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CHAPTER 1: VECTOR ANALYSIS
COMPETENCIES:
Student will be:
1. Understand about vector analysis, theorem for: gradient, curl, Stokes and
divergence in electric and magnetisme problems
2. Apply vector analysis, theorem for: gradient, curl, Stokes and divergence in
electric and magnetisme problems
A. VECTOR ALGEBRA
Vector Operations
a. Addition of to vector
A + B = B + A (addition is commutative)

B
  
A B C

A
b. Multiplication by a scalar
a (A+ B) = aA + aB
c. Dot product of two vectors
A . B = AB cos 
B
A
d. Cross product of twu vectors
 
A x B  AB sin  nˆ
n̂ = a unit vector
Vector Algebra: Component form
Basis vectors of vektor A:

A  Ax iˆ  Ay ˆj  Az kˆ
1
z
Az
A
Ay
y
Ax
x
 
A  B  ( Ax  Bx )iˆ  ( Ay  By ) ˆj  ( Az  Bz )kˆ
iˆ ˆj kˆ
 
b. A x B  Ax A y Az
a.
Bx B y Bz
Example : Find the angle between the face diagonals of a cube!
1
1
1
Solution: from the figure we have:

A 1iˆ  0 ˆj  1kˆ

B  0iˆ  1 ˆj  1kˆ
A =
 
A. B 1. 0  0 .11.11;
12  02 12
than
=
2 B =
 
A. B  A B cos
1 2 . 2 cos
In class problem :
Find the angle between the body diagonals of a cube!
2
02 12 12 =
so   60o
2
B. GRADIENT T 
Suppose, we have tree varible of a fuction temperature T (x, y, z) in a room. Partial
derivatives of T;
 T 
 T 
 T 
dT    dx    dy    dz ; i.e we take
 x 
 z 
 y 
dl  dx iˆ  dy ˆj  dz kˆ
so :
 T ˆ T ˆ T
dT  
i
j
y
z
 x
 T . dl 
T 


kˆ  . dx iˆ  dy ˆj  dz kˆ

T ˆ T ˆ T ˆ
i
j
k
x
y
z

is the gradient of T.
C. OPERATOR 𝛁



  iˆ  ˆj  kˆ
x
y
z
Three ways the operator 𝛁 can act:
a). On a scalar function T: 𝛁T (the “gradient”)
b). On a vector function v, via the dot product: 𝛁. 𝐯 (the “divergence”)
c). On a vector function v, via the cross product: 𝛁 𝐱 v (the “curl”)
D. DIVERGENCE (  . v )
Divergence is a scalar. From definition of  we have:
  v  v y  vz 

 . v   x 

z 
 x y
Geometrical interpretation, the divergence  . v is a measure of how much the vector
v spreads out (diverges) from the point in question.
3

E. THE CURL  x v

Curl is a vector. So, we have:
iˆ ˆj kˆ
  
 xv 
x y z
vx v y vz


Geometrical interpretation, the curl  x v is a measure of how much the vektor v
curls around the point in question.
F. THEOREM FOR GRADIENTS
From point B, we have
dT  T . dl .
Suppose we have a scalar function of three variables T (x, y, z). Starting at point




a a x, a y , a z we move a small distance dl to b bx by bz . The total change in T:
a t . dl  t b   t a 
b
This is the fundamental theorem for gradient. Apparently, then:
b
1. a t  . dl is independent of path taken from a to b
2.
 t .dl  0
if the integration is a closed loop or the beginning and end points
are identical.
G. THEOREM FOR DIVERGENCES
Theorem for divergences:
  . v d   v . da
vol
surf
This theorem has at least three special names : Gauss’s theorem, Green’s theorem,
or the divergence theorem. d = element volum, for cartecian coordinantes, d = dx
dy dz.
4
Example:

Check
the
divergence

theorem
using
the
function
V  y 2 iˆ  2 xy  z 2 ˆj   2 yz  kˆ and the unit cube situated at the origin.
5
2
1
3
4
1
1
1
6
We have  . v  2  x  y     . v
 d   2x  y  d
v
111
 2    x  y dx dy dz  2
000
Solution with surface integration :
11
1
2
v
.
da


  y dy dz 
3
s
00
Surface (1) :
11
1
2
Surface (2):  v . da     y dy dz  
3
s
00
11
2
Surface (3):  v . da     2 x  z  dx dz
s
00



4
3
11
1 2
1
2
Surface (4) :  v . da     z dx dz    z dz  
3
s
00
0
Surface (5) :
1
00
0

21
v
.
d
a

2
y
dx
dy

2
y
dy

y



0 1
s
Surface (6) :
11
 v . da  0
s
5
Total fluks :
That true
1
1
4
1
 v . da  3  3  3  3  1  0  2
surf
  .v d   v . da
vol
=2
surf
H. THEOREM FOR CURLS
The fundamental theorem for curls, which special name Stokes’ theorem:
  x v. da   v.dl
suf
boundary
line
Corollary 1:
  x v . da
depends only on the boundary line, not on the particular
suf
surface used.
Corollary 2:
  x v . da  0 for any closed surface, since the boundary line, like the
suf
mouth of a balloon.
Problem::
1. Find the angle between the body diagonals of a cube!
2. Test Stokes’ theorem for the function v = (𝑥𝑦)𝑖̂ + (2𝑦𝑧)𝑗̂ + (3𝑧𝑥)𝑘̂, using the
triangular shaded area of figure:
z
2
2
x
3.
y
Compute the gradient and Laplacian of the function T = r (cos  + sin  cos ).
Check the Laplacian by coverting T to Cartesian Coordinates and using
∂2 T ∂2 T ∂2 T
∇ ∙ (∇T) =
+
+
∂x 2 ∂y 2 ∂z 2
6
CHAPTER 2: ELECTROSTATICS
COMPETENCIES:
Student will be:
1.
Understand the laws in electric field and their application.
2. Understand about electric potential, work and energy, and conductor in electrostatics.
A. THE ELECTROSTATIC FIELD
The fundamental problem which electromagnetic theory hope to solve is this:
q1
Q
Test charge
q3
q2
q
q
q
Source Charges qi
Figure 2.1
How interaction between source charges qi and test charge Q? How interaction
between source charges and test charge Q if the source volum, surface or line form?
1. Coulomb Law
Consider a system of two point charges, and , separated by a distance in
vacuum. The force exerted by on is given by Coulomb's law:
Fi 
1
qi Q
rˆi
2
4  o ri
Figure 2.2 Coulomb interaction between two charges
2. Electric Field
Electric field is an area which still be influenced by electrics force. In matematically:

 F
E
Q
7
Electric field with discrit charge distributions:
If we have many charges q1, q2, ..., qn at distances r1, r2, ... rn from Q, then according to
the principle of superposition the total force on Q is:
F = F1 + F2 + .... + Fn = QE
Where

E 
1
4  o

qi
rˆi
2
ri
Class Problem:
a. find the electric field (magnitude and direction) a distance z above the midpoint
between two equal charges q a distance d apart. Check that your result is conistent
with what you’d expect when z >> d.
b. repeat part a, only this time make the right hand charge -q instead of +q.
Electric field with continuous charge distributions:
- For a line charge.
The electric field of a line charge is:

rˆ
E ( p)   2  dl   = the charge per unit lenght.
r
- for a surface charge:

rˆ
E ( p)   2  da  σ = the charge per unit area.
r
- for a volume charge :

rˆ
E ( p)   2  d  ρ = the charge per unit volume
r
Example: Find the electric field a distance z above the midpoint of a straight line segment
of lenght 2L, which carries a uniform line charge  (see figure below).

P
r
z
dq=  dx
x
8
From the figure that the horizontal components of the two field cancel, and the net field of
the pair is:
  dx 
2  2  cos kˆ ,
4  o  r 
1
dE 
here  = z / r ,
direction. E 
r  z 2  x 2 and the integral runs from 0 to L. it aims in the z1
2 L
4  o z z 2  L2
1
- For z >> L  E 
4  o
1
- for L    E 
4  o
2 L
z2
2
z
Problem take home : Find electric field a distance z from the center of a spherical surface
of radius R (fig. 2.4), which carries a uniform
z
r
charge density σ. Treat the case z < R (inside) as
well as z > R (outside). Express your answers in
 R
terms of the total charge q on the sphere.
(Suggestion: use the law of cosines to write r in
terms of R and . Be sure take the positive square
R 2  z 2  2 R z  R  z  if R > z but
root:
it’s (z – R) if R< z).
9
B. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS
1. Electric Flux
The strength of an electric field is proportional to the number of field lines per area.
The number of electric field lines that penetrates a given surface is called an “electric
flux,” which we denote as ΦE. The electric field can therefore be thought of as the
number of lines per unit area.
Figure 3.1 Electric field lines passing through a surface of area A.
In mathematics:
E 


 E  da
surf
2. Gauss’s Law
Consider a positive point charge Q located at the center of a sphere of radius r, as

shown in Figure 3.2. The electric field due to the charge Q is E 
1
4  o
(
Q
) rˆi , which
2
ri
points in the radial direction. We enclose the charge by an imaginary sphere of radius r
called the “Gaussian surface”.
Figure 3.2 A spherical Gaussian surface enclosing a charge Q .
10
In mathematics:
  1
E  da 
Qs for continuous charge :

o
surf

 E . da     . E  d
surf .

and Q s 
  d
 . E  1
Vol
Vol
o 
Divergence of E (Gauss’s law in differential form).
3. Applications of Gauss’s Law
Tric applications of Gauss’s law:
-
first known the problem
-
draw the Gaussian surface from point of question, correspond to selected
coordinate.
-
Calculate with Gauss equestion.
Example 1: find electric field outside a uniformly charged sphere of radius a.
Solution: draw a spherical surface at radius r:
Gauss’s law:
E
 
1
E
Qs ; Qs =
  da 
Perm.
r
o
q = total charge in sphere, and E direction
a
 
2
E
  da  EA  E  r so, E 
surf .
same with da that is direction r̂
q
o  r2
rˆ ; if q = ρ 4/3 π a3 so E 
4  a3
rˆ
3 o r 2
Example 2: A long cylinder carries a charge dencity which is proportional to the distance
from the axis: ρ = k r. Find the electric field inside this cylinder.
Gaussian
surface
r
L
11
Solution: Selected coordinate is cylinder coordinate. From Gauss’s law

  1
E  da 
Qs
o
surf .
r l 2
with Qs    d   
2
2
3
 k r d dz dr  3  kL r
00 0

k r2
rˆ
 k Lr  E 
3 o
3
 
2
1
E  da  EA  E 2 r L 
Qs  E 2  r L 
o
surf .
3
4. The Curl of E
What the result of  x E = ?
As describe we calculate integrati of electric field line as point charge from a to b
b
 E . dl
z
in spherical coordinates :
a
dl  dr rˆ  r d ˆ  r sin  d ˆ
rb
x
E . dl 
y
q
ra
b
q
dr
4  0 r 2
1
 E . dl 
a
if
Applying Stokes’ theorem :
   E . da   E . dl
surf .
1
rb
q
1
 dr  4  
4   o ra r 2
o
ra = rb   E .dl  0
q q
  
 ra rb 
the result is :  x E = 0
line
C. ELECTRIC POTENTIAL
1. Introduction to Potential
b
U (r )
V (r ) 
with U(r) electric potential energy U (b)  U (a)   q  E . dl
q
a
then relation between electrics potential by electrics field is :
b
b
a
a
V (b)  V (a) (V ) . dl    E . dl  E = -  V
12
2. Poisson’s dan Laplace’s Equation
E
We have got E = -  V. how about
 . E = .(-V) = -2V

and  x E = 0 relation to V?
0
2
So, the divergence of E is laplace’s of V, and Gauss’s law became:  V 


0
Poisson’s equation. If no charge inside of curve, so ρ = 0 and Laplace’s equation
2
 V 0
From equation  x E =  x (-V) = 0
Curl of gradient always equal to zero. So, to determine V only use divergence of
gradient cannot curl of gradient., except to determine E can both divergence and curl.
3. Potential of charges Distributions
Potential of point charges at infinity:
1 r q
1 q
V r   
dr 

4 0  r 2
4 0 r
If a charge, the potential is  V  p  
1
4 0
q
r
If many point charges distributions  V  p  
1
n q
i

4 0 i 1 ri
For continuous charges distributions:
- Volume  V  p  
- Line  V  p  
- surface 
1
4 0
1
4 0

V  p

r


r
d
dl
1
4 0


r
da
Example: find the potential of a uniformly charged spherical shell of radius R (see the
figure).
13
Solution:
we use equation of surface:
z
V  p
p
1
4 0


r
da
Cosines law:
r
r2 = R2 + z2 – 2Rz cos 

R
y
da  R 2 sin  d d
Finally, the result of this problem:
x
V ( p) 
V ( p) 
R
o
 inside
R2 
 outside
o z
D. WORK AND ENERGY IN ELECTROSTATICS
1. The Work Done in Moving a Charge
Suppose we have a stationary configuration of source charges, and we want move a test
charge Q from point a to point b. The electric force on Q is F = Q • E. The total work we
must do is
b
b
a
a
W   F . dl   Q  E . dl  Q V b   V a   V b   V a  
W
Q
In word, the potential difference between points a and b is equal to the work per unit
charge it takes to carry a particle from a to b. In particular, if you want to bring the charge
Q in from far away and stick it at point P, the work you must do is
W  Q V  p   V   = QV(p).
14
2. The Energy of a Point Charge Distribution
How much work was requared to assemble a collection of point charges? Let’s we bring
the charges in, one by one, from far away. No work to bring in the first charge (q1)
becouse no field in place. Next bring q2, now the work is q2V1(p2), where V1 is the
potential due to q1, and p2 is the place we’re putting q2 :
W2 
q 
q 2  1  with r12 = distance q1 to q2.
4 o  r12 
1
If q3 bring in, the work is W3  q3V12  p3  
q3  q1 q 2 

and so on. Then we
4 o  r13 r23 
bring in charge number 4, total work for four charges in configuration is:
1 n n qi q j
stopping calculation if j = i.
W
 
4 o i 1 j 1; j i rij
In Class problems:
a. Three charges are situated at the corners of a square (side s) as shown in figure below.
How much work does it take to bring in another charge, +q, from far away and place it
at fourth corner?
-q
+q
-q
b. How much work does it take to assemble the whole configuration of four charges?
3. The Energy of a Continuous Charge Distribution
-
For line charge, the energy is: W  12   V dl
-
For surface charge, the energy is: W  12   V da
-
For volume charge, the energy is: W  12   V d
15
There is a lovely way to rewrite equation the energy for volume charge, in which  and V
are eliminated in favor of E. From Gauss’ law:
W
o 

VE.da   E 2 d 


2  Perm.
Vol

In fact, at large distances from the charge, E goes like 1/r2 and V like 1/r. then the surface
integral goes to zero, and we are left with:
W
o
2
E
2
d
all pace
Example: Find the energy of a uniformly charged spherical shell of total charge q and
radius R.
Solution: E inside of sphere = 0.
 1 q 


ˆ
E

r
For outside, E 

2 
4


4  o r 2
r
o


2
d  r sin  dr d d  r: R
q
1
W
o
2
 E d
2
W
vol
2
2
o
2   
24   o 
2
q2  2
1 q2


    4  r sin  dr d d  8   R
0 0 R r 
0

E. CONDUCTOR
1. The basic electrostatic properties of ideal conductor
-
E = 0 inside a conductor.
-
Inside a conductor  = 0 and E = 0.
-
Any net charge resides on the surface.
-
V is constan, throughout a conductor.
-
E is perpendicular to the surface, just outside a conductor..
2. Force on a Surface Charge
The average field E 
1
2o
 nˆ
16

The force per unit area  f 
1
2o
 2 nˆ
This force due electrostatic pressure on the surface:
P
o
2
E2
3. Capacitors
Since E is proportional to Q, so also is V. The constant of proportionally is called
the capacitance of the arrangement: C 
Q
V
Example :  Capacitance of a parallel-plate capacitor C 
A o
d
A = surface of area and d = distance of plate
a. Capacitance
C
of two concentric spherical metal shells, (radii a < b)
Q
ab
 4  o
V
ba
CV
Q
q
b. Total work of charge q = 0  q = Q: W    dq  12

C
2
0C 
Q
2
2
Problem: Find the capacitance per unit length of two coaxial metal cylindrical tubes, of
radii a and b (fig.)!
b
a
17
CHAPTER 3: SPECIAL TECHNIQUES FOR CALCULATING
POTENTIALS
COMPETENCIES:
Student will be:
1.
Applying Laplace’s equation for calculating potential
2. Applying the method of image for calculating potential
A. LAPLACE’S EQUATION
Electric potential equation  V  p  
1
4 0


r
d
Any cases like conductor, is difficult to use this equation. Becouse charge dencity
() its self may not be known (charge is free to move around). So we use special
techniques like Poisson’s equation :  V  
2
1
o
 and Laplace’s equation:  2V  0
Laplace’s equation in one Dimentions only depend on one variable :
d 2V
 0 the solution is V  mx  b  V is a straight line..
dx 2
Laplace’s equation in two Dimentions
 2V  2V

0
x 2 y 2
The solution is V ( x, y ) 
1
4 R
 V dl
Average value of V at point (x,y). R is radii at point (x,y).
Laplace’s equation in three Dimentions
 2V  2V  2V


0
x 2 y 2 z 2
Explicit solution: V at point (p) is average of V a bove a spherical surface which radius R:
V ( p) 
1
 V da
4 R 2
a
R

r
18
p
Boundary Conditions and Uniqueness Theorems
Laplace’s equation not be solved no boundary conditions. For see the boundary conditions
used uniqueness theorem.
First uniqueness Theorem: Solution of Laplace equation at some part uniquely be
determined if value V is specific function at all of condition of the shares boundary.
Solution of first unique theorem  according to Laplace’s equation
 have corect value in boundary
Potential can be determined by unique if according to:
a. charge density known at all area.
b. Value of V be known at all boundary
Second uniqueness theorem: if in any conductors have a conductor that charge density ρ,
the electric field can be determined uniquely if total charge at every conductor be known.
on prove : suppose two electric fields E1 and E2 that first solution of Laplace’s equation.
So, both the electric fields must be according to:
Qi
E
.
da

 1
o
Si
and
.E3  . E1  . E 2 
Qi
E
.
da

 2
o
Si
so, if E3 = E1 – E2 must be according
 

 0 and  E 3 . da  0  E3 = 0
o o
S
19
B. THE METHOD OF IMAGES
1. The Classical Image Problem
Suppose a point charge q at distance d a bove conductor surface by garounded for
V= 0 (fig) and we want to determine the electric potential at z>0.
z
q
d
y
V=0
x
At Z > 0 and q at point (0,0,d) the boundary conditions is:
1. V = 0 if z = 0
2. V  0 far away from charge (x2 + y2 + z2 >> d2)
To solve this problem, suppose we have charge +q (0,0,d) and –q (0,0-d), the electric
potential at point (x,y,z) is :

q

V ( x, y , z ) 

4   o  x 2  y 2  z  d 2



2
2
2
x  y  z  d  
q
1
The equation this area z > 0 for charge +q at point (0,0,d); for area z < 0 the magnitude of
electric potential is equal  Uniqueness Theorem
2. The Induced Surface Charge
At cases, befor, the normal direction is the z direction, so relation of surface charge density
with electric potential:
   o

V
z z 0

q z  d 
V
1 

z 4   o  x 2  y 2   z  d 2



3
2
20



2 32 
2
2
x  y  z  d 


q z  d 

 ( x, y ) 

qd

2 x  y  d
2
2
2

3
 Charge of induction is –q and the biggest at
2
x = y =0.
3. Force and Energy
Force due by a charge q coused –q is:
F 
1
q2
4   o 2d 2
kˆ
q2
 Nonconductor matter
W 
4   o 2d 
1
q2
 conductor matter
W 
4   o 4d 
1
4. Other Image Problems
Image method can be limited by describing the mirror of charge.
Example: a charge q, at distance s from center of sphere with radius R. Find electric
potential for all point! (Electric potential inside of sphere is zero (0), so we can find
electric potential at outside of sphere).
P
R
q


a
s
r
’
q
q’
V= 0
s
Solution: first, we find image of charge q with condition: q '  
a
1  q q' 
R2
so, potential at point P : V 
  
4  o  r r' 
s
21
R
q , distance of q’ :
s
C. SEPARATION OF VARIABLES
1. Cartecian Coordinates
From
2
 V
equation
 2V
x 2

 2V
y 2
of
laplace’s
2
dimention
in
cartecius
coordinate
0
Suppose the solutions of times two x and y functions by variable sparations method. So
Laplace’s equations be came:
Y
d2X
dx 2
First
X
part
d 2Y
dy 2
only
0
1 d 2 X 1 d 2Y

0
X dx2 Y dy2

depend
with
X
dan
second
part
only
depend
1 d 2Y
1 d2X
 C2 ,  C1 + C2 = 0
f x  
 C1 and q y 
Y dy2
X dx 2
If C1 = k and C2 = -k That differential be come:
d2X
dx 2
2
 k X and
d 2Y
dy2
Solution  X x   A e
  k 2Y
kx
 B e kx and Y  y  C sin ky  D cos ky

solution V(x,y)  V x, y   A e
kx

 B e kx C sin ky  D cos ky 
with this boundary condition, general solution can be write:

V  x, y    C k e kx sin ky
k 1
2. Sperical Coordinates
The equation of spherical coordinates is:
 2V 
1   2 V 
1   V 
1
 2V
r

sin

0




2 r 
2
2
2

r
sin







 r sin  
r
22
with
Y,
The problems allways azimut, so V not depend with  and the equations given by
  2 V 
1  
V 
r

 sin 
0
r 
r  sin   
 
Solution of this equation is  V(r,) = R(r) ()
  2 V  d  2 R 
d  2 dR 
r
  r
   r

r 
r  dr 
r 
dr 
dr 
1  
V 
R d 
d 
 sin 

 sin 

sin   
  sin  d 
d 

d  2 dR 
R d 
d 
r

 sin 
0
dr 
dr  sin  d 
d 
1 d  2 dR 
1
d 
d 
r

 sin 
0
R dr 
dr   sin  d 
d 
Both of part, r and  not depend and both is constant.
1 d  2 dR 
r
  l l  1  (Radial equation)
R dr 
dr 
1
d 
d 
 sin 
  l l  1  angular equation
 sin  d 
d 
For radial equations :
d  2 dR 
B
l
r
  l l  1R basic solutions Rr   A r  l 1
dr 
dr 
r
For angular equations:
d
d
d 

 sin 
   l l  1sin    and the solutions is Legendre
d



Polinomial Equation: () = Pl (cos ) so,

B 
 P cos   l = 0, 1, 2……………
V r ,     Ar l 
l 1  l

r 

B 
V r ,      Al r l  l  Pl cos  linier.
r l 1 
l 1
Problem:
At the example, find electric potential outside of sphere by assumting not charge outside of
sphere!
23
CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
COMPETENCIES:
Student will be:
1. Analysis the physical electric field problem in matter and their interaction with polarized
substanceis
A. POLARISATION
1. Dielectric
every matters have different respons to the electrostatic field. Most everyday
objects belong, to one of two large classes: conductors and isolators (or dielectrics).
Conductors are subtances that contain an “unlimited” supply of charges that are free to
move about through the material. In dielectrics, all charges are attached to specific
atoms or molecules, they are on a tight leash, and all they can do to move a bit within
the molecules.
2. Induced Dipoles
what happens to a neutral atom when it is placed in an electric field E? Within neutral
atom have positively charged (the nucleus) and negatively charged (the electron), what
cloud surrounding it. These two regions of charge within the atom are influenced by
the field: the nucleus is pushed in the direction of the field, and the electrons the
opposite way. So that form dipole of charge polarization.
The dipole moment :
P=E
Where  is atomic polarizability.
Example:
A primitive model for an atom consists of a point nucleus (+q) surrounded by a
uniformly charged spherical cloud (-q) of radius a (fig). Calculate the atomic
polarizability of such an atom.
-q
a
+q
24
Solution:
If there are external field E, the nucleus will be shifted slightly to the right and the
electron cloud to the left:
d
E
+q
-q
1
qd
4 o a 3
P = qd = 4oa3E = E
E
3. Polarization
If there are external field around of matter, the charge within the matter will be
polarized. The charge is pushed in the direction of the field.
P ≡ dipole moment per unit volume.
B. THE FIELD OF A POLARIZED OBJECT
1.
Bound Charges
Suppose we have a piece of polarized material, with the potential is:
V
ˆ . p
1
4 o  2
where p= qd and  is direction from dipole to point of potential.
If the material form is volume, so the dipole moment, p = P d in each volume element
d so the total potential is:

p
V
1

p.ˆ
4 o vol  2
d
25
ˆ
1
   where (fA) = f (A) + A(f)

2
So :
V
1
1
1
1
p.da 

 . p  d
4 o surf 
4 o vol 
The first term looks just like the potential of a surface charge b = P. n̂ and the second
term looks like the potential of a volume charge b = -.P.
So :
V
1
1
1
1
  b da  4    b d
4 o surf 
o vol
Example:
Find the electric field produced by a uniformly polarized sphere of radius R.
Solution:
We may as well choose the z axis to coincide with the direction of polarization (see the
p is constant, but
figure). The volume bound charge density b = 0, since ~
 b  ~p.nˆ  p cos 
z
n̂
 is the usual spherical coordinate, so the
potential is

p
R
 p

r cos , for r  R
 3 o
V r ,   
3
 p R
 3 2 cos , for r  R
 o r
Where r cos  = z and the field inside the sphere is uniform,
E   V  
p
1
zˆ  
p for r  R and for outside the sphere r  R :
3 o
3 o
26
V
1
prˆ
4 o r 2
with p 
4 3
R p
3
2. Physical Interpretation of Bound Charge
The bound charge is acculations of charge that the field of a polarized object is
identical. The bound charge in piles up at the right end of the tube is therefore,
q = PA
if the ends have been sliced off perpendicularly, the surface charge density is
q
P
A
b 
A
A
For an oblique cut, the charge is still the same, but A= Aend cos , so
n
P
A
A end
b 
q
 P cos  P .nˆ  over the surface of the material
A end
If the polarization is nonuniform, we have equation,

b
d  
vol

P .da  
Surf
 .P d
vol
So, for any volume we have b = -. P
3. The Field Inside a Dielectric
suppose we want to calculate the macroscopic field at some point P within a
dielectric (see fig. below). The average field over the sphere due to all charges outside plus
the average due to all charges inside:
E = Eout + Ein
R
27
P
The potential outside of the sphere:
Vout 
P.ˆ
1
4o

2
d  Eout = 0
out
The average field of the dipole inside the sphere is:
Ein  
P
1
4

P   R3  P and Ein  
P
3
4o R
3 o
3

1
The macroscopic field given by the potential :
V
1
P.ˆ
d  the integral runs over the entire volume of the dielectric.

4 o  2
C. THE ELECTRIC DISPLACEMENT
From Gauss’s law, we designated by the letter D is known as the electric
displacement:
D oE  P ,
And Gauss’s law reads
.D = f 
 D.da  Q fenc
where Qfenc = total free charge enclosed in the volume.
Example: a long straight wire, carrying uniform line charge , is surrounded by rubber
insulation out to a radius R. Find the electric displacement!
R

r
L
Solution:

 D . da  Da  D 2rL  , Qfenc = L dan D  2r rˆ becouse outside of wire is insulator,
Surf
P = 0, so
E
1
o
D

rˆ, r  R
2 o r
28
D. LINEAR DIELECTRIC
In fact, many substances the polarization is proportional to the field,
P o e E
(e = electric susceptibility)  the materials called linear dielectric.
Electric displacement  D = o ( 1 + e) E =  E
Dielectric constant  K 

1   e
o
Incidentally, the volume bound charge in a homogeneous linear dielectric is proportional to
the density of free charge:
 e 
  f
1


e

 b   
Example: a metal sphere of radius a carries a charge Q (see figure). It is surrounded, out to
radius b, by linear dielectric material of permittivity . Find the potential at the center
(relative to infinity)
b
a
Solution: For compute V, we need to know E; to find E, we begin by calculating D using:
D
Q
4r
2
rˆ  r > a. for r < a E  p  D  0 so;
 Q rˆ a  r  b
 4  r 2
E Q
rˆ r  b

2
4


r

o
The potential at the center is:
29
a
0

Q 
Q



V   E .dl   
dr  
dr   0  dr
 4  r 2 
 4 r 2 

a
b
a

o

0

Q
4
b
 1
1
1 





b

a

b
 o

30