1.
The divergence of E
If the charge fills a volume, with charge per unit
volume .
'
dq d
 

Where d is an element of
volume.
For a volume charge:
'
'
2
0
1 ( )
ˆ
( )
4 v
r
E R rd
r



 
R

2.
Thus:
3 ' ' '
0
1
. 4 ( ) ( )
4 v
E r r r d
  

  

0
1
. ( )
E r


 
Gauss’s law in
differential form.
3
2
ˆ
. 4 ( )
r
r
r

 
 
 
 
 
' '
2
0
ˆ
1
. . ( )
4 v
r
E r d
r
 

 
 
  
 
 


3.
Spherical polar coordinates (r, , )
r: the distance from the origin
: the angle down from the z axis is called polar angle
: angle around from the x axis is called the azimuthal
angle
sin cos
x r  

sin sin
y r  

cos
z r 

ˆ ˆ ˆ ˆ
sin cos sin sin cos
r x y z
    
  
ˆ ˆ ˆ ˆ
cos cos cos sin sin
x y z
     
  
ˆ ˆ ˆ
sin cos
x y
  
   ˆ ˆ
ˆ sin
dl drr rd r d
  
  

4.
The Curl of E
For a point charge situated at origin:
2
0
1
ˆ
4
q
E r
r


Line integral of the field from some point a to some
other point b:
In spherical polar coordinates,
ˆ ˆ
ˆ sin
dl drr rd r d
  
  
2
0
1
.
4
q
E dl dr
r



5.
2
0
1
.
4
b b
a a
q
E dl dr
r


 
0
1
4 a b
q q
r r

 
 
 
 
True for electrostatic
field.
. 0
E dl 

Apply stokes theorem:
0
E
 
The integral around a closed path:

6.
Electric Potential
Basic concept:
The absence of closed lines is the property of vector
field whose curl is zero.
E is such a vector whose curl is zero.
Using this special kind of it’s property we can reduce a
vector problem: using V, we can get E very easily.
Vector whose curl is zero, is equal to the gradient of
some scalar function
E=0  the line integral of E around any closed loop
is zero (due to Stokes' theorem).
E V
 

7.
otherwise you could go out along path (i) and return
along path (ii) and
Because the line integral is independent of path, we
can define a function
O is some standard reference point.
. 0
E dl 

Therefore the line integral of E from
point a to point b is the same for all
paths.
. 0
E dl 

( ) .
r
o
V r E dl
 
is called electric
potential

8.
The potential difference between two points a and b:
( ) ( ) . .
b a
o o
V b V a E dl E dl
   
 
. .
b o
o a
E dl E dl
  
 
.
b
a
E dl
 
Using fundamental theorem for gradients:
( ) ( ) ( ).
b
a
V b V a V dl
  

So ( ). .
b b
a a
V dl E dl
  
  E V
 
Electric field is the gradient of a scalar potential.

9.
Electric Potential at an arbitrary point
•Electric potential at a point is given as the work done
in moving the unit test charge (q0) from infinity (where
potential is taken as zero) to that point.
• Electric potential at any point P is
Note that Vp represents the potential difference dV
between the point P and a point at infinity.
S.I. unit J/C defined as a volt (V) and 1 V/m = 1 N/C
.
p
p
V E ds

 
0
p
W
V
q


10.
Potential Difference in Uniform E field
• Electric field lines always point in the direction of
decreasing electric potential.
Example: Uniform field along –y
axis (E parallel to dl)
.
B B
B A
A A
V V V E dl Edl
      
 
B
A
V Edl Ed
    

• When the electric field E is directed downward,
point B is at a lower electric potential than point A. A
positive test charge that moves from point A to point
B loses electric potential energy.

11.
Potential Diff. in Uniform E field
Charged particle moves from A to B in uniform E
field.

.
b
a
V E ds
 

12.
Potential Diff. In Uniform E field
(Path independence)
Show that the potential difference between point A and
B by moving through path (1) and (2) are the same as
expected for a conservative force field.
By path (1), . cos
B
A
V E dl El 
    


13.
path (2)
= 0 since E is  to dl
. cos
C
A
V E dl Eh El 
      

. .
C B
A C
V E dl E dl
    
 

14.
Equipotential Surfaces (Contours)
VC = VB ( same potential)
In fact, points along this
line has the same
potential. We have an
equipotential line.
. 0
B
C
V E dl
   

No work is done in moving a test charge between any two
points on an equipotential surface.
The equipotential surfaces of a uniform electric field
consist of a family of planes that are all perpendicular to the
field.

15.
Equipotential Surface
Equipotential Surfaces (dashed blue lines) and electric field lines
(orange lines) for (a) a uniform electric field produced by infinite
sheet of charge, (b) a point charge, and (c) an electric dipole. In all
cases, the equipotential surfaces are perpendicular to the electric
field lines at every point.

16.
16
Electrostatic Potential of a Point
Charge at the Origin
Q
P
r
 
 
 
'
2
'
0
2
'
0 0
4
4 4
r r
r
Q
V r E dl dr
r
Q dr Q
r
r

 
 

     
 
 


17.
17
Electrostatic Potential Resulting from
Multiple Point Charges
Q1
P(R,,)
r 1
R
1
r
O
Q2
2
r
 
1 0
4
n
k
k k
Q
V r
R


 
2
R

18.
18
Electrostatic Potential Resulting from
Continuous Charge Distributions
 
 
 
0
0
0
1
4
1
4
1
4
L
S
V
dl
V r
R
ds
V r
R
dv
V r
R












 line charge
 surface charge
 volume charge

19.
19
Charge Dipole
• An electric charge dipole consists of a pair of equal
and opposite point charges separated by a small
distance (i.e., much smaller than the distance at
which we observe the resulting field).
d
+Q -Q

20.
Dipole Moment
• Dipole moment p is a measure of the strength of the
dipole and has its direction.
p Qd

+Q
-Q
d
p is in the direction from the
negative point charge to the
positive point charge

21.
21
Electrostatic Potential Due to Charge Dipole
observation
point
d/2
+Q
-Q
z
d/2

P
ˆz
p a Qd

R
R
r

22.
22
d/2
d/2



cos
)
2
/
(
cos
)
2
/
(
2
2
2
2
rd
d
r
R
rd
d
r
R








R
r
P
   
0 0
,
4 4
Q Q
V r V r
R R

 
 
  
R

23.
• first order approximation from geometry:


cos
2
cos
2
d
r
R
d
r
R






d/2
d/2

lines approximately
parallel
R
R
r

24.
24
• Taylor series approximation:








































cos
2
1
1
1
cos
2
1
1
cos
2
1
1
cos
2
1
1
1
r
d
r
R
r
d
r
r
d
r
d
r
R
  1
,
1
1
:
Recall



 x
nx
x
n
 
2
0
0
4
cos
2
cos
1
2
cos
1
4
,
r
Qd
r
d
r
d
r
Q
r
V






























25.
25
• In terms of the dipole moment:
2
0
ˆ
4
1
r
a
p
V
r




26.
Electric Potential Energy
of a System of Point Charges
1
0
1
4
q
V
r


2 ( )
W q V r

1 3 2 3
1 2
12 13 23
0 12 13 23
1
( )
4
q q q q
q q
W W W W
r r r

     
2
W F r q E r
   
q1
q2
2
b b
a a
W F dl q E dl
    
 
2[ ( ) ( )]
W q V b V a
 
2[ ( ) ( )]
W q V r V
  
and we know

27.
The Energy of a Continuous Charge Distribution
For a volume charge density p,
1
2
W Vd
 
 
0 .E
 
 
Using Gauss’s Law:
0
( . )
2
W E Vd


 

So:
0
.( ) .
2
W E V d VE da


 
   
 
 
By doing integration by part:
and so,
V E
   2
0
.
2 v s
W E d VE da


 
 
 
 
 
If we take integral over all space:
2
0
2 allspace
W E d


 

28.
Poisson’s and Laplace’s Equation
E V
 
The fundamental equations for E:
0
. ;
E


  0
E
 
2
. .( )
E V V
     
Gauss’s law then says that:
2
0
V


   This is known as
Poisson’s equation.
In regions where there is no charge: 0
 
Poisson’s equation reduces to Laplace’s equation.
2
0
V
 
This is known as
Laplace’s equation.