Welcome…
…to Physics 104.
Prologue
Some things you will recall from “last”* semester…
 Newton’s Laws
 F  ma
 energy and its conservation
1
KE  mv 2
2
1
2
U spring  k x  x o 
2
E  KU
U grav  mgy
Ef  Ei   Wother if
*or whenever you took your previous physics class
 momentum and its conservation (linear and angular)
p  mv
Lz  Iz
Pf  Pi
L z,f  L z,i
These “things” aren’t going to go away!
Electric Charge
Static Electricity
There are two kinds of charge.
+
-
Properties of charges
 like charges repel
 unlike charges attract
 charges can move but charge is conserved
“Law” of conservation of charge: the net amount of electric
charge produced in any process is zero.
Coulomb’s Law
Coulomb’s “Law” quantifies the magnitude of the electrostatic
force.
Coulomb’s “Law” gives the force (in Newtons) between charges
q1 and q2, where r12 is the distance in meters between the
charges, and k=9x109 N·m2/C2.
q1q 2
F k 2
12
r12
Force is a vector quantity. The equation on the previous slide
gives the magnitude of the force. If the charges are opposite in
sign, the force is attractive; if the charges are the same in sign,
the force is repulsive. Also, the constant k is equal to 1/40,
where 0=8.85x10-12 C2/N·m2.
I could write Coulomb’s “Law” like this…
q1q 2
F  k 2 , attractive for unlike
12
r12
repulsive for like
Remember, a vector has a magnitude and a direction.
The equation is valid for point charges. If the charged objects
are spherical and the charge is uniformly distributed, r12 is the
distance between the centers of the spheres.
r12
-
+
If more than one charge is involved, the net force is the vector
sum of all forces (superposition). For objects with complex
shapes, you must add up all the forces acting on each separate
charge (turns into calculus!).
+
+
+
-
-
-
We could have agreed that in the formula for F, the symbols q1
and q2 stand for the magnitudes of the charges. In that case,
the absolute value signs would be unnecessary.
However, in later equations the sign of the charge will be
important, so we really need to keep the magnitude part.
On your diagrams, show both the magnitudes and signs of q1
and q2.
Your starting equation is this version of the equation:
q1q 2
F k 2 ,
12
r12
which gives you the magnitude F12 and tells you that you need
to figure out the direction separately.
Solving Problems Involving Coulomb’s “Law”
and Vectors
You may wish to review vectors (on your own).
Example: Calculate the net electrostatic force on charge Q3
due to the charges Q1 and Q2.
y
30 cm
Q3=+65C
=30º
Q1=-86C
Q2=+50C
52 cm
x
Step 0: Think!
This is a Coulomb’s “Law” problem (all we have to work with, so
far).
We only want the forces on Q3. Don’t worry about other forces.
Forces are additive, so we can calculate F32 and F31 and add
the two.
If we do our vector addition using components, we must resolve
our forces into their x- and y-components.
Step 1: Diagram
y
Draw and label relevant
quantities—done.
30 cm
F32
Draw a representative
sketch—done.
Q3=+65C
F31
Draw axes, showing
origin and directions— Q2=+50C
done.
=30º
Q1=-86C
52 cm
Draw and label forces (only those on Q3).
Draw components of forces which are not along axes.
x
Step 2: Starting Equation
y
F32
30 cm
Q3=+65C
F31
=30º
Q1=-86C
Q2=+50C
52 cm
q1q 2
F k 2
12
r12
<complaining> “Do I have to put in the absolute value signs?”
Yes. Unless you like losing points.
x
Step 3: Replace Generic Quantities by Specifics
y
repulsive
Q3 Q 2
F
k 2
32, y
r32
Q3=+65C
r32=30 cm
Q3Q 2
F k 2 ,
32
r32
F32
F31
=30º
Q1=-86C
Q2=+50C
x
52 cm
F
 0 (from diagram)
32, x
Can you put numbers in at this point? OK for this problem. You
would get F32,y = 330 N and F32,x = 0 N.
Step 3 (continued)
attractive
Q3Q1
F
  k 2 cos 
31, x
r31
(+ sign comes from
diagram)
Q3Q1
F
 k 2 sin 
31, y
r31
y
F32
Q3=+65C
r32=30 cm
Q3Q1
F k 2 ,
31
r31
F31
=30º
Q1=-86C
Q2=+50C
x
52 cm
(- sign comes from diagram)
Can you put numbers in at this point? OK for this problem. You
would get F31,x = +120 N and F31,y = -70 N.
Step 3: Complete the Math
y
Q3=+65C
30 cm
The net force is the
vector sum of all the
forces on Q3.
F32
F3
F31
=30º
Q1=-86C
Q2=+50C
52 cm
F3x = F31,x + F32,x = 120 N + 0 N = 120 N
F3y = F31,y + F32,y = -70 N + 330 N = 260 N
You know how to calculate the magnitude F3 and the angle
between F3 and the x-axis. (If not, holler!)
x
I did a sample Coulomb’s law calculation using three point
charges.
How do you apply Coulomb’s law to objects that contain
distributions of charges?
We’ll use another tool to do that…
Today’s agendum:
The electric field.
You must be able to calculate the force on a charged particle in an electric field.
Electric field of due to a point charge.
You must be able to calculate electric field of a point charge.
Motion of a charged particle in a uniform electric field.
You must be able to solve for the trajectory of a charged particle in a uniform electric field.
The electric field due to a collection of point charges.
You must be able to calculate electric field of a collection of point charges.
The electric field due to a continuous line of charge.
You must be able to calculate electric field of a continuous line of charge.
Coulomb’s “Law”:
The Big Picture
Coulomb's “Law” quantifies the interaction between charged
particles.
1 q1q 2
F =
,
2
12 4πε 0 r12
r12
+
-
Q1
Q2
Coulomb’s “Law” was discovered through decades of
experiment.
By itself, it is just “useful." Is it part of something bigger?
Gravitational Fields
You experienced gravitational fields in Physics 103.
m1m 2
F =G 2 , attractive
G
r12
FG
g(r) =
m
Units of g are
actually N/kg!
g(r) is the local gravitational field. On earth, it is 9.8 N/kg,
directed towards the center of the earth. What we called
g = 9.8 m/sec2 is the magnitude of the gravitational field.
The Electric Field
Coulomb's “Law” (demonstrated in 1785) shows that charged
particles exert forces on each other over great distances.
How does a charged particle "know" another one is “there?”
Action At A Distance Viewpoint
Electric, magnetic, and gravitational forces result of direct and
instantaneous interaction between particles.
Relativity theory shows why this viewpoint is wrong. Faraday
developed the correct explanation.
Faraday, beginning in 1830's, was the leader in developing the
idea of the electric field. Here's the idea:
F12
 A charged particle emanates a "field"
into all space.
 Another charged particle senses the field,
and “knows” that the first one is there.
+
+
F21
like
charges
repel
F13
F31
unlike
charges
attract
The idea of an electric field is good for a number of reasons:
 It makes us feel good, like we’ve
actually explained something.
OK, that was a flippant remark. There are serious reasons
why the idea is “good.”
 We can develop a theory based on this
idea. From this theory may spring
unimagined inventions.
If the theory explains past observations and leads to new
predictions, the idea was “good.”
F12
+
+
F21
like
charges
repel
F13
F31
unlike
charges
attract
We define the electric field by the force it exerts on a test
charge q0:
F0
E=
q0
The subscript “0” reminds you the force is on the
“test charge.” I won’t require the subscripts when
you use this equation for boardwork or on exams.
This is your second starting equation. By convention the direction of the electric field
is the direction of the force exerted on a POSITIVE test charge. The absence of
absolute value signs around q0 means you must include the sign of q0 in your work.
If the test charge is "too big" it perturbs the electric field, so the
“correct” definition is
F0
E = lim
q 0 0 q
0
You won’t be required to use
this version of the equation.
Any time you know the electric field, you can use this equation to calculate the force
on a charged particle in that electric field.
The units of electric field are
newtons/coulomb.
 F0  N
 E  =   =
q0  C
Subsequently, you will learn that the units of electric field can
also be expressed as volts/meter:
N V
E = =
C m
The electric field exists independent of whether there is a
charged particle around to “feel” it.
Remember: the electric field direction is the
direction a + charge would feel a force.
+
A + charge would be repelled by another + charge.
Therefore the direction of the electric field is away from positive
(and towards negative).
http://regentsprep.org/Regents/physics/phys03/afieldint/default.htm
Today’s agendum:
The electric field.
You must be able to calculate the force on a charged particle in an electric field.
Electric field of due to a point charge.
You must be able to calculate electric field of a point charge.
Motion of a charged particle in a uniform electric field.
You must be able to solve for the trajectory of a charged particle in a uniform electric field.
The electric field due to a collection of point charges.
You must be able to calculate electric field of a collection of point charges.
The electric field due to a continuous line of charge.
You must be able to calculate electric field of a continuous line of charge.
The Electric Field
Due to a Point Charge
Coulomb's law says
q1q 2
F =k 2 ,
12
r12
... which tells us the electric field due to a point charge q is
E q =k
q
r
2
, away from +
…or just…
This is your third starting equation.
q
E=k 2
r
We define r̂ as a unit vector from the source point to the field
point:
source point
r̂ +
field point
The equation for the electric field of a point charge then
becomes:
q
E=k 2 rˆ
r
You may start with either
equation for the electric field
(this one or the one on the
previous slide).
Example: calculate the electric field at the electron’s distance
away from the proton in a hydrogen atom (5.3x10-11 m).
To be worked at the blackboard.
For comparison, air begins to break down and conduct
electricity at about 30 kV/cm, or 3x106 V/m.
Today’s agendum:
The electric field.
You must be able to calculate the force on a charged particle in an electric field.
Electric field of due to a point charge.
You must be able to calculate electric field of a point charge.
Motion of a charged particle in a uniform electric
field.
You must be able to solve for the trajectory of a charged particle in a uniform electric field.
The electric field due to a collection of point charges.
You must be able to calculate electric field of a collection of point charges.
The electric field due to a continuous line of charge.
You must be able to calculate electric field of a continuous line of charge.
Motion of a Charged Particle
in a Uniform Electric Field
A charged particle in an electric field experiences a force, and if
it is free to move, an acceleration.
If the only force is due to the
electric field, then
 F  ma  qE.
- - - - - - - - - - - - -
-
F
E
+ + + + + + + + + + + + +
If E is constant, then a is constant, and you can use the
equations of kinematics* (remember way back to the beginning
of Physics 103?).
Example: a proton and an electron enter a region of uniform
electric field. Describe their motion.
Direction of forces?
Magnitudes of accelerations?
Shape of trajectories?
Example: an electron moving with velocity v0 in the positive x
direction enters a region of uniform electric field that makes a
right angle with the electron’s initial velocity. Express the
position and velocity of the electron as a function of time.
y
- - - - - - - - - - - - -
x
-
v0
E
+ + + + + + + + + + + + +
To be worked at the blackboard.
Today’s agendum:
The electric field.
You must be able to calculate the force on a charged particle in an electric field.
Electric field of due to a point charge.
You must be able to calculate electric field of a point charge.
Motion of a charged particle in a uniform electric field.
You must be able to solve for the trajectory of a charged particle in a uniform electric field.
The electric field due to a collection of point charges.
You must be able to calculate electric field of a collection of point charges.
The electric field due to a continuous line of charge.
You must be able to calculate electric field of a continuous line of charge.
Example: calculate the electric field at position P due to the two
protons shown.
P
+
D
+
D
To be worked at the blackboard.
Example: field of an electric dipole.
No time to work today. Will work in Lecture 3. Study example
21.9 in your text.
Today’s agendum:
The electric field.
You must be able to calculate the force on a charged particle in an electric field.
Electric field of due to a point charge.
You must be able to calculate electric field of a point charge.
Motion of a charged particle in a uniform electric field.
You must be able to solve for the trajectory of a charged particle in a uniform electric field.
The electric field due to a collection of point charges.
You must be able to calculate electric field of a collection of point charges.
The electric field due to a continuous line of charge.
You must be able to calculate electric field of a continuous line of charge.
Electric Field Due To A Line of Charge
+ + + + + + + + + + + + + + +
+
Think of a line of charge as a collection of very tiny point
charges all lined up. The “official” term for “as tiny as you can
imagine” is “infinitesimal.”
We get the electric field for the line of charge by adding the
electric fields for all the infinitesimal point charges.
These words are meant to remind you of things you have
learned in calculus.
Consider charge uniformly distributed along a line (e.g.,
electrons on a thread).
 is the linear density of charge (amount of charge per unit
length).  may be a function of position.
Think    length.  times the length of line segment is the
total charge on the line segment.

 dx
x
dx
If charge is distributed along a straight line segment parallel to
the x-axis, the amount of charge dq on a segment of length dx
is  dx.
I’m assuming positively charged objects
in these “distribution of charges” slides.
P
dE
r’
r'
dq
The electric field at point P due to the charge dq is
1 dq
1 dx
dE =
r' =
r'
2
2
4πε 0 r'
4πε 0 r'
x
P
E
r’
r'
x
dq
The electric field at P due to the entire line of charge is
1
λ(x) dx
E=
r'
.
2

4πε 0
r'
The integration is carried out over the entire length of the line, which need
not be straight. Also,  could be a function of position, and can be taken
outside the integral only if the charge distribution is uniform.
Example: calculate the electric field due to an infinite line of
positive charge.
There are two approaches to the mathematics of this problem.
One approach is that of example 21.11, where an equation for
the electric field for an infinite line of charge is derived.
Thus, if this were given for homework, you would need to
repeat this derivation!
If you use the text’s approach, you must
evaluate this indefinite integral, which is in
appendix B (page A4) of your text:

dx
x
2
a
2

3
2
If you need it, you can look this integral up; your instructor will give it to or let you look it up in
the text.
Example: calculate the electric field due to an infinite line of
positive charge.
Today’s agendum:
Review.
The electric field of a dipole.
You must be able to calculate the electric field of a dipole.
The electric field due to a collection of point charges
(continued).
You must be able to calculate the electric field of a collection of point charges.
Electric field lines.
You must be able to draw electric field lines, and interpret diagrams that show electric field
lines.
A dipole in an external electric field.
You must be able to calculate the moment of an electric dipole, the torque on a dipole in
an external electric field, and the energy of a dipole in an external electric field.
The Big Picture, Part I
In Lecture 1 you learned Coulomb's Law:
1 q1q 2
F =
,
2
12 4πε 0 r12
r12
+
-
Q1
Q2
Coulomb’s Law quantifies the force between charged particles.
The Big Picture, Part II
In Lecture 2 you learned about the electric field.
There were two kinds of problems you had to solve:
1. Given an electric field, calculate the force on a charged
particle.
2. Given one or more charged particles, calculate the electric
field they produce.
The Big Picture, Part II
1. Given an electric field, calculate the force on a charged
particle.
F
E=
q
F= qE
-
F
E
You may not be given any information about
where this electric field “comes from.”
The Big Picture, Part II
2. Given one or more charged particles, calculate the electric
field they produce. 2 slides from now we’ll do this for a dipole.
source point
r̂ +
q
E=k 2 rˆ
r
E
field point
Example: electric field
of a point charge.
Today’s agendum:
Review.
The electric field of a dipole.
You must be able to calculate the electric field of a dipole.
The electric field due to a collection of point charges
(continued).
You must be able to calculate the electric field of a collection of point charges.
Electric field lines.
You must be able to draw electric field lines, and interpret diagrams that show electric field
lines.
A dipole in an external electric field.
You must be able to calculate the moment of an electric dipole, the torque on a dipole in
an external electric field, and the energy of a dipole in an external electric field.
A Dipole
A combination of two electric charges with equal magnitude and
opposite sign is called a dipole.
- -q
+q +
d
The charge on this dipole is q (not zero, not +q, not –q, not
2q). The distance between the charges is d. Dipoles are
“everywhere” in nature.
This is an electric dipole. Later in the course we’ll study magnetic dipoles, which, as you might
guess, have a north and a south magnetic pole.
The Electric Field of a Dipole
Example: calculate the electric field at point P, which lies on the
perpendicular bisector a distance L from a dipole of charge q.
P
qi
E=k  rˆ 2
ri
i
to be worked at the blackboard
L
qd
E
4 o r 3
- -q
+q +
d
P
qd
E
4 o r 3
L
- -q
+q +
d
Caution! The above
equation for E applies
only to points along
the perpendicular
bisector of the dipole.
Today’s agendum:
Review.
The electric field of a dipole.
You must be able to calculate the electric field of a dipole.
The electric field due to a collection of point charges
(continued).
You must be able to calculate the electric field of a collection of point charges.
Electric field lines.
You must be able to draw electric field lines, and interpret diagrams that show electric field
lines.
A dipole in an external electric field.
You must be able to calculate the moment of an electric dipole, the torque on a dipole in
an external electric field, and the energy of a dipole in an external electric field.
The Electric Field
Due to a Collection of Point Charges
Hang on to your seats, because the next 11 slides will go by
very fast!
In today's lecture I show how to calculate the electric
field due to a distribution of charges.
Matter is made of discrete atoms, but appears
"continuous" to us, and in Physics 103 we treated
matter as being a continuous entity.
Similarly, a charge distribution is made of individual
charged particles, but we can treat it as if the charge
were continuous.
The electric field due to a small "chunk" q of charge is
1 q
E =
r
2
4πε 0 r
unit vector from q
to wherever you
want to calculate E
The electric field due to collection of "chunks" of charge is
1
E =  E i =
4πε 0
i
q i
i r 2 r i
i
As qdq0, the sum becomes an integral.
unit vector from qi
to wherever you
want to calculate E
If charge is distributed along a straight line segment parallel to
the x-axis, the amount of charge dq on a segment of length dx
is  dx.

 dx
x
dx
 is the linear density of charge (amount of charge per unit
length).  may be a function of position.
Think    length.  times the length of line segment is the
total charge on the line segment.
I’m assuming positively charged objects
in these “distribution of charges” slides.
P
dE
r’
r'
dq
The electric field at point P due to the charge dq is
1 dq
1 dx
dE =
r' =
r'
2
2
4πε 0 r'
4πε 0 r'
x
P
E
r’
r'
x
dq
The electric field at P due to the entire line of charge is
1
λ(x) dx
E=
r'
.
2

4πε 0
r'
The integration is carried out over the entire length of the line, which need
not be straight. Also,  could be a function of position, and can be taken
outside the integral only if the charge distribution is uniform.
If charge is distributed over a two-dimensional surface, the
amount of charge dq on an infinitesimal piece of the surface is
 dS, where  is the surface density of charge (amount of
charge per unit area).
y

charge dq =  dS
x
area = dS
y
P
dE
r’
r'
x
The electric field at P due to the charge dq is
1 dq
1  dS
dE =
r' =
r'
2
2
4πε 0 r'
4πε 0 r'
y
P
E
r’
r'
x
The net electric field at P due to the entire surface of charge is
1
(x, y) dS
E=
r'

4πε 0 S
r'2
After you have seen the above, I hope you believe that the net
electric field at P due to a three-dimensional distribution of
charge is…
z
E
P
r’
r'
x
y
1
(x, y, z) dV
E=
r'
.
2

4πε 0 V
r'
Summarizing:
Charge distributed along a line:
1
λ dx
E=
r' 2 .

4πε 0
r'
Charge distributed over a surface:
1
 dS
E=
r' 2 .

4πε 0 S r'
Charge distributed inside a volume:
1
 dV
E=
r' 2 .

4πε 0 V r'
If the charge distribution is uniform, then , , and  can be taken outside
the integrals.
“Hold it right there! These equations…
1
λ dx
E=
r' 2

4πε 0
r'
1
 dS
E=
r' 2

4πε 0 S r'
1
 dV
E=
r' 2

4πε 0 V r'
…are not on my starting equation sheet. That’s not fair!”
1 q
1 dq
r or dE =
Just start with E =
.
2
2
4πε 0 r
4πε 0 r
The Electric Field
Due to a Continuous Charge Distribution
(worked examples)
Example: A rod of length L has a uniform charge per unit length
 and a total charge Q. Calculate the electric field at a point P
along the axis of the rod at a distance d from one end.
y
P
x
d
L
Let’s put the origin at P. The linear charge density and Q are
related by
Q
=
and Q = L
L
Let’s assume Q is positive.
y
dE
x
P
d
dQ =  dx
dx
x
L
The electric field points away from the rod. By symmetry, the
electric field on the axis of the rod has no y-component. dE
from the charge on an infinitesimal length dx of rod is
dq
 dx
dE = k 2  k 2
x
x
Note: dE is in the –x direction. dE is the magnitude of dE. I’ve
used the fact that Q>0 (so dq=0) to eliminate the absolute
value signs in the starting equation.
y
dE
x
P
d
E =
d+L
d
dQ =  dx
dx
x
L
dE x = -k 
d+L
d
 dx ˆ
i = -k 
2
d
x
d+L
dL
dx ˆ
 1 ˆ
i = -k    i
2
x
 x d
 d  d  L  ˆ
1
1 ˆ
L ˆ
kQ ˆ

E = -k  
  i = -k 
i=
-k
i=
i

d d  L
d d  L
 dL d
 d d  L 
Example: A ring of radius a has a uniform charge per unit
length and a total positive charge Q. Calculate the electric field
at a point P along the axis of the ring at a distance x0 from its
center.
dQ
r
a

x0
P

dE
x
By symmetry, the y- and zcomponents of E are zero,
and all points on the ring
are a distance r from point
P.
dQ
dQ
dE=k 2
r
r
a

x0
P

x
No absolute value
signs because Q is
positive.
dQ
dE x =k 2 cos 
r
dE
r = x a
2
0
2
x0
cos  
r
x0
 dQ  x 0
E x   dE x    k 2   k 3
r  r
r
ring
ring 
For a given x0, r is a constant
for points on the ring.
x0
kx 0Q
ring dQ  k r3 Q  x 2  a 2 3/ 2


Or, in general, on the ring axis E x,ring 
0
kxQ
2
x
 a

2 3/ 2
.
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
dQ
r
P
R
x
x0
The disc is made of
concentric rings. The
area of a ring at a
radius r is 2rdr, and
the charge on each ring
is (2rdr).
We can use the equation on the previous slide for the electric
field due to a ring, replace a by r, and integrate from r=0 to
r=R.
dE ring 
kx 0 2rdr
x
2
0
r

2 3/ 2
.
Caution! I’ve switched
the “meaning” of r!
dQ
r
P
x
x0
R
Ex 
 dE
disc
x


disc
kx 0 2rdr
x
 x2  r 
0
E x  kx 0  
 1/ 2

2
0
r
2 1/ 2

2 3/ 2
R
 kx 0  
R
0
x
2r dr
2
0
r

2 3/ 2



x
x
0

  2k  0 
 x 0  x 2  R 2 1/ 2 

0
0


Example: Calculate the electric field at a distance x0 from an
infinite plane sheet with a uniform charge density .
Treat the infinite sheet as disc of infinite radius.
1
Let R and use k 
to get
4 0
Esheet 

20
.
Interesting...does not depend on distance from the sheet.
Today’s agendum:
Review.
The electric field of a dipole.
You must be able to calculate the electric field of a dipole.
The electric field due to a collection of point charges
(continued).
You must be able to calculate the electric field of a collection of point charges.
Electric field lines.
You must be able to draw electric field lines, and interpret diagrams that show electric field
lines.
A dipole in an external electric field.
You must be able to calculate the moment of an electric dipole, the torque on a dipole in
an external electric field, and the energy of a dipole in an external electric field.
Electric Field Lines
Electric field lines help us visualize the electric field and predict
how charged particles would respond the field.
-
+
Example: electric field lines for isolated +2e and -e charges.
Here’s how electric field lines are related to the field:
 The electric field vector E is tangent to the field lines.
 The number of lines per unit area through a surface
perpendicular to the lines is proportional to the electric field
strength in that region
 The field lines begin on positive charges and end on
negative charges.
 The number of lines leaving a positive charge or
approaching a negative charge is proportional to the
magnitude of the charge.
 No two field lines can cross.
Example: draw the electric field lines for charges +2e and -1e,
separated by a fixed distance.
Today’s agendum:
Review.
The electric field of a dipole.
You must be able to calculate the electric field of a dipole.
The electric field due to a collection of point charges
(continued).
You must be able to calculate the electric field of a collection of point charges.
Electric field lines.
You must be able to draw electric field lines, and interpret diagrams that show electric field
lines.
A dipole in an external electric field.
You must be able to calculate the moment of an electric dipole, the torque on a dipole in
an external electric field, and the energy of a dipole in an external electric field.
Electric Dipole in an
External Electric Field
An electric dipole consists of two charges +q and -q, equal in
magnitude but opposite in sign, separated by a fixed distance d.
q is the “charge on the dipole.”
Earlier, I calculated the electric field along the perpendicular
bisector of a dipole.
qd
E
.
3
4 o r
The electric field depends on the product qd.
q and d are parameters that specify the dipole; we define the
"dipole moment" of a dipole to be the vector
p  qd,
caution: this p is not momentum!
where the direction of p is from negative to positive (NOT away
from +).
+q
-q
p
To help you remember the direction of p, this is on the equation
sheet:
p  q d, from  to plus
A dipole in a uniform electric field experiences no net force, but
probably experiences a torque…
p
F-
+q
F+

-q
There is no net force on the dipole:
F  F

 F  qE  qE  0.
E
p
½ d sin
F-
-q
+q
F+
E
½ d sin

If we choose the midpoint of the dipole as the origin for
calculating the torque, we find
d sin 
d sin 
       2 qE  2 qE  qdE sin ,
and in this case the direction is into the plane of the figure.
Expressed as a vector,
  p  E.
Recall that the unit of torque is
N·m, which is not a joule!
p
½ d sin
F-
-q
+q
F+
E
½ d sin

The torque’s magnitude is p E sin and the direction is given by
the right-hand rule.
Energy of an Electric Dipole in an
External Electric Field
p
F-
-q
+q
F+
E

If the dipole is free to rotate, the electric field does work* to
rotate the dipole.
W  pE(cos initial  cos final ).
The work depends only on the initial and final coordinates, and
not on how you go from initial to final.
*Calculated using
W   z d , using torque times angular displacement.
Does that awaken vague memories of Physics 103?
If a force is conservative, you can define a potential energy
associated with it.
What kinds of potential energies did you learn about in Physics
103?
Because the electric force is conservative, we can define a
potential energy for a dipole. The equation for work
W  pE(cos initial  cos final )
suggests we should define
U dipole  pE cos .
U dipole   pE cos 
p
F-
-q
+q
F+
E

With the definition on the previous slide, U is zero when =/2.
U is maximum when cos=-1, or = (a point of unstable
equilibrium).
U is minimum when cos=+1, or =0 (stable equilibrium).
It is “better” to express the dipole potential energy as
Udipole  p  E.
Recall that the unit of energy is the
joule, which is a N·m, but is not the
same as the N·m of torque!
Today’s agendum:
Electric flux.
You must be able to calculate the electric flux through a surface.
Gauss’ “Law.”
You must be able to use Gauss’ “Law” to calculate the electric field of a high-symmetry
charge distribution.
Conductors in electrostatic equilibrium.
You must be able to use Gauss’ “Law” to draw conclusions about the behavior of charged
particles on, and electric fields in, conductors in electrostatic equilibrium.
Gauss’ “Law”
Electric Flux
We have used electric field lines to visualize electric fields and
indicate their strength.
We are now going to count* the
number of electric field lines passing
through a surface, and use this
count to determine the electric field.
*There are 3 kinds of people in this world: those who can count, and those who can’t.
E
The electric flux passing through a surface is the number of
electric field lines that pass through it.
Because electric field lines are drawn
arbitrarily, we quantify electric flux
like this: E=EA.
If the surface is tilted, fewer lines cut
the surface.
Later we’ll learn about magnetic flux, which is
why I will use the subscript E on electric flux.
A
E
E

We define A to be a vector having a
magnitude equal to the area of the
surface, in a direction normal to the
surface.
A

E

The “amount of surface” perpendicular
to the electric field is A cos .
Because A is perpendicular to the surface, the amount of A
parallel to the electric field is A cos .
A = A cos  so E = EA = EA cos .
Remember the dot product from Physics 103? E  E  A
If the electric field is not uniform, or the surface is not flat…
divide the surface into
infinitesimal surface
elements and add the
flux through each…
A
dA
E
 E  lim
Ai  0
 E  A
i
i
 E   E  dA
i
If the surface is closed (completely encloses a volume)…
…we count* lines going
out as positive and lines
going in as negative…
E
dA
 
 E   E  dA
a surface integral, therefore a
double integral 
*There are 10 kinds of people in this world: those who can count in binary, and those who can’t.
What is this

thing?
Nothing to panic about!
The circle just reminds you
to integrate over a closed
surface.
Question: you gave me five different equations for electric flux.
Which one do I need to use?
Answer: use the simplest (easiest!) one that works.
 E  EA
Flat surface, E  A, E constant over surface. Easy!
 E  EA cos 
Flat surface, E not  A, E constant over surface.
E  E  A
Flat surface, E not  A, E constant over surface.
 E   E  dA
Surface not flat, E not uniform.
 
 E   E  dA
Closed surface. Most general. Most complex.
If the surface is closed, you may be able to “break it up” into
simple segments and still use E=E·A for each segment.
Example: Calculate the electric flux through a cylinder with its
axis parallel to the electric field direction.
E
To be worked at the blackboard…
+
-
E
If there were a + charge inside the cylinder, there would be
more lines going out than in.
If there were a - charge inside the cylinder, there would be
more lines going in than out…
…which leads us to…
Today’s agendum:
Electric flux.
You must be able to calculate the electric flux through a surface.
Gauss’ “Law.”
You must be able to use Gauss’ “Law” to calculate the electric field of a high-symmetry
charge distribution.
Conductors in electrostatic equilibrium.
You must be able to use Gauss’ “Law” to draw conclusions about the behavior of charged
particles on, and electric fields in, conductors in electrostatic equilibrium.
Gauss’ “Law”
Mathematically*, we express the idea two slides back as
  q enclosed
 E   E  dA 
o
Gauss’ “Law”
We will find that Gauss’ “Law” gives a simple way to calculate
electric fields for charge distributions that exhibit a high degree
of symmetry…
…and save more complex and realistic charge distributions for
advanced classes.
*“Mathematics is the Queen of the Sciences.”—Karl Gauss
To see how this works, let’s do an example.
Example: use Gauss’ “Law” to calculate the electric field from
an isolated point charge q.
To apply Gauss’ “Law”, we construct a “Gaussian Surface”
enclosing the charge.
The Gaussian surface should mimic the symmetry of the charge
distribution.
For this example, choose for our Gaussian surface a sphere of
radius r, with the point charge at the center.
I’ll work the rest of the example on the blackboard.
Strategy for Solving Gauss’ “Law” Problems
 Select a Gaussian surface with symmetry that matches the
charge distribution.
 Draw the Gaussian surface so that the electric field is either
constant or zero at all points on the Gaussian surface.
 Use symmetry to determine the direction of E on the Gaussian
surface.
 Evaluate the surface integral (electric flux).
 Determine the charge inside the Gaussian surface.
 Solve for E.
Example: calculate the electric field outside a long cylinder of
finite radius R with a uniform volume charge density  spread
throughout the volume of the cylinder.
The cylinder being “long” and the radius “finite” are “code
words” that tell you to neglect end effects from the cylinder
(i.e., assume it is infinitely long).
Know how to interpret “code words” when exam time comes!
Let’s go through this a step at a time
(work to be shown at board)
.
 Select a Gaussian surface with symmetry that matches the
charge distribution.
Pick a cylinder of length L and radius r, concentric with the
cylinder of the problem.
 Draw the Gaussian surface so that the electric field is either
constant or zero at all points on the Gaussian surface.
Already done!
 Use symmetry to determine the direction of E on the Gaussian
surface.
Electric field points radially away from cylinder, and magnitude
does not depend on direction.
 Evaluate the surface integral (electric flux).
Surface integral is just E times the curved area.
 Determine the charge inside the Gaussian surface.
The charge inside is just the volume of charged cylinder inside
the Gaussian surface, times the charge per volume.
 Solve for E.
R 2
E
20 r
Example: use Gauss’ “Law” to calculate the electric field due to
a long line of charge, with linear charge density .
Example: use Gauss’ “Law” to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
These are easy using Gauss’ “Law” (remember what a pain they
were in the previous chapter). Study these examples and others
in your text!
E line


.
20 r
E sheet


.
20
Today’s agendum:
Electric flux.
You must be able to calculate the electric flux through a surface.
Gauss’ “Law.”
You must be able to use Gauss’ “Law” to calculate the electric field of a high-symmetry
charge distribution.
Conductors in electrostatic equilibrium.
You must be able to use Gauss’ “Law” to draw conclusions about the behavior of charged
particles on, and electric fields in, conductors in electrostatic equilibrium.
Conductors in Electrostatic Equilibrium
Electrostatic equilibrium means there is no net motion of the
charges inside the conductor.
The electric field inside must be zero.
If this were not the case, charges would move.
Any excess charge must reside on the outside surface of the
conductor.
Apply Gauss’ “Law” to a Gaussian surface just inside the
conductor surface. The electric field is zero, so the net charge
inside the Gaussian surface is zero.
The electric field just outside a charged conductor must be
perpendicular to the conductor’s surface.
Otherwise, the component of the
electric field parallel to the surface
would cause charges to accelerate.
The magnitude of the electric field just outside a charged
conductor is equal to /0, where  is the local surface charge
density.
A simple application Gauss’ “Law”, which I will show if time
permits.
If there is an empty nonconducting cavity inside the conductor,
Gauss’ “Law” tells us there is no net charge on the interior
surface of the conductor.
Construct a Gaussian surface that includes the inner surface of
the conductor. The electric field at the Gaussian surface is zero,
so no electric flux passes through the Gaussian surface. Gauss’
Law says the charge inside must be zero. The conductor does
not have to be symmetric, as shown.
If there is a nonconducting cavity inside the conductor, with a
net charge inside the cavity, Gauss’ “Law” tells us there is an
equal and opposite induced charge on the interior surface of
the conductor.
Construct a Gaussian surface that includes the inner surface of
the conductor. The electric field at the Gaussian surface is zero,
so no electric flux passes through the Gaussian surface. Gauss’
Law says the charge inside must be zero. There must be a –Q
on the inner surface. The conductor does not have to be
symmetric, as shown.
+Q
-Q
Example: a conducting spherical shell of inner radius a and
outer radius b with a net charge -Q is centered on point charge
+2Q. Use Gauss’s “Law” to show that there is a net charge of
-2Q on the inner surface of the shell, and a net charge of +Q
on the outer surface of the shell.
-Q
  q enclosed
 E  dA  o
a
+2Q
b
Example (if time permits): an insulating sphere of radius a has
a uniform charge density ρ and a total positive charge Q.
Calculate the electric field at a point inside the sphere
r
Q
a
  q enclosed
 E  dA  o