Transcript PowerPoint Presentation - Lecture 1 Electric Charge*

Lecture 3 Gauss’s Law Ch. 23
• Physlet
http://webphysics.davidson.edu/physletprob/ch9_problems/
ch9_2_gauss/default.html
• Topics
–Electric Flux
–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, electrometer
• Elmo
1
–Problems 29
Examples of
Field Lines
2
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
r
where A  An̂
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r r
Electric Flux    (E cos )A  E  A
1.
 
E A
So,
0
  EA cos 0  EA
A
2.
 
E A
r
where A  An̂
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
5
Derivation of Gauss’ Law from a point
charge and using Coulombs law:
Summary of steps
• Start with an isolated point charge.
• Choose a sphere around the charge.
• Calculate
net
r r
—
 E  dA
• Show that
 net 
qenc
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
kq
For a Point charge E  2
r
En
cos 0  1
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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Valid even when charges are not at the center
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
9
Remember
E
nˆ
  EA
A

nˆ
E
A
  En A  0  A  0
En  E cos 


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
2
What would be the flux if the
cylinder were vertical ?
Suppose it were any
shape?
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Flux from a. + b. + c. 0as expected
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 these things
1
2
3
4
5
6
7
8
9
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 σ.
Find electric field of an infinitely long uniformly line of charge λ.
Find E inside and outside of a long non-conducting solid cylinder
of uniform charge density ρ.
Find E for a thin cylindrical shell of surface charge density σ
Find E inside and outside a solid non-conducting sphere of
uniform charge density ρ.
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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
surface.  0
• Draw a small pill box that extends
into the conductor. Since there is
no field inside, all the flux comes
out through the top.
EA 
q
0

E
0

A
0
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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
length L that is uniformly charged.
n̂
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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
 4 3, r R
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 nonconducting
solid sphere filled with a uniform charge
distribution
Er  r
Er
Er 
1
r2
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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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Problem 23 -4
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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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More questions to ask your self
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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