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ELECTRIC POTENTIAL
September 13, 2006
Goings On For the Next Few Days
Quiz Today – Gauss/Electric Field
 Today – Begin Chapter 25 – Potential
 Monday - Potential
 Wednesday – EXAMINATION #1
 Friday – More Potential

Picture a Region of
space Where there is an Electric
Field
Imagine there is a particle of charge q at
some location.
 Imagine that the particle moves to
another spot within the field.
 Work must be done in order to accomplish
this.

Electric Potential

We will be dealing with



Work
Energy
We do work if we try to move a charge in an
electric field.


Does the FIELD do work?
Does the FIELD have a brain? Muscles?
Energy Methods



Often easier to apply than to solve
directly Newton’s law equations.
Only works for conservative forces.
One has to be careful with SIGNS.

VERY CAREFUL!
I need some help.
THINK ABOUT THIS!!!

When an object is moved from one point to
another in an Electric Field,



It takes energy (work) to move it.
This work can be done by an external force (you).
You can also think of this as the
FIELD
negative of
doing the
this amount of work on the particle.
Move It!


Move the charge at constant velocity so it is
in mechanical equilibrium all the time.
Ignore the acceleration at the beginning
because you have to do the same amount of
negative work to stop it when you get there.
And also remember:
The net work done by a conservative (field)
force on a particle moving
around a closed path is
ZERO!
A nice landscape
Work done by external force = mgh
How much work here by
gravitational field?
h

mg
The gravitational case:

Someone else’s path

IMPORTANT

The work necessary for an external
agent to move a charge from an initial
point to a final point is INDEPENDENT
OF THE PATH CHOSEN!
The Electric Field

Is a conservative field.





No frictional losses, etc.
Is created by charges.
When one (external agent) moves a test charge
from one point in a field to another, the external
agent must do work.
This work is equal to the increase in potential
energy of the charge.
It is also the NEGATIVE of the work done BY
THE FIELD in moving the charge from the same
points.
A few things to remember…


A conservative force is NOT a Republican.
An External Agent is NOT 007.
Electric Potential Energy

When an electrostatic force acts between
two or more charged particles, we can
assign an ELECTRIC POTENTIAL ENERGY
U to the system.
Example: NOTATION U=PE
HIGH U
LOWER U
E

q
F
A
B
d
Work done by FIELD is Fd
Negative of the work done by the FIELD is -Fd
Change in Potential Energy is –Fd.
The charge sort-of “fell” to lower potential energy.
Another way to look at it:




Consider an external agent.
The external agent moves a charge from one
point to the other.
The work done by the external agent in
moving the charge between these points is
the change in potential energy.
This is the negative of the work that the
FIELD does!
Electrons have those *&#^
negative signs.





Electrons sometimes seem to be more
difficult to deal with because of their
negative charge.
They “seem” to go from low potential
energy to high.
They DO!
They always fall AGAINST the field!
Strange little things. But if YOU were
negative, you would be a little strange too!
AN IMPORTANT DEFINITION

Just as the ELECTRIC FIELD was defined as
the FORCE per UNIT CHARGE:
F
E
q
VECTOR
We define ELECTRICAL POTENTIAL as the
POTENTIAL ENERGY PER UNIT CHARGE:
U
V
q
SCALAR
UNITS OF POTENTIAL
U
Joules
V 
 VOLT
q Coulomb
Let’s move a charge from one point to
another via an external force.




The external force does
work on the particle.
The ELECTRIC FIELD
also does work on the
particle.
We move the particle
from point i to point f.
The change in kinetic
energy is equal to the
work done by the applied
forces. Assume this is
zero for now.
K  K f  K i  Wapplied  W field
if
K  0
Wapplied  W field
also
U  U f  U i  Wapplied
V 
Wapplied
q
Furthermore…
U Wapplied
V 

q
q
so
Wapplied  qV
If we move a particle through a potential
difference of V, the work from an external
“person” necessary to do this is qV
Example
Electric Field = 2 N/C

1 mC
d= 100 meters
Work done by EXTERNAL agent
 Change in potential Energy.
PE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
One Step More
Work done by EXTERNAL agent
 Change in potential Energy.
PE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
PE
Change in POTENTIAL 
q
2 104 Joules
J
V 
 200  200 Volts
6
110 C
C
Consider Two Plates
OOPS …
Look at the path issue
The difference in potential between the
accelerating plates in the electron gun of a
TV picture tube is about 25 000 V. If the
distance between these plates is 1.50 cm,
what is the magnitude of the uniform
electric field in this region?
An ion accelerated through a potential
difference of 115 V experiences an increase
in kinetic energy of 7.37 × 10–17 J. Calculate
the charge on the ion.
Important




We can define an absolute level of potential.
To do this, we need to define a REFERENCE
or ZERO level for potential.
For a uniform field, it doesn’t matter where
we place the reference.
For POINT CHARGES, we will see shortly
that we must place the level at infinity or the
math gets very messy!
An Equipotential Surface is defined as a
surface on which the potential is constant.
V  0
It takes NO work to move a charged particle
between two points at the same potential.
The locus of all possible points that require NO
WORK to move the charge to is actually a surface.
Example: A Set of Equipotenital Surfaces
Back To Yesteryear
Field Lines and Equipotentials
Electric
Field
Equipotential
Surface
Components
Enormal
Electric
Field
x
Eparallel
Work to move a charge a distance
x along the equipotential surface
Is Q x Eparallel X x
Equipotential
Surface
BUT




This an EQUIPOTENTIAL Surface
No work is needed since V=0 for such a
surface.
Consequently Eparallel=0
E must be perpendicular to the equipotential
surface
Therefore
E
E
E
V=constant
Field Lines are Perpendicular to the
Equipotential Lines
Equipotential
Work external  q0 (V f  Vi )
Consider Two Equipotential
Surfaces – Close together
Work to move a charge q from a to b:
dWexternal  Fapplied  ds  qEds
also
dWexternal  q(V  dV )  V   qdV
b
a
E
ds
 Eds  dV
V+dV
V
and
dV
E
ds
Vector...E  V
Where



  i  j k
x
y
z
Over a certain region of space, the electric
potential is V = 5x – 3x2y + 2yz2. Find
the expressions for the x, y, and z
components of the electric field over this
region. What is the magnitude of the field
at the point P that has coordinates (1, 0, –
2) m?
Typical Situation
Keep in Mind
W  F d


Force and Displacement are
VECTORS!
Potential is a SCALAR.
UNITS




1 VOLT = 1 Joule/Coulomb
For the electric field, the units of N/C can be
converted to:
1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM)
Or
1 N/C = 1 V/m

So an acceptable unit for the electric field is now
Volts/meter. N/C is still correct as well.
In Atomic Physics




It is sometimes useful to define an energy
in eV or electron volts.
One eV is the additional energy that an
proton charge would get if it were
accelerated through a potential
difference of one volt.
1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6
x 10-19 Joules.
Nothing mysterious.
Coulomb Stuff:
A NEW REFERENCE: INFINITY
Consider a unit charge (+) being brought from
infinity to a distance r from a
Charge q:
q
x
r
1
q
E
2
40 r
To move a unit test charge from infinity to the point
at a distance r from the charge q, the external force
must do an amount of work that we now can
calculate.
Just Do It!
OK, doing it!
VB  VA    E  ds
q
   k 2 runit  ds
r
rB
1 1
dr
 kq  2  kq  
r
 rB rA 
rA
Set the REFERENCE LEVEL OF POTENTIAL
at INFINITY so (1/rA)=0.
For point charges
qi
V

40 i ri
1
For a DISTRIBUTION of charge:
dq
V   k
r
volume
Ponder –
What is the potential a distance d from
an infinite plane whose charge per unit
area is s?
Given two 2.00-μC charges, as shown in Figure P25.16, and a
positive test charge q = 1.28 × 10–18 C at the origin, (a) what
is the net force exerted by the two 2.00-μC charges on the test
charge q? (b) What is the electric field at the origin due to the
two 2.00-μC charges? (c) What is the electrical potential at the
origin due to the two 2.00-μC charges?
The three charges in
Figure P25.19 are at the
vertices of an isosceles
triangle. Calculate the
electric potential at the
midpoint of the base,
taking q = 7.00 μC.
A disk of radius R has a non-uniform surface
charge density σ = Cr, where C is a constant
and r is measured from the center of the disk.
Find (by direct integration) the potential at P.
Example: Find potential at P
q1
q2
1 1
V
(q1  q2  q3  q4 )
40 r
d  1.3m
d
r
 0.919m
2
d
P
r
q3
q4
q1=12nC q2=-24nC q3=31nC q4=17nC Sq=36 x 10-9C
V=350 Volts (check the arithmetic!!)
An Example
finite line of charge
x
dx
r
1
dx
dV 
40 (d 2  x 2 )1/ 2
d
1
dx
V
40 0 (d 2  x 2 )1/ 2
L
P
and
What about a rod
that goes from –L to +L??

1
L  ( L2  x 2 )1/ 2
V
ln
40
d
At P
Using table of integrals

Example
z
R
disk
s=charge
per
unit
area
s
V
2 0
z
2

d

z
dz
R z
2
V
s
Ez  

z
2 0
s 
z
1 
Ez 
2 0 
z 2  R2
2
 R2  z




Which was the result we obtained earlier

In the figure, point P is at the center of the
rectangle. With V = 0 at infinity, what is the net
electric potential in terms of q/d at P due to the six
charged particles?
Continuing
1
2
3
5
6
s
4
qi
  2q  2q 3q  3q  5q  5q 
 k


r
d
/
2
1
.
12
d


i
i
q
q
  8q 10q 


V  k


k

8

8
.
93

0
.
93
k

d
1
.
12
d
d
d


q
V  8.35 x109
d
V  k
HRW gets 8.49 … one of us is right!
2
d 2 5d 2
d 
2
2
2
s  d    d 

2
4
4
 
d
s
5  1.12d
2
Derive an expression in terms of q2/a for the work required to set
up the four-charge configuration in the figure, assuming the
charges are initially infinitely far apart.
3
1
diagonal  a 2  1.71a
4
2