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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
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
 

 
 
  
 
 

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
  
  
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


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:
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
 
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
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.
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

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.
Potential Diff. in Uniform E field
Charged particle moves from A to B in uniform E
field.

.
b
a
V E ds
 
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 
    

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
    
 
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.
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
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
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
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
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
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
Electrostatic Potential Due to Charge Dipole
observation
point
d/2
+Q
-Q
z
d/2

P
ˆz
p a Qd

R
R
r
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
• 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
• 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
• In terms of the dipole moment:
2
0
ˆ
4
1
r
a
p
V
r



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
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


 
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.

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228602.ppt

  • 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
  • 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.