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,
E
max 6 6
30 kV/cm
Ex.
d
2mm
V
doorknob
V
finger Arc discharge equalizes the potential
What is ΔV?
E
max
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.
r 1 r 2 Ball 1 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.
r 1 r 2 Ball 1 Ball 2
As V is increased, which ball will induce breakdown first?
(a) Ball 1 (b) Ball 2 (c) Same Time
E
surface
Ex.
V
k Q r
2 \
100 kV
V E
surface
Q k V r r
Smaller
r
higher
E
closer to breakdown
r
3 0.03m
3cm
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 10 8 m/s
Factoids:
V
~ 200 M volts
I
~ 40, 000 amp
t
~ 30ms
P
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)
V
A
> V
B b)
V
A
= V
B c)
V
A
< V
B
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
. Equipotential Last time…
E r
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
Therefore,
B A
E
d l
V
We can conclude then, that
E
0
d l
is zero. (it’s an equipotential!) 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
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. How do the potentials at the conductor surfaces compare now ? a)
V
A
> V
B b)
V
A
= V
B c)
V
A
< V
B
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)
Q
A increases
b) Q
A decreases
c) Q
A doesn’t change
Charge on Conductors?
•
How is charge distributed on the surface of a conductor?
–
KEY: Must produce surface .
E
=0 inside the conductor and
E
normal to the 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.
1A
Lecture 6, ACT 1
An uncharged spherical conductor has 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
(b)
Exactly q
(c)
More than q
1B 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
1A
Lecture 6, ACT 1
An uncharged spherical conductor has 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
(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). -q
1B
Lecture 6, ACT 1
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 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 non spherical conductor?? Claim largest charge density at smallest radius of curvature.
2 spheres, connected by a wire, “far” apart Both at same potential But:
r
S
r
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
Black ghost knife fish Dipole-like equipotentials More info: Prof. Mark Nelson, Beckman Institute, UIUC -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
V r
ˆ
dx V
+d
V dV
E
ˆ
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
V
A = 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
1A (a)
Lecture 6, ACT 2
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
V a
at the inner surface of the spherical shell?
(b) (c) 1B
Q a b
The electric potential in a region of space is given by (a) The
x
-component of the electric field
E x
at
x
= 2 is
E x
=
0
(b)
E x
> 0
(c)
E x
< 0
1A (a)
Lecture 6, ACT 2
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
V a
at the inner surface of the spherical shell?
(b) (c)
Q E
out
a b
•
How to start?? The only thing we know about the potential is its definition:
•
To calculate
• •
V a
, we need to know the electric field Outside the spherical shell: Apply Gauss’ Law to sphere:
E
•
Inside the spherical shell:
E
= 0
1B
Lecture 6, ACT 2
The electric potential in a region of space is given by (a) The
x
-component of the electric field
E x
at
x
= 2 is
E x
= 0
(b)
E x
> 0
(c)
E x
< 0
We know
V (x)
“everywhere” To obtain
E x
“everywhere”, use
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:
E
A B C 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
3A 3B
Lecture 6, ACT 3
Two test charges are brought separately to the vicinity of a positive charge
Q
.
– – –
Q
charge +
q
is brought to pt A , a distance
r
charge +2
q
from
Q
.
is brought to pt B , a distance 2
r
from
Q
.
Compare the potential energy of
q
(
U
A ) to that of 2
q
(
U
B ):
Q
(a)
U
A <
U
B (b)
U
A =
U
B
r
A (c)
U
A >
U
B
q
2
r
•
Suppose charge 2
q
has mass
m
and is released from rest from the above position (a distance 2
r
What is its velocity
v
f as it approaches
r
= from
Q
).
?
B 2
q
(a) (b) (c)
3A
Lecture 6, ACT 3
•
Two test charges are brought separately to the vicinity of positive charge
Q
.
Q
–
charge +
q
distance
r
is brought to pt A , a from
Q
.
–
charge +2
q
distance 2
r
is brought to pt B , a from
Q
.
(a)
–
Compare the potential energy of
q
to that of 2
q
(
U
B ): (
U
A )
U
A <
U
B (b)
U
A =
U
B (c)
Q r U
A >
U
B A
q
2
r
B 2
q
• • •
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
/r .
The potential energy of 2
q
is proportional to Q( 2
q
)/(2
r
) .
Therefore, the potential energies
U
A and
U
B are EQUAL!!!
3B
Lecture 6, ACT 3
•
Suppose charge 2
q
has mass
m
and is released from rest from the above position (a distance
2 r
What is its velocity
v
f as it approaches
r
= from
Q
).
?
(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) =
U
B .
Finally:
•
The charge has no potential energy (
•
U
The charge does have kinetic energy =
KE
1/
R
)
MKS:
U
=
QV
Energy Units
1 coul-volt for particles (e, p, ...) 1 e V = 1 joule = 1.6x10
-19 joules
Accelerators
•
Electrostatic: Van de Graaff electrons
100 ke V ( 10 5 e V )
•
Electromagnetic: Fermilab protons
1Te V ( 10 12 e V )
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.
+q a a
q
z -q
•
Rewrite this for special case
r
>>
a
:
r r
1
r
2
r
2 -r 1 Now we can use this potential to calculate the
E
field of a dipole (after a picture) (remember how messy the direct calculation was?)
Appendix A: Electric Dipole
+
q z a
q
a
-
q
•
Calculate
E
in spherical coordinates:
r r
1
r
2 the dipole moment
z
+
q a
q
a
-
q r
Appendix A: Dipole Field
y = Etot 0 E
q
E r 0
/
/
x =
•
Sample Problem
Consider the dipole shown at the right.
– –
Fix
r
=
r
0 >> a Define
q
max such that the polar component of the electric field has its maximum value (for r = r
0
).
What is
q
max ?
(a)
q
max = 0 (b)
q
max = 45
(c)
q
+
q
-
a a q
max
q
z
= 90
r r
1
r
2
•
The expression for the electric field of a dipole (
r
>>
a
) is:
•
The polar component of
E
is maximum when sin
q
is maximum.
•
Therefore,
E
q
has its maximum value when
q
= 90
.
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 field that is zero inside conductor!
+ + + + + by source charge: Induced charge distribution is “real” and sources
E
+
– –
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
E normal
(
r surface
)
E
(
r surface
)
s
(
r surface
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
determines
s .
E normal
(
r
surface
)
E
(
r
surface
)
s
(
r surface
o
)