Transcript L2 Gauss

Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Irina Segel
Week 2. Gauss’s law – Scalar and vector fields • why fields? • ring
of charge • disk of charge • lines of electric field • electric dipole in
a constant field • flux • Gauss’s law • applications
Sources: Halliday, Resnick and Krane, 5th Edition, Chaps. 26-27;
Purcell (Berkeley course 2), Chap. 1, Sects. 1.8 – 1.13.
Scalar and vector fields
Here are two examples of scalar fields:
Time-dependent wave field ·
Instantaneous temperature field ·
In how many spatial dimensions
do these scalar fields “live”?
Scalar and vector fields
Here are two examples of vector fields:
Electric dipole field ·
Wind field ·
A vector field has a direction and a
magnitude at every point in space.
Why fields?
F21
m1
F12 
Gm1m2
r12 
2
F12
m2
Why fields?
“That gravity should be innate, inherent
and essential to matter so that one body
may act upon another at a distance
through a vacuum without the mediation
of anything else, by and through which
their action or force may be conveyed
from one to another, is to me so great an
absurdity that I believe no man who has
in philosophical matters any competent
faculty of thinking can ever fall into it.”
– Sir Isaac Newton
F21
m1
F12 
Gm1m2
r12 
2
F12
m2
Why fields?
Why fields?
“Newton himself was better aware of the
weaknesses inherent in his intellectual
edifice than the generations of learned
scientists who followed him. This fact
has always aroused my deep admiration.”
– Albert Einstein
Ring of charge
What is the electric field of a uniform ring of radius R and linear
charge density λ along its axis of symmetry?
z
θ
0
Ring of charge
What is the electric field of a uniform ring of radius R and linear
charge density λ along its axis of symmetry?
z
θ
0
Answer: E( z )  zˆ
1
2R
cos
4 0 R  z
1
Rz
 zˆ
2 0 [ R 2  z 2 ]3 / 2
2
2
.
Disk of charge
What is the electric field of a uniform disk of radius R, and
surface charge density λ, along its axis of symmetry?
z
θ
0
Disk of charge
What is the electric field of a uniform disk of radius R, and
surface charge density λ, along its axis of symmetry?
R
z
θ
0
d
z
Answer: E( z )  zˆ
2 0 [  2  z 2 ]3 / 2
0

.
Disk of charge
What is the electric field of a uniform disk of radius R, and
surface charge density λ, along its axis of symmetry?
R
z
d
z
Answer: E( z )  zˆ
2 0 [  2  z 2 ]3 / 2
0

θ
0
Substitute ρ = z tan θ
dρ = z dθ/cos2 θ
0 ≤ θ ≤ tan–1R/z
.
Disk of charge
What is the electric field of a uniform disk of radius R, and
surface charge density λ, along its axis of symmetry?
z
Answer:
E( z ) 
θ

0
zˆ
2 0
zˆ
2 0
tan 1 R / z

 d tan cos
0
tan 1R/z
0
  cos 
 
z
1 

2
2
2 0 
R

z

Substitute ρ = z tan θ
dρ = z dθ/cos2 θ
0 ≤ θ ≤ tan–1R/z

 zˆ .

Disk of charge
What is the electric field of a uniform disk of radius R, and
surface charge density λ, along its axis of symmetry?
z
θ
 
z
1 
E( z ) 
2
2
2 0 
R

z


 zˆ

For R → ∞ we get

E( z ) 
zˆ
2 0
0
while for R → 0 we get back the electric
field of a point charge q. (We must take
the limit with nonvanishing q = λπR2.)
Lines of electric field
The great experimental physicist
Faraday, in his revolutionary papers on
electromagnetism, did not write down a
single equation! How did he do it?
He visualized the electric (and
magnetic) fields.
Particles…elementary particles…
I used to study them…calculate
them…theorize about them…but
now I can see them!
based on drawing and text
Lines of electric field
The great experimental physicist
Faraday, in his revolutionary papers on
electromagnetism, did not write down a
single equation! How did he do it?
He visualized the electric (and
magnetic) fields.
Lines of electric field
The great experimental physicist
Faraday, in his revolutionary papers on
electromagnetism, did not write down a
single equation! How did he do it?
He visualized the electric (and
magnetic) fields.
Lines of electric field
The great experimental physicist
Faraday, in his revolutionary papers on
electromagnetism, did not write down a
single equation! How did he do it?
He visualized the electric (and
magnetic) fields.
Lines of electric field
Some basic rules and observations about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
Lines of electric field
Some basic rules and observations about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
Lines of electric field
Some basic rules and observations about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
E
density
Electric dipole in a constant field
A constant electric field is the same E everywhere. So the force
on a charge q is everywhere Fq = qE.
Suppose we have a rigid dipole – two equal and opposite charges
of magnitude q separated by a rigid (assumed massless) rod of
length d. Since the dipole has a direction, we can represent it by
a vector p which points in the direction of the positive charge and
has magnitude p = qd.
What happens to the dipole in a constant field E?
q
E
θ
–q
E
Electric dipole in a constant field
A constant electric field is the same E everywhere. So the force
on a charge q is everywhere Fq = qE.
Suppose we have a rigid dipole – two equal and opposite charges
of magnitude q separated by a rigid (assumed massless) rod of
length d. Since the dipole has a direction, we can represent it by
a vector p which points in the direction of the positive charge and
has magnitude p = qd.
What happens to the dipole in a constant field E?
q
Fq
θ
F-q
–q
Electric dipole in a constant field
A constant electric field is the same E everywhere. So the force
on a charge q is everywhere Fq = qE.
Suppose we have a rigid dipole – two equal and opposite charges
of magnitude q separated by a rigid (assumed massless) rod of
length d. Since the dipole has a direction, we can represent it by
a vector p which points in the direction of the positive charge and
has magnitude p = qd.
What happens to the dipole in a constant field E?
1. Since the charges are equal and opposite, the force is zero.
2. The torque τ on the dipole is τ = p × E.
3. The potential energy of the dipole is U = –p·E.
Electric dipole in a constant field
3. The potential energy of the dipole is U = –p·E:
θ
F-q
Fq
–q
Let’s start at θ = 90°. The displacement parallel to the force is
(d/2) cos θ, so the work is 2F (d/2) cos θ = qEd cos θ = pE cos θ
= p·E. The potential energy U is minus the work, so U = –p·E.
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 6:
+
R
y
θ
E
+
+
+
+
+
+
+
+ +
+
+
This semi-infinite bar has charge λ per unit
length. Prove that θ = 45° regardless of R.
x
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 6:
+
R
y
θ
E
+
+
+
+
+
+
+
+ +
+
+
This semi-infinite bar has charge λ per unit
length. Prove that θ = 45° regardless of R.
1
Solution: E x 
4 0
Ey 
1

xdx
 [ R 2  x 2 ]3 / 2
0

Rdx
4 0  [ R 2  x 2 ]3 / 2
0
,
.
x
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 6:
+
R
y
θ
E
+
+
+
+
+
+
+
+ +
+
+
This semi-infinite bar has charge λ per unit
length. Prove that θ = 45° regardless of R.
1
Solution: E x 
4 0
Ey 
1

xdx
 [ R 2  x 2 ]3 / 2
0

Rdx
4 0  [ R 2  x 2 ]3 / 2
,
.
0
We substitute x = R tanφ, dx = R dφ/cos2φ and integrate.
x
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 6:
+
R
y
+
+
+
+
+
+
+ +
+
+
x
This semi-infinite bar has charge λ per unit
length. Prove that θ = 45° regardless of R.
θ
E
+
Solution: E x 
Ey 

4 0 R

4 0 R
 /2

sin d 
0
 /2

4 0 R

 cos d  4 0 R
0
,
.
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 14:
Find the frequency of small oscillations of a dipole of strength p
and moment of inertia I around its equilibrium position in a
constant electric field E.
θ
F-q
–q
Fq
Halliday, Resnick and Krane, 5th Edition, Chap. 26, Prob. 14:
Find the frequency of small oscillations of a dipole of strength p
and moment of inertia I around its equilibrium position in a
constant electric field E.
θ
F-q
Fq
–q
Solution: The equilibrium position is θ = 0. The magnitude of
the torque τ = p × E is pE sinθ ≈ pEθ so
 I
d 2
dt 2
  pE and θ = A sin (ωt+ θ0) where ω =
pE
.
I
Flux
dA
E
Definition of an element of electric flux: E·dA
Electric field E
Oriented area element dA
The direction of dA is always perpendicular to the surface.
In this example, E·dA is (E)(dA) cos 145°.
Flux
E
E
dA
E
dA
dA
E
Definition of an element of electric flux: E·dA
Electric field E
Oriented area element dA
The direction of dA is always perpendicular to the surface.
In this example, 0 ≤ E·dA ≤ (E)(dA).
Flux
The total flux through a surface, denoted ФE, is the sum of all the
Electric field E
flux elements: ФE =
E  dA .
Oriented area element dA
Flux through back wall: –E(10 cm)(30 cm).
Flux through upper wall: E(cos 60º)(30 cm)(10 cm )/(sin 30º).
Gauss’s law

 E  E  dA 
q
0
The total flux ФE (the sum of all the fluxes) through a closed
surface equals the total charge inside the surface divided by ε0.
Gauss’s law

 E  E  dA 
q
0
The total flux ФE (the sum of all the fluxes) through a closed
surface equals the total charge inside the surface divided by ε0.
If there is no charge inside a
closed surface, then all the field
lines entering must also leave!
Then the total flux is zero. ·
Gauss’s law

 E  E  dA 
q
0
ФE > 0
Gauss’s law

 E  E  dA 
q
0
ФE = 0
Gauss’s law

 E  E  dA 
q
0
ФE < 0
Applications
Let’s use Gauss’s law to derive the electric field of a point charge
q. Consider a sphere of radius r centered on the point charge. By
symmetry, E must be radial and E must be constant and so

 E  E  dA  ( E )(4 r 2 )
.
Now we apply Gauss’s law,
q
0
  E  4E r 2 ,
and immediately obtain E = q/4πε0r2, which is Coulomb’s law.
Applications
Gauss’s law makes it easy to obtain the electric field when there
is symmetry to help us. For example, what is the electric field of
an infinite rod with uniform charge density λ? Let the rod be the
axis of a cylinder of radius ρ and height h; by symmetry, we have
h q

  E  E  dA  ( E )(2h) ,
0 0

so E 
.
2 0
E

E
Applications
Similarly, we can use Gauss’s law to calculate the electric field of
a plane with uniform charge density σ. If we take a box of any
cross-sectional area A and enclose part of the plane in the box, the
electric flux will be zero on the sides of the box and 2EA through
the top and bottom of the box. Thus
E
A q


  E  E  dA  2 EA , so E 
0 0
2 0

.
Gauss’s law
• Gauss’s law is often more convenient than Coulomb’s law.
• Gauss’s law implies Coulomb’s law.
• Coulomb’s law is true only for static charges (just as Newton’s
law for the gravitational force is true only for static masses) but
Gauss’s law is true also for moving charges.
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 6:
P
This infinite plane has a uniform charge density σ, except that a
disk of radius R is missing. The point P is on the symmetry axis
at a height z above the plane. What is the electric field EP at P?
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 6:
P
z

zˆ .
Answer: Including the disk, we would have E plane 
2 0
The field due to the disk is E disk


2 0

z
1 

R2  z2

 zˆ .

Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 6:
P
z
The electric field is a vector field, hence EP + Edisk = Eplane and
the field at the point P is EP = Eplane – Edisk :

 
z
1 
EP 
zˆ 
2
2
2 0
2 0 
R

z



 zˆ 
2 0

z
R2  z2
zˆ .
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 19:
A conducting sphere carrying a charge Q is surrounded by a
spherical conducting shell.
(a) What is the total charge on the inner surface of the shell?
(b) A point charge q is placed outside the shell. Now what is the
total charge on the inner surface of the shell?
(c) Now the point charge q is between the shell and the sphere.
What is the total charge on the inner surface of the shell?
(d) Are the answers the same if the sphere and shell are not
concentric?
Q
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 19:
(a) What is the total charge on the inner surface of the shell?
Answer: E vanishes everywhere inside a conductor. (Otherwise
electrons would rearrange themselves in order to make it vanish.)
Hence the flux through the surface shown vanishes. Hence the
total charge inside the surface vanishes. Hence the charge on the
inner surface of the shell must be –Q.
Q
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 19:
(b) A point charge q is placed outside the shell. Now what is the
total charge on the inner surface of the shell?
Answer: E still vanishes everywhere inside a conductor. Hence
the flux through the surface shown still vanishes. Hence the total
charge inside the surface still vanishes. Hence the charge on the
inner surface of the shell must still be –Q. The point charge
outside can’t change these facts.
q
Q
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 19:
(c) Now the point charge q is between the shell and the sphere.
What is the total charge on the inner surface of the shell?
Answer: E still vanishes everywhere inside a conductor. Hence
the flux through the surface shown still vanishes. Hence the total
charge inside the surface still vanishes. But now the charge
inside the inner surface of the shell is Q + q. Hence the charge
on the inner surface of the shell must be –Q – q.
Q
q
Halliday, Resnick and Krane, 5th Edition, Chap. 27, Prob. 19:
(d) Are the answers the same if the sphere and shell are not
concentric?
Answer: Yes – we never assumed that the sphere and shell were
concentric.
Q