We often draw two or more E field vectors at one point, when
there are two or more charges causing the field.
Each vector then represents the field that would exist if a
particular source charge were the only charge.
The total field is still a single, unique, vector, of course.
It is found by the vector sum of the contributions from each
charge.
http://phet.colorado.edu/en/simulation/electric-hockey
A] 3,2, 4&1 tie (both = 0)
B] 3,2,4,1 (none zero)
C] 3,4,2,1
D] 3&4 tie, 2, 1
E] 4,3,2,1
Please Note: Field lines are NOT
trajectories of charged particles.
Force (which is directed along the field
line) is NOT proportional to v, but to a!
http://phet.colorado.edu/en/simulation/electric-hockey
VECTOR Superposition
Is the magnitude of the electric field bigger at 1 or at 2?
A] 1
B] 2
C] same
D] there’s no way to tell
VECTOR Superposition
Is the magnitude of the electric field bigger at 1 or at 2?
A] 1
B] 2
C] same
D] there’s no way to tell
VECTOR Superposition
Is the magnitude of the electric field bigger at 1 or at 2?
A] 1
B] 2
C] same
D] there’s no way to tell
VECTOR Superposition
What is your answer if we change ONE charge to -?
Is the magnitude of the electric field bigger at 1 or at 2?
A] 1
B] 2
C] same
D] there’s no way to tell
VECTOR Superposition
Is the magnitude of the electric field bigger at 1 or at 2?
A] 1
B] 2
C] same
D] there’s no way to tell
VECTOR Superposition
At the point shown, is the y-component of the electric field
up or down?
A] up
B] down
C] there’s no way to tell
Symmetry
What is the direction of the field at dot?
A)
B)
C)
D)
E)
horizontal, left
Horizontal, right
Vertical, down
Vertical, up
At an angle, with both horizontal & vertical components
Symmetry
What is the direction of the field at dot?
A)
B)
C)
D)
E)
horizontal, left
Horizontal, right
Vertical, down
Vertical, up
At an angle, with both horizontal & vertical components
Integration to find Fields
We wish to write the field of a disk as the vector sum (integral) of
the fields from annular rings of infinitessimal width, from which
we can construct the disk.
What is the area of one such ring, with radius r?
A] r2
B]  r2
C]  r dr
D] 2  r dr
E]  r2 dr
Field from a point ~ 1/r2
Field from a line ~ 1/r
Field from a plane ~ constant
Let’s draw field lines for some infinite planes of charge.
--------You should be able to calculate E fields from a couple point
charges (add components!)
You should be able to calculate E fields from a distributed source
(using integration)(Also with a given charge density!)
You should be able to draw field lines for a couple points, a line, or
any number of infinite planes
And now… a tale of invisible fish.
All the fish in the river are dead, so they float with the
current.
The opening on the net is a flat surface of area A.
How should the net be oriented to catch the most fish?
In orientation A, you catch 10 fish per hour.
How many f.p.h. will you catch in orientation C?
A] 0
B] about 2
C] 5
D] about 7
E] 10
In orientation A, you catch 10 fish per hour.
How many f.p.h. will you catch in orientation B?
A] 0
B] about 2
C] 5
D] about 7
E] 10
The number of f.p.h. is directly proportional to the
projected area of the net opening.
Projected area means “how big the net looks if you sight
along the flow direction.”
The projected area is Acos, where  is the angle
between the actual net orientation and the best
orientation, with the opening perpendicular to the flow.
Suppose the flow of fish is
v
F = density of fish per unit volume x velocity of fish
Suppose the flow of fish is defined as
v
F = density of fish per unit volume x velocity of fish
With the net opening of area A, oriented perpendicular
to the flow, how many fish per hour will be caught?
(Density is in fish per m3, velocity is in m/hr)
A] FA/2
B] FA
C] 60*FA
D] none is correct - the answer is a vector
Suppose the flow of fish is defined as
v
F = density of fish per unit volume x velocity of fish
With the net opening of area A, oriented perpendicular
to the flow, how many fish per hour will be caught?
(Density is in fish per m3, velocity is in m/hr)
FA fish per hour will be caught.
All the fish in a volume given by
area x (velocity x time t) will be caught in time t.
Suppose the net opening had the same area, but were
a different shape (say, a flat circle, rather than a flat
square).
Would this change the number of fish per hour caught?
A] yes
B] no
Suppose the net opening had the same area, but were
a different shape (say, a flat circle, rather than a flat
square).
Would this change the number of fish per hour caught?
NO.
Suppose the net opening had the same area, but had
been bent in the middle, as shown.
Would this change the number of fish per hour caught?
A] yes
B] no
Suppose the net opening had the same area, but had
been bent in the middle, as shown.
Would this change the number of fish per hour caught?
YES. The “area” here is not flat, so the projected area
(perpendicular to the flow) is smaller.
For FLAT areas, it is useful to define a vector area,
which is perpendicular to the area.
The most fish are caught when the vector area lines up
with the flow.
The f.p.h. will be equal to F times the projected area,
Ap = Acos
f.p.h. = FAcos
cos is the angle by which the net deviates from its
optimal orientation.
It is also the angle between the vector area and the
flow.
So, for any flat net,
f.p.h. =

v v
FA
Suppose we have the bent net shown. It has two flat parts. So it
would catch the same f.p.h. as two smaller nets, each oriented as
shown.
In terms of the original (unbent) area of the net, the top flat part has
area A/2 and the bottom also has A/2.
In terms of the original (unbent) area of the net, how many fish
per hour will be caught with this net in this orientation?
A] 0
B] FA/4
C] FA/2
D] FA
E] FA√3/4
If you rotate the net as shown, how many f.p.h will you
catch?
A] 0
B] FA/4
C] FA/2
D] FA
E] FA√3/4
You need to add up the f.p.h. for each flat surface.
If a surface is curved, we use calculus to break it up into
small areas, each of which is so small it may be taken
as flat.
Positive charges are sources of small invisible fish.
Negative charges are sinks of small invisible fish.
(By the way, the fish are massless.)
When a + charge is sitting in a flow of fish, it feels a
force downstream. Force = charge x flow rate.
Since we use F for force, we choose to use E for flow
rate (invisible fish only.)
(We also call the flow of invisible fish an “electric field”
so people won’t call us crazy.)
Field lines are just invisible fish trajectories.
There is nothing in Coulomb’s law that argues against
a “fish picture” of electric fields.
If we count fish flowing from a + charge, we should get
the same number of fish if we capture them 1 m away,
or 10 m away. Coulomb’s law says we do.
(In fact, all of electromagnetic theory and experiment is
consistent with the invisible fish being quite real, and
traveling at the speed of light!)
Electric field flux is a measure of how many invisible fish
will be caught per unit time… but we need to be careful
about signs.
A surface vector can have either orientation, and that
will change the sign of the flux.
By convention, with closed surfaces, we take the area
vector(s) to be outward.
Today:
Some clicker quizzes on flux
What use is Gauss’ Law?
• It’s beautiful
• It allows finding fields
(when there is symmetry)
• It allows finding charges
on conductors


If the dot product of E and the vector area of plate 1 is EA1, what
is the dot product of E and the area of plate 2?

A] same
B] EA1 cos
C] EA1 /cos
D] EA1 /sin 

What would happen to the flux if we were to double the field?

A] same
B] double
C] half
D] quadruple

The flux is proportional to the field, and to the perpendicular area.
We know the density of field lines (#lines per perpendicular area) is
proportional to the field.


So the flux is proportional to the number of field lines
penetrating a surface area.
Note: the sign depends on how we “orient” the
surface!
Does the total flux through a spherical surface at radius
R change if we move the charge off-center?
• A] yes
• B] no
Does the total flux through a closed surface surrounding a
point charge change is we distort the surface (while keeping
it closed)?
• A] yes
• B] no
Gauss’ Law is useful for finding E fields when there is
symmetry.
Sphere of charge
Line of charge
Sheet of charge
Gauss’ law also shows that excess charge on a conductor is
all at the surface!
A small sphere and a short rod have the same charge Q.
Gauss’ Law allows us to conclude that the electric field a
distance r away from each (if looked at separately) would be
the same.
• A] True
• B] False
FALSE! There is not enough symmetry to use
Gauss with the finite rod!!!!!!
What is the charge on the inner surface of the conducting shell?
A] 0
B] +Q
C] -Q
D] +2Q
E] cannot determine
If I put no excess charge on the conductor, the net charge on
its inner surface will be:
a] 0
b] +Q
c] -Q
d] -2Q
e] cannot determine
Because there is zero E field in a conductor, the inner
surface must hold a charge -Q.
What is the charge on the outer surface?
a] 0
b] +Q
c] -Q
d] -2Q
e] cannot determine
With no net charge on the outer conductor, the inner surface
gets -Q and so the outer surface gets +Q.
Suppose, instead, I put +8Q on the outer conductor. What
then is the charge on its inner surface?
a] 0
b] +Q
c] -Q
d] +3Q
e] +4Q