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.

1
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.
2
How should the net be oriented to catch the most
fish?
3
In orientation A, you catch 10 fish per hour.
How many f.p.h. will you catch in orientation
C?A 0B about 2C 5D about 7E 10
4
In orientation A, you catch 10 fish per hour.
How many f.p.h. will you catch in orientation
B?A 0B about 2C 5D about 7E 10
5
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
density of fish per unit volume x velocity of
fish
6
Suppose the flow of fish is defined as
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/2B FAC
60FAD none is correct - the answer is a vector
7
Suppose the flow of fish is defined as
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.
8
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 yesB
no
9
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.
10
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 yesB no
11
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.
12
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.
13
So, for any flat net,f.p.h.
14
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
0B FA/4C FA/2D FAE FAv3/4
Each half is bent 60 back
15
If you rotate the net as shown, how many f.p.h
will you catch?A 0B FA/4C FA/2D FAE
FAv3/4
16
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.
17
Positive charges are sources of small invisible
fish. Negative charges are sinks of small
invisible fish. 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 wont call us
crazy.)Field lines are just invisible fish
trajectories.
18
There is nothing in Coulombs 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. Coulombs 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!)
19
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.