Transcript electric potential

POTENTIAL
February 1, 2009
We complete Gauss’s Law
 We begin the topic of POTENTIAL –
Chapter 25.
 Problem Session Wednesday Morning
 Examination #1 is on Friday – 2/13/09

◦ Exam includes Chapters 23,24 and as much of
25 as we cover today (Monday)

Next Week – More Potential
This Week
A particle with charge Q is located at the center of a
cube of edge L. In addition, six other identical negative
charged particles -q are positioned symmetrically
around Q as shown in the figure below. Determine the
electric flux through one face of the cube.
A slab of insulating material has a nonuniform positive charge density ρ
= Cx2, where x is measured from the center of the slab, as shown in
the figure below, and C is a constant. The slab is infinite in the y and z
directions.
Derive expressions for the electric field for
the following regions.
A solid insulating sphere of radius a carries a net positive charge 3Q,
uniformly distributed throughout its volume. Concentric with this sphere is
a conducting spherical shell with inner radius b and outer radius c, and
having a net charge -Q, as shown in the figure below.
mc
A solid insulating sphere of radius a carries
a net positive charge 3Q, uniformly
distributed
throughout
its
volume.
Concentric with this sphere is a conducting
spherical shell with inner radius b and
outer radius c, and having a net charge -Q,
as shown in the figure below.
(a) Construct a spherical Gaussian surface of radius r > c and find the net charge
enclosed by this surface.
(b) What is the direction of the electric field at r > c?
(c) Find the electric field at r > c.
(d) Find the electric field in the region with radius r where c > r > b.
(e) Construct a spherical Gaussian surface of radius r, where c > r > b, and find the
net charge enclosed by this surface.
(f) Construct a spherical Gaussian surface of radius r, where b > r > a, and find the
net charge enclosed by this surface.
(g) Find the electric field in the region b > r > a.
(h) Construct a spherical Gaussian surface of radius r < a, and find an expression for
the net charge enclosed by this surface, as a function of r. Note that the charge inside
this surface is less than 3Q.
(i) Find the electric field in the region r < a.
(j) Determine the charge on the inner surface of the conducting shell.
(k) Determine the charge on the outer surface of the conducting shell.
(l) Make a plot of the magnitude of the electric field versus r.
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B
q
A
We have a region in space where
there is an Electric Field
• There is a particle of charge q at some
location.
• The particle must be moved to another
spot within the field.
• Work must be done in order to
accomplish this.
Electric Potential

We will be dealing with



Work
Energy & Conservation
Work must be done to move a charge in
an electric field.

Let’s do a weird demo ….
I need some help.
What we will do ….
E
+
charge
Mrs. Fields




Mr. External
For the moment, assume the charge has MASS. (It may not.)
Assume the charge is initially stationary.
The charge is to be moved to the left.
The charge is to be moved at CONSTANT velocity.
Start and Stop



ENERGY is required to bring the charge up
to speed (if it has mass).
ENERGY is required to bring the particle
back to rest (if it has mass).
The sum of these two is ZERO.
Each does the negative
amount of work than the
other does.
So, when we move a charge in an
Electric Field ..


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.
Summary-
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.
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.
The change in potential energy of a charge
is the amount of work that is done by an
external force in moving the charge from its
initial position to its new position.
It is the negative of the work done by the
FIELD in moving the particle from the initial
to the final position.
Definition – Potential Energy


PE or U is the work done by an external
agent in moving a charge from a
REFERENCE POSITION to a different
position.
A Reference ZERO is placed at the most
convenient position

Like the ground level in many gravitational
potential energy problems.
Example:
Work by External Agent
Wexternal = F  d = qEd= U
E
d
Zero Level
q
F
Work done by the Field
is:
Wfield= -qEd
= -Wexternal
A uniform electric field of magnitude 290 V/m is directed
in the positive x direction. A +13.0 µC charge moves from
the origin to the point (x, y) = (20.0 cm, 50.0 cm).(a) What
is the change in the potential energy of the charge field
system?
[-0.000754] J
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 defined an absolute level of potential.
To do this, we needed to define a
REFERENCE or ZERO level for potential.
For a uniform field, it didn’t matter where we
placed 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
V+dV
V
 Eds  dV
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
2
2 0 
z R
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