Transcript Phys132 Lecture 5

Physics 1402: Lecture 5
Today’s Agenda
• Announcements:
– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #2:
– On Masterphysics today: due next Friday
– Go to masteringphysics.com and register
– Course ID: MPCOTE92051
• Labs: Begin next week
Today’s Topic :
• Chapter 23: Electric potential
– Definition
– How to compute it: E from V
» Example: dipole
– Equipotentials and Conductors
– Electric Potential Energy
» of Charge in External Electric Field
Electric Potential
V
Q
4pe0 r
Q
4pe0 R
R
r
R
C
R
B
r
B
q
r
A
A
path independence
equipotentials
Electric Potential
• Suppose charge q0 is moved from pt
A to pt B through a region of space
described by electric field E.
q0
A
E
B
• Since there will be a force on the charge due to E, a certain
amount of work WAB will have to be done to accomplish this
task. We define the electric potential difference as:
• Is this a good definition?
• Is VB - VA independent of q0?
• Is VB - VA independent of path?
Independent of Charge?
Fwe supply = -Felec
• To move a charge in an E field, we must
supply a force just equal and opposite to Felec
q0
that experienced by the charge due to
E
the E field.
A
B

1
Lecture 5, ACT 1
•
A single charge ( Q = -1mC) is fixed at the origin.
Define point A at x = + 5m and point B at x = +2m.
– What is the sign of the potential difference
between A and B? (VAB  VB - VA )
(a) VAB < 0
(b) VAB = 0
B

-1mC
(c) VAB > 0
A

x
Independent of Path?
Felec
-Felec
q0
E
A
B
• This equation also serves as the definition for the potential
difference VB - VA.
•The integral is the sum of the tangential (to the path)
component of the electric field along a path from A to B.
•The question now is: Does this integral depend upon the exact
path chosen to move from A to B?
•If it does, we have a lousy definition.
• Hopefully, it doesn’t.
• It doesn’t. But, don’t take our word, see appendix and
following example.
Does it really work?
• Consider case of constant
field:
– Direct: A - B
B
hQ
A
C
r
E
dl
• Long way round: A - C - B
• So here we have at least one example of a case in which the integral
is the same for BOTH paths.
Electric Potential
• Define the electric potential of a point in space as the potential
difference between that point and a reference point.
• a good reference point is infinity ... we typically set V = 0
• the electric potential is then defined as:
• for a point charge, the formula is:
Potential from charged spherical shell
V
• E Fields (from Gauss' Law)
• r < R:
E=0
• r > R:
E=
• Potentials
• r > R:
• r < R:
1 Q

4 pe0 r 2
Q
4pe0 R
Q
4pe0 r
R
R
R
r
Potential from N charges
r1
The potential from a collection of N
charges is just the algebraic sum of
the potential due to each charge
separately.

x
q1
q2
r2
r3
q3
Electric Dipole
The potential is much easier to
calculate than the field since it is an
algebraic sum of 2 scalar terms.
z
+q
aq
a
r1
r
r2-r1
-q
• Rewrite this for special case r>>a:

Can we use this potential somehow to calculate
the E field of a dipole?
(remember how messy the direct calculation was?)
r2
Appendix: Independent of Path?
• We want to evaluate potential difference
from A to B
• What path should we choose to
evaluate the integral?.
• If we choose straight line, the
integral is difficult to evaluate.
• Magnitude different at each pt
along line.
• Angle between E and path is
different at each pt along line
E
B
r
B
q
A
C
E
A
.
• If we choose path ACB as shown,
our calculation is much easier!
• From A to C, E is perpendicular
to the path. ie
• From A to C, E is perpendicular to
the path. ie
r
B
r
B
q
r
A
A
Appendix: Independent of Path?
• Evaluate potential difference from
A to B along path ACB.
by definition:
Evaluate the integral:
C
E
B
r
B
q
r
A
A
Appendix: Independent of Path?
C
B
r
B
r
q
• How general is this result?
•
A
B
Consider the approximation to the straight
path from A->B (white arrow) = 2 arcs (radii =
r1 and r2) plus the 3 connecting radial pieces.
• For the 2 arcs + 3 radials path:
A
r2
q
r1
A
This is the same result as above!!
The straight line path is better
approximated by Increasing the number
of arcs and radial pieces.
Appendix: Independent of Path?
B
r2
q
r1
A
• Consider any path from A to B as being made up of
a succession of arc plus radial parts as above. The
work along the arcs will always be 0, leaving just the
sum of the radial parts. All inner sums will cancel,
leaving just the initial and final radii as above..
Therefore it's general!
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r<a)
• Where do we know the potential,
and where do we need to know it?
V=0 at r= ... we need r<a ...
• Determine E(r) for all regions
in between these two points
II
III
IV
uncharged
conductor
I
r
a
c
b
sphere with
charge Q
• Determine DV for each region by integration
... and so on ...
• Check the sign of each potential difference DV
DV > 0 means we went “uphill”
DV < 0 means we went “downhill”
(from the point of view
of a positive charge)
E from V?
• We can obtain the electric field E from the potential V by
inverting our previous relation between E and V:
• Expressed as a vector, E is the negative gradient of V
• Cartesian coordinates:
• Spherical coordinates:
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:
Electric Dipole
• Last time, we derived for r>>a:
z
+q
r1
r
aq
a
-q
• Calculate E in spherical coordinates:
the dipole moment

r2
The Bottom Line
If we know the electric field E everywhere,
allows us to calculate the potential function V everywhere
(define VA = 0 above)
If we know the potential function V everywhere,
1
allows us to calculate the electric field E everywhere.
Units for Potential! 1 Joule/Coulomb = 1 VOLT
z
Lecture 5, ACT 2
• Consider the dipole shown at the right.
– Fix r = r0 >> a
– Define qmax such that the polar
component of the electric field has its
maximum value (for r=r0).
1
What is qmax?
(a) qmax = 0
(b) qmax = 45
+q
r1
r
aq
a
-q
(c) qmax = 90
r2
Dipole Field
y=
Etot
z
+q
a q
r
E
q
0
a
Er
-q
0
p/
p
Q
p
p/
x=
Equipotentials
Defined as: The locus of points with the same potential.
•
Example: for a point charge, the equipotentials are
spheres centered on the charge.
• GENERAL PROPERTY:
– The Electric Field is always perpendicular to an
Equipotential Surface.
• Why??
The gradient (  ) says E is in the direction of
max rate of change.
Along the surface, there is NO change in V
(it’s an equipotential!)
So, there is NO E component along the surface
either… E must therefore be normal to surface
Dipole
Equipotentials
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!!
•
Note
Positive charges move from regions of higher potential to lower
potential (move from high potential energy to lower PE).
Equilibrium means charges rearrange so potentials equal.
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.
A Point Charge Near
Conducting Plane
q
+
a
V=0
- - - - - - -- - - - - - -- --- - - - - - - - - - - - - - - - - - -
A Point Charge Near
Conducting Plane
q
+
a
The magnitude of the force is
q2
F
4pe0 2a 2
1
Image Charge
The test charge is attracted to a conducting plane
Charge on Conductor
• 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
rL
rS
But:

Smaller sphere
has the larger
surface charge
density !
Equipotential Example
•
Field lines more closely
spaced near end with most
curvature .
•
Field lines ^ to surface near
the surface (since surface is
equipotential).
•
Equipotentials have similar
shape as surface near the
surface.
•
Equipotentials will look more
circular (spherical) at large r.
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.
St Elmo’s Fire
Van de Graaff Generator
Schematic view of a classical Van de
Graaff generator.
1) hollow metal sphere
2) upper electrode
3) upper roller (for example in acrylic
glass)
4) side of the belt with positive charges
5) opposite side of the belt with negative
charges
6) lower roller (metal)
7) lower electrode (ground)
8) spherical device with negative charges,
used to discharge the main sphere
9) spark produced by the difference of pot
Lightning
Energy Units
MKS:
U = QV 
for particles (e, p, ...)
1 coulomb-volt = 1 joule
1 eV
= 1.6x10-19 joules
Accelerators
• Electrostatic: VandeGraaff
electrons  100 keV ( 105 eV)
• Electromagnetic: Fermilab
protons  1TeV ( 1012 eV)