Transcript PHY 184 lecture 8

PHY 184
Spring 2007
Lecture 8
Title: Calculations on Electrostatics
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184 Lecture 8
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Announcements
 Homework Set 2 is due Tuesday morning, January
23, at 8:00 am.
 Honors Option students will provide help in the SLC
starting this week.
 Today we will finish the electric field and begin
electric potential.
 We will start clicker questions today. More details
later during the lecture.
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Review – Gauss’s Law
  q
 E  dA 
S
0
q = net charge enclosed by S
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Review - Electric Fields from Charge Distributions
 The electric field E at distance r
from a wire with charge density
 is

2k
E

2 0r
r
 The electric field E produced by
an infinite non-conducting plate
with charge density  is

E
2 0
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Review - Electric Fields from Charge Distributions (2)
 The electric field E produced by an
infinite conducting plane with
charge density  is

E
0
 The electric field inside a spherical shell
of charge q is zero
 The electric field outside a spherical
shell of charge q is the same as the field
from a point charge q.
q
E k 2
r
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Review - Spherical Charge Distributions
Non-conducting sphere
Conducting sphere
Q
Q
+
+ R +
+
+
+ +
+ + + +
+ + +
+
+
+
+
+
+
E
E
E=0
R
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r
184 Lecture 8
R
r
6
Review - Electric Fields from a Ring of Charge
 The electric field E resulting from a ring of
charge (radius R, charge density =q/(2R)) on
the axis
ds
 Strategy: Imagine the ring is divided into
differential elements of charge dq=ds. Use
the electric field of a point charge for every
one of them.
dq
 ds
dE  k 2  k 2
r
R  z2
dE z  dE cos
E z (z)   k
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R
and
 ds z
2
z

2 3/ 2
z
cos 

R
R2  z2
kqz
2
z

2 3/ 2
184 Lecture 8
 kq/z2 for large z
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Example - Charge in a Cube
 Q=3.76 nC is at the center of a
cube. What is the electric flux
through one of the sides?
 Gauss’ Law:
  Q / 0
Q
 Since a cube has 6 identical sides and the point charge is at the center
oneface
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 1Q
Nm2
 
 (substitute the numerical values)  70.8
6 6 0
C
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Example - E Field and Force


The figure shows the defecting plates of an ink-jet printer. A
negatively charged ink drop (q=1.5 x 10-13 C) enters the region
between the plates with a velocity of v=18 m/s along x. The
length L of each plate is 1.6 cm. The plates are charged to
produce an electric field at all points between them (E=1.4 x
106 N/C). The vertical deflection of the drop at x=L is 0.64
mm. What is the mass of the ink drop?
Idea: A constant electrostatic force
of magnitude qE acts upward on the drop.
F qE
ay  
m m
… constant acceleration
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Example - E Field and Forces (2)
 What is the
mass of the ink
drop?
 Idea: Let t be the
time required to
pass through the
plates. Then…
y  12 at2
and
L  vt
2y
2yv 2
which implies a 
 2
2
(L / v )
L
qE
So... m 
 (substitute thenumericalvalues)  1.3  1010 kg
a
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Clicker Quizzes Starting Today
 You need a registered HITT clicker.
 Get up to 5% (but not more) extra credit according to Clicker’s Law
(you can miss 20% of the quizzes and still get the full extra credit)
 You can expect clicker questions each lecture.
 If you missed the clicker registration, fill in the clicker sheet.
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Induction, Conduction and Polarization
Consider three neutral metal spheres
in contact and on insulating stands.
Which diagram best
represents the
charge distribution
on the spheres
when a positively
charged rod is
brought near the
leftmost sphere
(without touching it)?
+
+ ++ ++
++-
++A
+
+ -+
+ -+
+
-+ -+
-+ -+
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++ + ++ + -
--B
+
+
+
+
C
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+
+
+
+
---
-+
-+
++
++
D
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Clicker Question - Enclosed Charge
 Shown is an arrangement of five charged pieces of plastic
(q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is
indicated. What is the net electric flux through the surface?
A: =-6 x 10-9C/0= -678 Nm2/C
B: = x 10-9C/0= -1356 Nm2/C
C: =0
D: =  x 10-9C/0= 328 Nm2/C
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Clicker Question - Enclosed Charge
 Shown is an arrangement of five charged pieces of plastic
(q1=q4=3nC, q2=q5=-5.9nC and q3=-3.1nC). A Gaussian surface S is
indicated. What is the net electric flux through the surface?
A: =-6x10-9C/0= -678 Nm2/C
enclosed charge
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Clicker Question - Flux
 Shown is a Gaussian surface in the form of a
cylinder of radius R and length L immersed in a
uniform electric field E. What is the flux of the
electric field through the closed surface?
A:
B:
C:
D:
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=2R2E
=R2E
=0
=(2RL+2R2)E
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Clicker Checkpoint - Flux
 Shown is a Gaussian surface in the form of a cylinder of radius R
and length L immersed in a uniform electric field E. What is the
flux of the electric field through the closed surface?
C: =0
Fluxes:
…left end = R2
…right end = +R2
…around cylinder = 0
…full flux = 0
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The Electric Potential
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Electric Potential
 We have been studying the electric field.
 Next topic: the electric potential
 Note the similarity between the gravitational force and the
electric force.
 Gravitation can be described in terms of a gravitational
potential and we will show that the electric potential is
analogous.
 We will see how the electric potential is related to energy
and work.
 We will see how we can calculate the electric potential from
the electric field and vice versa.
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Electric Potential Energy
 The electric force, like the gravitational force, is a conservative
force. (‡)
 When an electrostatic force acts between two or more charges
within a system, we can define an electric potential energy, U, in
terms of the work done by the electric field, We, when the
system changes its configuration from some initial configuration
to some final configuration.
Change in electric potential energy = -Work done by electric field
U  U f  U i  We
U i is the initial electric potential energy
U f is the final electric potential energy
(‡) Conservative force: The work is path-independent.
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Electric Potential Energy (2)
 Like gravitational or mechanical potential energy, we
must define a reference point from which to define the
electric potential energy.
 We define the electric potential energy to be zero when
all charges are infinitely far apart.
 We can then write a simpler definition of the electric
potential taking the initial potential energy to be zero,
U  Uf  0  U  W
 The negative sign on the work:
• If E does positive work then U < 0
• If E does negative work then U > 0
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Constant Electric Field
 Let’s look at the electric potential energy when we move a
charge q by a distance d in a constant electric field.
 The definition of work is
 
W  F d
 For a constant electric field the
force is F = qE …
 … so the work done by the electric field on the charge is
 
W  qE  d  qEd cos
Note:  = angle between E and d.
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Constant Electric Field - Special Cases
 Displacement is in the same
direction as the electric field
W  qEd
so
U  qEd
• A positive charge loses potential energy when
it moves in the direction of the electric field.
W  qEd
so
U  qEd
 Displacement is in the direction opposite to
the electric field
W  qEd
so
U  qEd
• A positive charge gains potential energy when
it moves in the direction opposite to the
electric field.
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Definition of the Electric Potential
 The electric potential energy of a charged particle in an electric
field depends not only on the electric field but on the charge of
the particle.
 We want to define a quantity to probe the electric field that is
independent of the charge of the probe.
 We define the electric potential as
U
V
q
“potential energy per unit charge of a test
particle”
 Unlike the electric field, which is a vector, the electric potential
is a scalar.
• The electric potential has a value everywhere in space but has no
direction.
Units: [V] = J / C, by definition, volt
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