Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1502: Lecture 4
Today’s Agenda
• Announcements:
– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #1:
– On Masterphysics today: due next Friday
– Go to masteringphysics.com and register
– Course ID: MPCOTE33308
• Labs: Begin next week
Today’s Topic :
• End of Chapter 21: Gauss’s Law
– Motivation & Definition
– Coulomb's Law as a consequence of Gauss' Law
– Charges on Insulators:
» Where are they?
• Chapter 22: Electric potential
– Definition
– How to compute it
Karl Friedrich Gauss
(1777-1855)
0  E dS  0  qenclosed
Infinite Line of Charge
• Symmetry  E field
must be ^ to line and
can only depend on
distance from line
y
Er
Er
•
Therefore, CHOOSE
+ + +++++++ + +++++++++++++ + + + + + +
Gaussian surface to be a
x
cylinder of radius r and length
h aligned with the x-axis.
h
• Apply Gauss' Law:
• On the ends, E  dS  0
• On the barrel,
 E  dS  2prhE
AND q = h

E

2 p 0 r
NOTE: we have obtained here the same result as we did last lecture
using Coulomb’s Law. The symmetry makes today’s derivation easier!
Gauss’ Law: Help for the Problems
• How to do practically all of the homework problems
• Gauss’ Law is ALWAYS VALID!!
• What Can You Do With This??
If you have (a) spherical, (b) cylindrical, or (c) planar symmetry
AND:
• If you know the charge (RHS), you can calculate the
electric field (LHS)
• If you know the field (LHS: usually because E=0 inside
conductor), you can calculate the charge (RHS).
Application of Gauss’ Law:
• Spherical Symmetry: Gaussian surface = Sphere of radius r
LHS:
RHS: q = ALL charge inside radius r
• Cylindrical Symmetry: Gaussian surface = Cylinder of radius r
LHS:
RHS: q = ALL charge inside radius r, length L
• Planar Symmetry: Gaussian surface = Cylinder of area A
LHS:
RHS: q = ALL charge inside cylinder=sA
Insulators vs. Conductors
• Insulators – wood, rubber, styrofoam, most
ceramics, etc.
• Conductors – copper, gold, exotic ceramics, etc.
• Sometimes just called metals
• Insulators – charges cannot move.
– Will usually be evenly spread
throughout object
• Conductors – charges free to move.
– on isolated conductors all charges
move to surface.
Conductors vs. Insulators
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E
E
-
+ + +
- +
+
+ + - + - +
+
+ -
-
+
+
+
+
+
+
+
- Ein = 0
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
E
-+
-+
E
-+
-+
-+ -+
Ein < E
-+ -+
-+
-+
-
-
Hollow conductors
Conductors & Insulators
• Consider how charge is carried on macroscopic objects.
• We will make the simplifying assumption that there are only two kinds
of objects in the world:
• Insulators.. In these materials, once they are charged, the charges
ARE NOT FREE TO MOVE. Plastics, glass, and other “bad
conductors of electricity” are good examples of insulators.
• Conductors.. In these materials, the charges ARE FREE TO MOVE.
Metals are good examples of conductors.
• How do the charges move in a
conductor??
• Hollow conducting sphere
Charge the inside, all of
this charge moves to the
outside.
Conductors vs. Insulators
++++++
++++++
-
-+
+
-
+ + - +
- -
+
+
+
+
- -
++++++
++++++
-
-
-+
-+
-+
-+
-+
-+
-+
-+
-
-
-+
Charges on a Conductor
• Why do the charges always move to the surface of
a conductor ?
– Gauss’ Law tells us!!
– E = 0 inside a conductor when in equilibrium (electrostatics) !
» Why?
If E  0, then charges would have forces on them and
they would move !
• Therefore from Gauss' Law, the charge on a
conductor must only reside on the surface(s) !
+
+
+
++++++++++++
+
+
1
+
++++++++++++
+
+
Infinite conducting
Conducting
plane
sphere
Lecture 4, ACT 1
s2
Consider the following two topologies:
A) A solid non-conducting sphere
carries a total charge Q = -3 mC
distributed evenly throughout.
It is surrounded by an
uncharged conducting
spherical shell.
B) Same as (A) but conducting
shell removed.
1A
-Q
E
• Compare the electric field at point X in cases A and B:
(a) EA < EB
1B
s1
(b) EA = EB
(c) EA > EB
• What is the surface charge density s1 on the inner surface of
the conducting shell in case A?
(a) s1 < 0
(b) s1 = 0
(c) s1 > 0
Lecture 4, ACT 2
• A line charge  (C/m) is placed along the axis
of an uncharged conducting cylinder of inner
radius ri = a, and outer radius ro = b as
shown.
– What is the value of the charge density so
(C/m2) on the outer surface of the
cylinder?
(a)
(b)
b
a

(c)
Electric Potential
V
Q
4p0 r
Q
4p0 R
R
r
R
C
R
B
r
B
q
r
A
A
path independence
equipotentials
Overview
• Introduce Concept of Electric Potential
– Is it well-defined? i.e. is Electric Potential a property of the space
as is the Electric Field?
• Calculating Electric Potentials
– Charged Spherical Shell
– N point charges
– Electric Dipole
• Can we determine the Electric Field if we know the
Electric Potential?
Text Reference: Chapter 22
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 4, ACT 3
•
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 p0 r 2
Q
4p0 R
Q
4p0 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!