Transcript No Slide Title

Fields and Waves
Lesson 3.3
ELECTROSTATICS - POTENTIALS
Darryl Michael/GE CRD
MAXWELL’S SECOND EQUATION
Lesson 2.2 looked at Maxwell’s 1st equation:

D  
 
 D  ds     dv
Today, we will use Maxwell’s 2nd equation:

 E  0
or


 E  dl  0
Importance of this
equation is that it allows
the use of Voltage or
Electric Potential
POTENTIAL ENERGY
Work done by a force is given by:

dl

F
 
 F  dl
If vectors are parallel, particle gains
energy - Kinetic Energy
 
If,  F  dl  0
Conservative Force
Example : GRAVITY

F
• going DOWN increases KE, decreases PE
• going UP increases PE, decreases KE
POTENTIAL ENERGY
If dealing with a conservative force, can use concept of
POTENTIAL ENERGY
For gravity, the potential energy has the form mgz
Define the following integral:
 
  F  dl  Potential Energy Change
P2
P1
POTENTIAL ENERGY


 
Since  E  dl  0 and F  q  E
We can define:
Potential Energy =
 
  q  E  dl
P2
P1
Also define: Voltage = Potential Energy/Charge
 
V ( P2 )  V ( P1 )    E  dl
P2
P1
Voltage always needs
reference or use
voltage difference
POTENTIAL ENERGY
Example: Use case of point charge at origin and obtain potential
everywhere from E-field
Spherical
Geometry
Point charge
at (0,0,0)

E
q
4 0 r
2
 aˆ r
Integration Path
r

dl

infinity
Reference:
V=0 at infinity
POTENTIAL ENERGY
The integral for computing the potential of the point charge is:
 
V (r )  V (r  )    E  dl
r
r 
0
r
V (r )    E  dr
r 
r
 
r 
q
4 0 r
 dr
2
V (r ) 
q
4 0 r
POTENTIAL ENERGY - problems
Do Problem 1a
Hint for 1a:
R=a
R=b
R=r
Use r=b as the
reference Start here and
move away or
inside r<b
region
POTENTIAL ENERGY - problems
For conservative fields:
 
 E  dl  0
,which implies that:
 
   E  ds  0
, for any surface

  E  0
From vector calculus:
  f  0
,for any field f
Define:

E   V
Can write:

E  f
POTENTIAL SURFACES
Potential is a SCALAR quantity
Graphs are done as Surface Plots or Contour Plots
Example - Parallel Plate Capacitor
+V0/2
+V0
+V0
-V0
-V0
0
Potential Surfaces
-V0/2
E-Field
E-field from Potential Surfaces
From:

E   V
Gradient points in the direction of largest change
Therefore, E-field lines are perpendicular (normal) to
constant V surfaces
(add E-lines to potential plot)
Do problem 2
Numerical Simulation of Potential
In previous lesson 2.2, problem 3 and today in problem 1,
Given  or Q
E-field
derive
V
derive
Look for techniques so that , given  or Q
V
derive
Numerical Simulation of Potential
For the case of a point charge:
q
q

V

   V (r )
4 0 r 4 0 r  r 
Distance from charge

r
, is field point where we are measuring/calculating V

r
, is location of charge
Numerical Simulation of Potential
For smooth charge distribution:

 (r )  dv

V (r )  
 
4 0 r  r 
 
 (r )  dl

V (r )  
 
4 0 r  r 
Volume charge distribution
Line charge distribution
Numerical Simulation of Potential
Problem 3
 
 (r )  dl

V (r )  
 
4 0 r  r 
Setup for Problem 3a and 3b
Line charge:

r
origin
 
r  r

r
Location of
measurement of V
Line charge distribution
Integrate along charge means dl is dz
Numerical Simulation of Potential
Problem 3 contd...
Numerical Approximation
Break line charge into 4 segments
l

r
Charge
for each
segment
 
r  r
q   l  l
Segment
length
qi
V 
 
4 ch arg es 4 0 r  ri
Distance to charge
Numerical Simulation of Potential
Problem 3 contd...
For Part e….
Get V(r = 0.1) and V(r = 0.11)
Use:

V
V
E  V   aˆ r  
aˆ r
r
r
So..use 2 points to get V and r
• V is a SCALAR field and easier to work with
• In many cases, easiest way to get E-field is
to first find V and then use,

E   V