Transcript PowerPoint Presentation - Lecture 1 Electric Charge*

Make-up from last time
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Electric field from line of charge
Electric field from arc of charge
Van de Graaff Demos & cage, needle, electroscope
Smoke Remover
Problem of electron in an electric field
Todays Material
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What is the electric field from an infinitely long wire
with linear charge density of +100 nC/m at a point 10
away from it.
++++++++++++++++++++++++++++++++
y =10 cm
.
Ey 
2k
sin  0
y
Ey 

0=90 for an
infinitely long wire
Ey
2  8.99  10 9 Nm 2  100  10 9
0.1m
C
m sin 90  1.8 *10 4 N
C
Typical field for the electrostatic smoke remover
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Smoke Remover
Negatively charged central wire has electric field that varies as 1/r (strong electric
field gradient). Field induces a dipole moment on the smoke particles. The
positive end gets attracted more to the wire.
In the meantime a corona discharge is created. This just means that induced dipole
moments in the air molecules cause them to be attracted towards the wire
where they receive an electron and get repelled producing a cloud of ions
around the wire.
When the smoke particle hits the wire it receives an electron and then is repelled to
the side of the can where it sticks. However, it only has to enter the cloud of ions
before it is repelled.
It would also work if the polarity of the wire is reversed
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What is the field at the
center due to arc of charge
L /2
Ey 
 dE
y
0
L / 2
dEx= k dq cos /r2
dEx= k ds cos /r2
E x  k
L /2
 rd cos /r
2
 k /r
L / 2
Ex 
0
 d cos
s=r 
ds=r d
 0
2k
sin 0
r
What is the field at the center
of a circle of charge? 0=180
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Example illustrating the motion of a charged particle in
an electric field:
An electron is projected perpendicularly to a downward
electric field of E= 2000 N/C with a horizontal velocity
v=106 m/s. How much is the electron deflected after
traveling 1 cm.
V
y
•
E
d
E
y =1/2(qE/m)t2
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Lecture 3 Gauss’s Law Ch. 23
• Cartoon - Electric field is analogous to gravitational field
• Physlet /webphysics.davidson.edu/physletprob
• Topics
–Electric Flux
–Electric Flux example
–Gauss’ Law
–Coulombs Law from Gauss’ Law
–Isolated conductor and Electric field outside conductor
–Application of Gauss’ Law
–Conductor in a uniform field
• Demo
–Faraday Ice pail: metal cup, charge ball,teflon rod, silk,electroscope
• Elmo
–Problems 29
• Polling and Clickers
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Electric Flux
Electric flux is the number of Electric field lines penetrating a
surface or an area. Call it  .
r r
Electric Flux    (E cos )A  E  A


where A  A  n̂
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

where A  A  n̂
r r
Electric Flux    (E cos )A  E  A
1.
 
E A
So,
0
  EA cos 0  EA
A
2.
 
E A
E
0
Let   45 Then,
  EA cos 45  0.707EA
A

E
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Gauss’s Law - Powerful relationship between
charges and electric fields through closed surfaces
r qenc
—
 E  dA 
0
• Gauss’s law makes it possible to find the electric field easily in
highly symmetric situations.
•
If there is no charge inside, then the flux is 0
 net  0
• Let look at an example
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Find the electric flux through a cylindrical surface in
a uniform electric field E
Uniform electric field
Cylindrical surface choose because of symmetry
No charge inside
Flux should be 0
 
We want to calculate
  E  dA

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Remember
E
nˆ
  EA
A

nˆ
E
A
  En A  0  A  0


E
nˆ

A 
  En A  E cos  A
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
 
   E  dA
a.

 E cos180dA    EdA  ER
b.

 E cos90dA  0
c.
   E cos 0dA   EdA  ER 2
Flux from a. + b. + c.  as expected
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What would be the flux if the
cylinder were vertical ?
Suppose it were any
shape?
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Derivation of Gauss’ Law from a point charge
and using Coulombs law: Summary of steps
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Start with an isolated point charge.
Choose a sphere around the charge.
r r
Calculate   —
 E  dA
qenc
Show that net  
net
0
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Start with a point charge.
Note that the electric lines of flux are radial
for a point charge
Choose a spherical surface to
simplify the calculation of the flux
E
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• Calculate the net flux through the closed surface.
Net Flux =
net
r r
—
 E  dA 
 E cosdA   EdA
En
cos 0  1
kq
For a Point charge E  2
r

kq
   EdA   2 dA
r
nˆ
kq
kq
  2  dA  2 ( 4r 2 )
r
r
  4kq
4k 
1
0
net 
where  0  8.85x10
qenc
0
Gauss’ Law
dA
12
dA  nˆdA

C2
Nm2

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Gauss’ Law
 net 
qenc
0
This result can be extended to any shape surface
with any number of point charges inside and
outside the surface as long as we evaluate the
net flux through it.
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Approximate Flux
 
   E  A
Exact Flux
 
   E  dA
dA  nˆdA
Circle means you integrate
over a closed surface.

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You should know how to prove these
things
1
2
3
4
5
6
7
8
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The electric field inside a conductor is 0.
The total net charge inside a conductor is 0. It resides on the
surface.
Find electric field just outside the surface of a conductor.
Find electric field around two parallel flat conducting planes.
Find electric field of a large non-conducting sheet of charge.
Find electric field of an infinitely long uniformly line of charge.
Find E inside and outside of a long non-conducting solid
cylinder of uniform charge density.
Find E for a thin cylindrical shell of surface charge density .
Find E inside and outside a solid non-conducting sphere of
uniform charge density .
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1. Electric field inside a conductor is 0.
Why?
• Inside a conductor in electrostatic
equilibrium the electric field is always
zero.
(averaged over many atomic volumes)
• The electrons in a conductor move
around so that they cancel out any
electric field inside the conductor
resulting from free charges anywhere
including outside the conductor. This
results in a net force of 0 on any
particular charge inside the conductor.
r
F  qE
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2. The net charge inside a conductor is 0.
• Any net electric charge resides
on the surface of the conductor
within a few angstroms (10-10 m).
• Draw a Gaussian surface just
inside the conductor. We know
everywhere on this surface E=0.
• Hence, the net flux is zero. From
Gauss’s Law the net charge inside is
zero.
• Show Faraday ice pail demo.
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3. Find electric field just outside the surface
of a conductor
• The electric field just outside a conductor has
magnitude  and is directed perpendicular to the
0
surface.
– Draw a small pill box that extends
into the conductor. Since there is
no field inside, all the flux comes
out through the top.
A
EA  
0 0

E
0
q
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4. Find electric field around two parallel flat
conducting planes.
1
E
0
E
1
0
E
2 1
0
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6. Find the electric field for a very long wire of
q
length L that is uniformly charged.

L
q
—
 E dA  
n
Gauss's Law
0
endcaps+sides

En  E  n̂
n̂
sides
—
 E dA  E —
 dA  E  2 rh
n
  endcaps  side
h
 0
 E  2rh  
0
2k 
E
r̂
r
2k 
E
r

E  n̂
  90
Cos90  0

E
20r
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9. Electric field inside and outside a nonconducting solid uniformly charged sphere
• Often used as a model of the nucleus.
• Electron scattering experiments have shown that the charge
density is constant for some radius and then suddenly drops
off at about
2  3 1014 m.
For the nucleus,
26 C

  10
m3
  Charge
density per
unit volume
R
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 10 14 m
Electric Field inside and outside a uniformly charged sphere
Q= Total charge
Q
 4 3, r R
3 R
r̂
= Z  1.6  10-19C
Inside the sphere:
To find the charge at a distance r<R
Draw a gaussian surface of radius r
By symmetry E is radial and parallel to
normal at the surface. By Gauss’s Law:

q
—
 E dA  
n
0
3
4

r
q
E  4r 2   3
0
0

r
E
30
r
E
r̂
3 0
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Electric Field inside and outside a uniformly charged sphere

Q
, r R
3
4
3 R
Q= Total charge
= Z  1.6  10-19C
Outside the sphere:

3
4

R
q
E  4r 2   3
0
0

R 3
E
3 0 r 2
Same as a point charge q
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Electric field vs. radius for a conducting sphere
Er  r
Er
Er 
1
r2
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Comment on Sample Problem 23 -4
A charge of - 5uC is placed inside a neutral conducting shell at R/2.
What is the charge on the inner and outer surface of the shell?
How is it distributed?
What are the field lines inside and outside the shell?
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Comment on Sample Problem 23 -4
Charge on the inner of the shell is +5 uC? Outer surface is - 5 uC
How is it distributed? Uniformly on outer surface, but not on inner surface.
What are the field lines inside and outside the shell? As shown on the inside.
Like a point charge of 5 uC at the center of the sphere on the outside.
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Chapter 23 Problem 29 Elmo
A long straight wire has a fixed negative charge with a linear
charge density of magnitude 3.1 nC/m. The wire is to be
enclosed by a coaxial, thin walled, nonconducting cylindrical
shell of radius 1.8 cm. The shell is to have positive charge on its
outside surface with a surface charge density that makes the net
external electric field zero. Calculate the surface charge density.
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Here is a essay question you should be able to
answer based on our ideas so far
1) Suppose you put a neutral ideal conducting solid sphere in a
region of space in which there is, initially, a uniform electric
field.
a) Describe (as specifically as possible) the electric field inside the
conductor.
b) Describe the distribution of charge in and on the conductor.
c) Describe the electric field lines at the surface of the conductor.
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External Electric field penetrates conductor
E
Free charges inside conductor move around onto the
surface in such a way as to cancel out the field inside.
E
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E=0 inside conducting sphere
E
E
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Orientation of electric field is perpendicular to
surface
E
If the electric field were not perpendicular, then there would be a
component of electric field along the surface causing charges to
accelerate on the surface and it would not be in electrostatic equilibrium
contrary to experience.
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2) Suppose you put some charge on an initially-neutral, solid,
perfectly-conducting sphere (where the sphere is not in a
pre-existing electric field).
– Describe the electric field inside the conductor,
– Describe the electric field at the surface of the conductor, and outside
the conductor as a result of the unbalanced charge.
– Describe the distribution of the charge in and on the conductor.
3) Repeat questions 1-3 for the case of a hollow perfectlyconducting spherical shell (with the interior being vacuum).
4) How would your answers to questions 1-4 change if the
conductor had some shape other than spherical?
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