Transcript Lecture 3

Lecture 3: May 22nd 2009
Physics for Scientists and Engineers II
Physics for Scientists and Engineers II , Summer Semester 2009
Gauss’s Law (Karl Friedrich Gauss 1777-1855)
•
•
Example of surface of a cube in a constant electric field resulted in a net flux of zero
through that closed surface.
Today: We will look at a spherical closed surface (shell) concentric around a point
charge q and calculate the electric flux through that shell of radius r.
E parallel to d A everywhere on surface
E also has the same magnitude everywhere on the
dA
E
surface of the shell : E  k e
q
r2
 d E  Ed A  EdA
q
r
 E 
 Ed A   EdA  E  dA
shell
shell
 E 4r 2  k e
 E 
1
4o
4q 
shell
q
4r 2  ke 4q
2
r
q
o
fE is independent of r and proportional to q.
Physics for Scientists and Engineers II , Summer Semester 2009
Generalization
For non-spherical closed surfaces surrounding q, the net electric flux is the same as
for the spherical concentric shell (just harder to calculate). Example:
dA
E
r2
q
r1
 E 
 Ed A   EdA   EdA
surface
left half
righthalf
 Eleft 2r2  Eright 2r1  k e
2
 E 
Physics for Scientists and Engineers II , Summer Semester 2009
1
4o
4q 
2
q
o
q
q
2
2
2

r

k
2r1  ke 4q
2
e
2
2
r2
r1
Generalization
The electric flux through a closed surface surrounding a point charge q equals  E 
What if multiple charges are inside a surface?
 E d A   E
1

 E 2  E 3  E 4  ... d A
(superposi tion principle of E )
  E d A   E1 d A   E 2 d A   E 3 d A  ...

q1
0

q2
0

q3
0
 .... 
Physics for Scientists and Engineers II , Summer Semester 2009
qinside
0
q
o
Gauss’s Law
 E   Ed A 
Physics for Scientists and Engineers II , Summer Semester 2009
qin
o
Example: Electric Field due to an Infinitely Long Uniformly Charged
Rod…..doing it the hard way
y
dE
a
dq = l dx
P

r
x
x
l
Contributi on to E from dq :
l dx
sin  iˆ
2
r
l dx x
 ke 2 iˆ
r r
l x dx
 ke 3 iˆ
r
l x dx ˆ
 ke
i
3
2
2 2
a  x 
d E  ke
Physics for Scientists and Engineers II , Summer Semester 2009
l dx
cos  ˆj
2
r
l dx a ˆ
 ke 2
j
r r
l a dx
 ke 3 ˆj
r
l a dx ˆ
 ke
j
3
2
2 2
a  x 
 ke
y
dq = l dx
P
dE
a
r

x
x
l
d E  ke
Contributi on to E from dq :
a
l x dx
2
 x2

3
2
iˆ  ke
a
l a dx
2
 x2

3
ˆj
2

l x dx ˆ
l a dx
E     ke
i

k
e
3
3

2
2 2
2
2 2
 
a x
a x

Total electric field at point P :


E x  ke l 

a
x


x
2

2

3
dx
2
Physics for Scientists and Engineers II , Summer Semester 2009
E y  ke l a 

a

1
2
x
2

3
dx
2

ˆj 


….solving the integrals

E x  ke l 

a

x
x
2
2

3
2

Substituti on : x  a tan 
E y  ke l
90

90
a
E y  ke l 
dx  0
 dx 
a
2
 a tan  
2
2
3
2
a
d
cos 2 
a
d
cos 2 
 90
a
a
2
x
2

3
dx
2
(I wouldn' t expect you to know that)
90

ke l
1
1
d
3
2

2
2 cos 
a 90
1  tan 
 90
kl
ke l
1
1
 e 
d


 d
 cos
a 90  1  3 2 cos 2 
a 90



2
 cos  
kl
kl
2k l
 90
 e sin 90  e sin 90  sin  90  e
a
a
a
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Electric Field due to an Infinitely Long Uniformly Charged
Rod…..using Gauss’s Law
a
L
By symmetry : E x  0 and Eradial  const . at any point
that is radially a distance of " a" away from the line charge.
E is perpendicu lar to the cylindrica l round surface everywhere .
No flux goes through t he flat end surfaces of the cylinder.
Physics for Scientists and Engineers II , Summer Semester 2009
The charge enclosed in the cylinder is :
q inside  l L
Gauss' s Law applied :
E 

Ed A
cylinder surface
 E 2aL  
E 
 E  dA 
E 2aL  
qinside
0

lL
0
lL
0
lL
l
l

 2k e
2aL 0
2a 0
a
Question: Why would Gauss’s law be unpractical to use if we
picked a spherical surface in this particular problem?
Answer: The electric field vector is not constant along the spherical surface.
Physics for Scientists and Engineers II , Summer Semester 2009
Rules for Using Gauss’s Law to Evaluate the Electric Field
 E can be argued by symmetry t o be constant in magnitude and parallel to d A
over a part or all of the surface.
 Ed A  EdA
 All other surfaces have zero flux going through t hem ( E  d A  0).
This condition is fulfilled if either E  0, or if E  d A on those surfaces.
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Spherical Shell with Charge on it. Determine E(r)
Constant volume charge density.
Total charge = Q
Region 1: r < a
 E   Ed A 
qinside
0
0 E0
Region 2: a < r < b
a
b
Region 3: r > b
 E   Ed A 
E
qinside
0

Q
0
Q
Q

k
e
4r 2 0
r2
Physics for Scientists and Engineers II , Summer Semester 2009
 E   Ed A 
qinside
0
E
qinside
qinside

k
e
 0 4r 2
r2
4
4

qinside  Vinside     r 3   a 3 
3
3

Q
r 3  a3
4 3 4
3

  r   a Q 3
4
4
3
b  a3

3
33

 b  a 
3
3

Q r 3  a3
Q 
a3 
 r  2 
 E  ke 2 3
 ke 3
r b  a3
b  a 3 
r 
Conductors in Electrostatic Equilibrium
•
•
Electrostatic equilibrium in a conductor: The net motion of charges within the
conductor is zero.
Properties of conductors in electrostatic equilibrium:
 E = 0 inside the conductor (hollow or solid).
 Charged conductors: Charge is on the surface.
 Just outside the surface of the conductor:
E  surface
E

, where  is local surface charge density.
0
 Surface charge density is greatest where surface has smallest radius of
curvature. (…and therefore E is largest just outside those surface areas.)
Physics for Scientists and Engineers II , Summer Semester 2009
Conductors in Electrostatic Equilibrium
E = 0 inside the conductor (hollow or solid).
How do we know?
Consider an uncharged conductor
(or charged, it doesn’t matter)
…positive charge remains on the right side
as negative charge accumulates on the left…
…and place it into an electric field…
Fe
-
-
E
free electrons experience a force
(here, to the left side)
E extra
+
E
Charges keep separating and creating extra electric field.
….until extra field cancels the external electric field….
….and the separation of charges stops
(electrons on the inside no longer experience a net force.)
 Einside=0
Physics for Scientists and Engineers II , Summer Semester 2009
Conductors in Electrostatic Equilibrium
Charged conductors: Charge is on the surface.
How do we know?
Consider an charged conductor of some shape
We know : E  0 on the inside
Imagine a Gaussian closed surface
just beneath the actual surface.
Since E inside  0   E  0 through this surface
 no net charge inside of Gaussian surface  0
 entire net charge must reside on the surface
Physics for Scientists and Engineers II , Summer Semester 2009
Conductors in Electrostatic Equilibrium
E  surface
How do we know?
Suppose E was not perpendicular to surface
E
Then E would have a component parallel to the surface
 Motion of charges would occur along the surface (not yet in electrostatic equilibrium).
 In electrostatic equilibrium there cannot be a component of E parallel to the surface.
Physics for Scientists and Engineers II , Summer Semester 2009
Conductors in Electrostatic Equilibrium
E

, where  is the local surface charge density.
0
How do we know?
Consider a surface element of area A, small enough to be “flat”
E
Consider a Gaussian surface (e.g., cylinder with end caps parallel to surface, one end cap
On the inside, one just outside)
Calculate flux through the cylinder
 E  EA 
Physics for Scientists and Engineers II , Summer Semester 2009
qinside
0

A
0
 E

0
Chapter 25: Work Done by Electric Force – Small Displacement
•
Imagine a charge q0 is moved in an electric field by some small displacement:
E
Fe
ds
+ q0
Work done by electric force : WFe  F e  d s  q0 E  d s
Physics for Scientists and Engineers II , Summer Semester 2009
Chapter 25: Work Done by Electric Force – Along a Path
•
Imagine a charge q0 is moved in an electric field along some path
ds
E
Fe
+ q
Work done by electric force : WFe 
F
path
e
 d s  q0
 E ds
path
“path integral” or
“line integral”
Physics for Scientists and Engineers II , Summer Semester 2009
Work Done on Charge by Electric Force Along a Closed Path
•
Imagine a charge in an electric field is moved by external force as shown in a
closed path.
E
 x1
 x2
 x3
+
Fe
 x5
 x4
a
WFe 
F
e
 d s  F e   x1  F e   x 2  F e   x 3  F e   x 4  F e   x 5   Fe a  Fe a  0
path
(Work done by Fe  0 actually for any closed path)
•
In other words: Electrostatic force is a conservative force.
 An electric potential (a scalar quantity) can be defined.
(just like we did with the gravitational force / potential)
Physics for Scientists and Engineers II , Summer Semester 2009