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PHYS2012
EMP WORKSHOP 2
Question 1
A conducting sphere of radius a carrying a charge q is half-submerged into a nonconducting liquid of dielectric constant r.
Find the electric field outside the sphere (in the liquid and in the air) and the charge
density on the surface of the sphere.
ISEE approach
(1)
Draw a diagram of the sphere showing the air, non-conducting liquid, the radius a
and a Gaussian surface S of radius r > a.
By symmetry, what can you say about the electric field and electric displacement?
(2)
Apply Gauss’s Law to the surface S
 D
dA  qenclosed
S
(3)
Re-write Gauss’s Law applied to the surfaces Sair (Dair) and Sliquid (Dliquid).
(4)
Hence, show that
Dair  Dliquid 
q
2 r 2
(5)
Explain the why the following equations are correct
Dair   0 Eair
Diquid   r  0 Eliquid
(6)
At an interface between two dielectric materials, the tangential components of the
electric fields must be equal E1t = E2t. Show this fact on your diagram. Hence,
explain why E = Eair = Eliquid.
(7)
Show that the electric field is
E
q
2  0  r  1 r 2
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
W2.1
(8)
Show that
Dair 
q
2  r  1 r 2
Dliquid 
r q
2  r  1 r 2
(9)
Give expressions for the charged density on the surface of the sphere for the
exposed (air) and submerged (liquid) parts
 liquid = ?
 air = ?
(10)
Sketch the electric field and the electric displacement surrounding the charged
sphere.
(11)
Verify the answers are correct by substituting  r  1 .
(12)
What is the electric field inside the sphere?
This approach to studying more completed problems by considering the electric
displacement first then the electric field is a most useful one.
Question 2
(1)
Sketch three separate diagrams to explain how dipoles can be created in a
dielectric.
Consider the gas xenon ( = 3.5410-40 F.m2 Z = 54 p = 5 atm T = 300 K) in an
electric field of strength 1.45105 V.m-1.
(2)
What is meant by the symbol ? Does it depend on the state of matter of the
dielectric (gas, liquid, solid)?
(3)
How does the polarization of the xenon gas as occur in an electric field?
(4)
Calculate the dielectric constant of xenon.
(5)
Calculate the radius of a xenon atom.
(6)
Calculate the charge separation in the induced dipole of a xenon atom.
(7)
Calculate the electric dipole moment of the xenon atom.
(8)
Calculate the polarization of the xenon gas.
(9)
Are the above values sensible?
(10)
How would the values be different if the gas was helium?
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
W2.2
Solutions
Question 1
(1)
Gaussian surface S
conducting
sphere q
r
air
a
non-conducting
liquid
Symmetry  field lines must be radial
(2) (3) (4)
 D
dA  qenclosed
S
 D
air
Sair
dA 

Sliquid
Dair  Dliquid 
(5)
Dliquid dA  Dair  2 r 2   Dliquid  2 r 2   q
q
 2 r 
2
The relationship between the electric and displacement fields is given by D   E
Dair   0 Eair
ijcooper/physics/p2/em/wks_02.doc
Diquid   r  0 Eliquid
11 May 2017
W2.3
(6)
conducting
sphere q
air
Eairt
Eliquidt
non-conducting
liquid
Symmetry 
(7)
Eairt = Eliquidt  Eair = Eliquid = E
from parts (4) and (5)
0 E   r 0E 
E
q
 2 r 
2
q
(8)
2  0  r  1 r 2
from parts (5) and (7)
(9)
r q
q
Dliquid 
2
2  r  1 r
2  r  1 r 2
On the surface of the conducting sphere r = a
Dair 
 air  Dair (a ) 
q
2  r  1 a 2
 l iquid  Dliquid (a ) 
r q
2  r  1 a 2
Greater concentration of free charge on the bottom of the sphere compared to top.
Greater charge on bottom  increase in E, dielectric  decrease in E  effects
cancel – E uniform value around conducting sphere at any given radius r > a.
(10)
field lines of E
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
field lines of D
W2.4
(11)
r = 1
E
q
2  0  r  1 r
2

q
q

2
2  0 1  1 r
4  0 r 2
which is the electric field surrounding a point charge  ok
q
q
q


2
2
2  r  1 a
2 1  1 a
4 a 2
which is the charge divided by the surface area of the sphere  ok

(12)
Sphere is a conductor – electric field inside must be zero. The charge q is located
on the outer surface of the sphere.
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
W2.5
Question 2
(1)
E
+
-
+
-
+
-
+
shift in atoms
due to ionic nature of bond
induced dipoles due
to shift in electron cloud
rotation
orientation of polar molecules
(2)

e
polarizibility of molecule
electronic polarizibility
The polarizibiltiy is a property of the atom or molecule. The value of the polarizability is
approximately the same for solids, liquids and gasses.
(3)
P  n  Eloc
For xenon gas we can assume that the major contribution to the polarization is due
to the induced dipoles created by the shift in the electron cloud
  e
Eloc  E
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
W2.6
(4)
r  1 
n
0
For a gas pV  N k T
 n
N
p

V kT
(5)
radius of atom a
  4  0 a 3
(6)
charge separation
d
(7)
electric dipole moment
(8)
polarization
(9)
values ok eg radius of atom ~ 10-10 m
(10)
A helium atom compared with xenon atom
smaller radius
smaller dipole moment
smaller dielectric constant
P  n E
ijcooper/physics/p2/em/wks_02.doc
4  0 a 3
E
Ze
p  Z ed
P   r  1  0 E
11 May 2017
Pnp
d is smaller than the nucleus
W2.7
Using Matlab as a super calculator
%% physics 2 emp
% Workshop 2 - question 1
clear all; close all; clc
alpha = 3.54e-40
% alpha
Z = 54
% atomic number
pressure = 5*1.013e5
% pressure [Pa]
T = 300
% temperature
E = 1.45e5
% electric field [V/m]
e = 1.602e-19
% electron charge
k = 1.38e-23
% Boltzmann constant
eps0 = 8.85e-12
% permittivity of free space
n = p / (k * T)
% number density
er = 1 + n*alpha / eps0
% dielectric constant
a = (alpha / (4*pi*eps0))^(1/3) % radius of atom
d = 4*pi*eps0*a^3*E / (Z*e)
% charge separation
p = Z*e*d
% electric dipole moment
P1 = n*alpha*E
P2 = (er-1)*eps0*E
P3 = n*p
% polarization
Numerical answers
alpha = 3.5400e-040
Z = 54
pressure = 506500
T = 300
E = 145000
e = 1.6020e-019
k = 1.3800e-023
eps0 = 8.8500e-012
n = 1.2234e+026
er = 1.0049
a = 1.4710e-010
d = 5.9336e-018
p = 5.1330e-035
P1 = 6.2799e-009
P2 = 6.2799e-009
P3 = 6.2799e-009
ijcooper/physics/p2/em/wks_02.doc
11 May 2017
W2.8