Physics 1502: Lecture 5 Today’s Agenda • Announcements: – Lectures posted on: www.phys.uconn.edu/~rcote/ – HW assignments, solutions etc. • Homework #2: Homework #2: – On Masterphysics today: due next On Masterphysics today: due next Friday Friday – Go to masteringphysics.com and register – Course ID: MPCOTE33308 • Labs: Begin next week
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Physics 1502: Lecture 5Today’s Agenda
• Announcements:– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #2:Homework #2:– On Masterphysics today: due next FridayOn Masterphysics today: due next Friday
– Go to masteringphysics.com and register
– Course ID: MPCOTE33308
• Labs: Begin next week
Today’s Topic :• Chapter 22: 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
qA
C
B
rA
Br
path independence equipotentials
R
R
R r
VQ
4 rQ
4 R
Electric Potential• Suppose charge q0 is moved from pt
A to pt B through a region of space described by electric field E.
• 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?
A B
q0E
Independent of Charge?
• To move a charge in an E field, we must supply a force just equal and opposite to that experienced by the charge due to the E field.
Lecture 5, ACT 1• A single charge ( Q = -1C) 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 < (b) VAB = (c) VAB >
x-1C AB
Independent of Path?
A B
q0 EFelec
-Felec
• 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
• 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.
A
CB
Eh rdl
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
• E Fields (from Gauss' Law)
• Potentials
• r > R:
• r < R:
R
R
Q4 R
R r
VQ
4 r
E = 0• r < R:
E = 1
4 Qr 2
• r > R:
Potential from N charges
The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately.
xr1
r2 r3
q1
q3
q2
Electric Dipole z
a
a
+q
-q
r
r1r2
The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms.
• 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-r1
Appendix: Independent of Path?
qA
C
B
rA
Br
E
• We want to evaluate potential difference from A to B
qA
B
rA
Br
E
• 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.
• 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
Appendix: Independent of Path?• Evaluate potential difference from
A to B along path ACB.
qA
C
B
rA
Br
E
Evaluate the integral:
by definition:
Appendix: Independent of Path?
• How general is this result?• Consider the approximation to the straight
path from A->B (white arrow) = 2 arcs (radii = r1 and r2) plus the 3 connecting radial pieces.
qA
B
This is the same result as above!!The straight line path is better
approximated by Increasing the number of arcs and radial pieces.
qA
B
rA
Br
C
r1
r2
• For the 2 arcs + 3 radials path:
Appendix: Independent of Path?
• 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!
qA
B
r1
r2
a b
cunchargedconductor
sphere withcharge Q
Calculating Electric PotentialsCalculate the potential V(r)
at the point shown (r<a)
• Where do we know the potential, and where do we need to know it?
• Determine E(r) for all regions in between these two points
• Determine V for each region by integration
• Check the sign of each potential difference V
V > 0 means we went “uphill”V < 0 means we went “downhill”
III
IIIIV
r
... and so on ...
(from the point of viewof a positive charge)
V=0 at r= ... we need r<a ...
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
• 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:
• Consider the following electric potential:
Electric Dipole z
a
a
+q
-q
r
r1r2• Last time, we derived for r>>a:
• Calculate E in spherical coordinates:
the dipole moment
The Bottom LineIf 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,
allows us to calculate the electric field E everywhere.