nuclear dimensions
(10)
P  e  0 E  n p  n Z e d
d
e   r  1
n
 0  r  1 M E
Z e  NA
 NA
M
Solution Question 2
(1) add diagram
Assume vacuum filled capacitor to start.
Radius of inner conducting sphere (charge +Q), R1
Radius of outer conducting sphere (charge –Q), R2
Q
C
capacitance
V
2
potential
V    E.dl
Gauss’s Law
 E.dA 
electric field between spheres
E
1
qenclosed
0
Q
4 0 r2
(2)
potential from electric field
R2 dr
Q
Q 1 1 
Q R2  R1
V  |V |

 

2

R
4 0 1 r
4   0  R1 R2  4   0 R1 R2
(3)
Q 4   0 R1 R2

V
R2  R1
Capacitance depends only on the geometry
C
capacitance
p2/em/wks_3_09.doc
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4
(4)
When dielectric added between conducting spheres, can replace 0 by the permittivity of
the material  = r 0.
4   r  0 R1 R2
C
capacitance with dielectric
R2  R1
Capacitance increases by the factor r (r > 1) because the presence of the dielectric
decreases the electric field between the plates (E smaller, V smaller, Q constant, C larger)
(5)
For large radii of curvature and as R2  R1

R1 R2 = R2 and R2 – R1 = d A = 4 R2
  A
C r 0
capacitance of a parallel plate capacitor
d
(6)
Isolated conducting sphere capacitor
C
two spheres
4 
1
1

R1 R2
Take limits R1  R and R2  
capacitance of isolated sphere
C  4  R
where  is the permittivity of medium surrounding the sphere.
Capacitance of the Earth
1
  0 k 
 9  109
4 0
4 0 
1
9  109
RE  6.38  106 m
CEarth = 7.110-4 F
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Solution Question 3
Y
Current in Z direction
Current in – Z direction
X
R
r
B
d
ds
In the presence of a magnetic medium, Ampere’s Law becomes
H
dl  i
H
B
0
B  r 0 H
M
M   r  1 H
For a circular loop of radius r
H
 r =1500
Ni
2 r
B
N = 3000
r 0 N i
2 r
M   r  1
r = 0.40 m
i = 1.6 A
Ni
2 r
B = 3.6 T M = 2.9106 A.m-1
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