Transcript Document

Electric Potential
Equipotentials and Energy
Today…
•
•
•
•
Equipotentials and conductors
E from V
Calculate electric field of dipole from potential
Electric Potential Energy
– of charge in external electric field
– stored in the electric field itself (next time)
• Appendix:
– Example calculation of a spherical charge configuration
– Calculate electric field of dipole from potential
Text Reference: Chapter 24.3,5 and 25.1
examples: 24.7,11,13,15 and 25.1
Sparks
• High electric fields can ionize nonconducting materials
(“dielectrics”)
Dielectric
Insulator
Conductor
Breakdown
• Breakdown can occur when the field is greater than the
“dielectric strength” of the material.
– E.g., in air,
Emax  3 106 N/C  3 106 V/m  30 kV/cm
What is ΔV?
d  2mm
Ex.
Vdoorknob
Vfinger
Arc discharge equalizes
the potential
V  Emax  d
 30 kV/cm• 0.2 cm
 6 kV
Note: High humidity can also bleed the charge off  reduce ΔV.
Question 1
Two charged balls are each at the same potential V. Ball 2 is
twice as large as ball 1.
r1
Ball 1
r2
Ball 2
As V is increased, which ball will induce breakdown first?
(a) Ball 1
(b) Ball 2
(c) Same Time
Question 1
Two charged balls are each at the same potential V. Ball 2 is
twice as large as ball 1.
r1
Ball 1
r2
Ball 2
As V is increased, which ball will induce breakdown first?
(a) Ball 1
Esurface
Q
k 2
r
\
(b) Ball 2
(c) Same Time
Q
r
V
Smaller r  higher E  closer to breakdown

r
V k
Esurface
Ex. V  100 kV
100 103 V
r
 0.03m  3cm
6
3 10 V/m
High Voltage Terminals must be big!
Lightning!
+
_
+
_
+
_
Collisions produce
charged particles.
The heavier
particles (-) sit near
the bottom of the
cloud; the lighter
particles (+) near
the top.
Stepped
Leader
Negatively
charged
electrons
begin
zigzagging
downward.
Attraction
As the stepped
leader nears
the ground, it
draws a
streamer of
positive charge
upward.
Flowing
Charge
As the leader
and the
streamer come
together,
powerful
electric current
begins flowing
Contact!
Intense wave of
positive charge,
a “return stroke,”
travels upward
at 108 m/s
t ~ 30ms
Factoids: V ~ 200 M volts
I ~ 40,000 amp
P ~ 1012 W
Question 2
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has a
larger radius than conductor B.
A
B
Compare the potential at the surface of conductor A with
the potential at the surface of conductor B.
a) VA > VB
b) VA = VB
c) VA < VB
Potential from a charged sphere
• The electric field of the charged
sphere has spherical symmetry.
• The potential depends only on
the distance from the center of the
sphere, as is expected from
spherical symmetry.
• Therefore, the potential is
constant along a sphere which is
concentric with the point charge.
These surfaces are called
equipotentials.
• Notice that the electric field is
perpendicular to the equipotential
surface at all points.
Er
Equipotential
Last time…
(where V ()  0 )
Equipotentials
Defined as: The locus of points with the same potential.
•
Example: for a point charge, the equipotentials are spheres centered on
the charge.
The electric field is always perpendicular
to an equipotential surface!
Why??
From the definition of potential
Along the surface, there is NO change in V (it’s an equipotential!)
Therefore,
 
  E  dl  V  0
B
A
 
We can conclude then, that E  dl is zero.
If the dot product of the field vector and the displacement vector is zero,
then these two vectors are perpendicular, or the electric field is always
perpendicular to the equipotential surface.
Conductors
+
+
+
+
+
+
+
+
+
+
+
+
+
+
• Claim
The surface of a conductor is always an equipotential
surface (in fact, the entire conductor is an equipotential).
• Why??
If surface were not equipotential, there would be an electric
field component parallel to the surface and the charges
would move!!
Question 3
B
A
The same two conductors that were in Question 2 are now
connected by a wire, before they each carried the same positive
charge Q. How do the potentials at the conductor surfaces
compare now ?
a) VA > VB
b) VA = VB
c) VA < VB
Question 4
A
B
The same two conductors that were in Question 2 are now
connected by a wire, before they each carried the same positive
charge Q. What happens to the charge on conductor A after it is
connected to conductor B ?
a) QA increases
b) QA decreases
c) QA doesn’t change
Charge on Conductors?
• How is charge distributed on the surface of a conductor?
– KEY: Must produce E=0 inside the conductor and E normal to the
surface .
Spherical example (with little off-center charge):
+ + +
+
- -- +
- +
+ -+q - +
+ - +
+ - +
+
+ + +
E=0 inside conducting shell.
charge density induced on
inner surface non-uniform.
charge density induced on
outer surface uniform
E outside has spherical
symmetry centered on spherical
conducting shell.
Lecture 6, ACT 1
An uncharged spherical conductor has
1A a weirdly shaped cavity carved out of
it. Inside the cavity is a charge -q.
How much charge is on the cavity wall?
(a) Less than< q
1B
(b) Exactly q
(c) More than q
How is the charge distributed on the cavity wall?
(a) Uniformly
(b) More charge closer to –q
(c) Less charge closer to -q
1C
How is the charge distributed on the outside of the sphere?
(a) Uniformly
(b) More charge near the cavity
(c) Less charge near the cavity
-q
Lecture 6, ACT 1
An uncharged spherical conductor has
1A a weirdly shaped cavity carved out of
it. Inside the cavity is a charge -q.
-q
How much charge is on the cavity wall?
(a) Less than< q
(b) Exactly q
(c) More than q
By Gauss’ Law, since E=0 inside the conductor, the
total charge on the inner wall must be q (and
therefore -q must be on the outside surface of the
conductor, since it has no net charge).
Lecture 6, ACT 1
1B
How is the charge distributed on the cavity wall?
(a) Uniformly
(b) More charge closer to -q
(c) Less charge closer to -q
-q
The induced charge will distribute itself nonuniformly to
exactly cancel E everywhere in the conductor. The surface
charge density will be higher near the -q charge.
Lecture 6, ACT 1
1C
How is the charge distributed on the outside of the
sphere?
(a) Uniformly
(b) More charge near the cavity
(c) Less charge near the cavity
-q
As in the previous example, the charge will be uniformly
distributed (because the outer surface is symmetric).
Outside the conductor the E field always points directly to
the center of the sphere, regardless of the cavity or
charge.
Note: this is why your radio, cell phone, etc. won’t
work inside a metal building!
Charge on Conductor Demo
• How is the charge distributed on a nonspherical conductor?? Claim largest charge
density at smallest radius of curvature.
• 2 spheres, connected by a wire, “far” apart
• Both at same potential
r
rS
But:

L
Smaller sphere
has the larger
surface charge
density !
Equipotential Example
• Field lines more closely
spaced near end with most
curvature – higher E-field
• Field lines ^ to surface
near the surface (since
surface is equipotential).
• Near the surface,
equipotentials have similar
shape as surface.
• Equipotentials will look
more circular (spherical) at
large r.
Electric Dipole Equipotentials
•First, let’s take a look at the equipotentials:
Electric Fish
Some fish have the ability to
produce & detect electric fields
• Navigation, object detection,
communication with other electric
fish
• “Strongly electric fish” (eels) can
stun their prey
Dipole-like equipotentials
More info: Prof. Mark Nelson,
Beckman Institute, UIUC
Black ghost knife fish
-Electric current flows down the voltage gradient
-An object brought close to the fish alters the
pattern of current flow
E from V?
• We can obtain the electric field E from the
potential V by inverting our previous relation
between E and V:

r

r  xˆ dx
V
V+dV

dV   E  xˆ dx   E x dx
• Expressed as a vector, E is the negative gradient of V
• Cartesian coordinates:
• Spherical coordinates:
Preflight 6:
This graph shows the electric
potential at various points
along the x-axis.
8) At which point(s) is the electric field zero?
A
B
C
D
E from V: an Example
• Consider the following electric potential:
• What electric field does this describe?
... expressing this as a
vector:
• Something for you to try:
Can you use the dipole potential to obtain the
dipole field? Try it in spherical coordinates ...
you should get (see Appendix):
The Bottom Line
If we know the electric field E everywhere,

allows us to calculate the potential function V everywhere (keep
in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere,
allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT
2
Lecture 6, ACT 2
1A
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
a
Q
b
the inner surface of the spherical shell?
(b)
(a)
1B
(c)
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
(c)
Ex < 0
Eout
Lecture 6, ACT 2
1A
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
the inner surface of the spherical shell?
(a)
(b)
a
Q
b
(c)
• How to start?? The only thing we know about the potential is
its definition:
• To calculate Va, we need to know the electric field E
• Outside the spherical shell:
• Apply Gauss’ Law to sphere:
• Inside the spherical shell:
E=0
Lecture 6, ACT 2
1B
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
We know V(x) “everywhere”
To obtain Ex “everywhere”, use
(c)
Ex < 0
Electric Potential Energy
• The Coulomb force is a CONSERVATIVE force
(i.e. the work done by it on a particle which
moves around a closed path returning to its
initial position is ZERO.)
• Therefore, a particle moving under the influence of
the Coulomb force is said to have an electric
potential energy defined by:
this “q” is the ‘test charge”
in other examples...
• The total energy (kinetic + electric potential) is then
conserved for a charged particle moving under the
influence of the Coulomb force.
3
Preflight 6:
A
E
C
B
6) If a negative charge is moved from point A to point B, its
electric potential energy
a) increases
b) decreases
c) doesn’t change
Lecture 6, ACT 3
3A
Two test charges are brought
separately to the vicinity of a
positive charge Q.
Q
r
q
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
Q
distance 2r from Q.
– Compare the potential energy of q
(UA) to that of 2q (UB):
(a) UA < UB
3B
(b) UA = UB
A
2r
2q
(c) UA > UB
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ?
(a)
(b)
B
(c)
Lecture 6, ACT 3
3A
• Two test charges are brought
separately to the vicinity of positive
charge Q.
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
distance 2r from Q.
Q
r
q
A
Q
2r
2q
B
– Compare the potential energy of q (UA)
to that of 2q (UB):
(a) UA < UB
(b) UA = UB
(c) UA > UB
•Look familiar?
•This is ALMOST the same as ACT 2 from the last lecture.
• In that ACT, we discovered that the potential at A was TWICE the
potential at B. The point was that the magnitudes of the charges at A and
B were IRRELEVANT to the question of comparing the potentials.
• The charges at A and B are NOT however irrelevant in this ACT!!
• The potential energy of q is proportional to Qq/r.
• The potential energy of 2q is proportional to Q(2q)/(2r).
• Therefore, the potential energies UA and UB are EQUAL!!!
Lecture 6, ACT 3
3B
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ?
(a)
(b)
(c)
• What we have here is a little combination of 111 and 112.
• The principle at work here is CONSERVATION OF ENERGY.
• Initially:
• The charge has no kinetic energy since it is at rest.
• The charge does have potential energy (electric) = UB.
• Finally:
• The charge has no potential energy (U  1/R)
• The charge does have kinetic energy = KE
Energy Units
MKS:
U = QV
1 coul-volt
for particles (e, p, ...)
1 eV
= 1 joule
= 1.6x10-19 joules
Accelerators
• Electrostatic: Van de Graaff
electrons  100 keV ( 105 eV)
• Electromagnetic: Fermilab
protons  1TeV ( 1012 eV)
Summary
• Physically, V is what counts
• The place where V=0 is “arbitrary” (at infinity)
• Conductors are equipotentials
• Find E from V:
• Potential Energy


E   V
U  qV
• Next time, capacitors:
Reading assignment: 25.2, 4
Examples: 25.2,3,5,6 and 7
Appendix A: Electric Dipole
The potential is much easier to
calculate than the field since it is
an algebraic sum of 2 scalar
terms.
z
+q
a
r1
r
q
a
r2-r1
-q
• Rewrite this for special case r>>a:

Now we can use this potential to calculate the E
field of a dipole (after a picture)
(remember how messy the direct calculation was?)
r2
Appendix A: Electric Dipole
z
+q
aq
a
-q
• Calculate E in spherical coordinates:
the dipole
moment

r1
r
r2
Appendix A: Dipole Field
y=
z
+q
a q
a
Etot
r
E
q
0
Er
-q
0
/



/
x=
Sample Problem
• Consider the dipole shown at the
right.
– Fix r = r0 >> a
z
+q
– Define qmax such that the polar
component of the electric field has its
maximum value (for r = r0).
a q
a
What is qmax?
-q
(a) qmax = 0
(b) qmax = 45
r1
r
(c) qmax = 90
• The expression for the electric
field of a dipole (r >> a) is:
• The polar component of
E is maximum when sinq is maximum.
• Therefore, Eq has its maximum value when q = 90.
r2
Appendix B: Induced charge distribution on
conductor via
“method of images”
• Consider a source charge
brought close to a
conductor:
+
-
+
+
+
+
• Charge distribution
- +
“induced” on conductor
by source charge:
• Induced charge distribution is “real” and sources Efield that is zero inside conductor!
– resulting E-field is sum of field from source charge
+
and induced charge distribution
+
-
– E-field is locally perpendicular to surface
+
+
+
– just like the homework problem.
• With enough symmetry, can solve for s on conductor
– how? Gauss’ Law



Enormal ( rsurface )  E ( rsurface ) 
+
+
- +
- +
- +

s ( rsurface )
o
Appendix B: Induced charge distribution on
conductor via
“method of images”
• Consider a source charge brought
close to a planar conductor:
-
• Charge distribution “induced” on
conductor by source charge
– conductor is equipotential
+
-
-
– E-field is normal to surface
– this is just like a dipole
• Method of Images for a charge (distribution) near a flat
conducting plane:
– reflect the point charge through the surface and put a
charge of opposite sign there
– do this for all source charges
– E-field at plane of symmetry - the conductor surface 
 
s ( rsurface )

determines s.
Enormal ( rsurface )  E ( rsurface ) 
o