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 = density of fish per unit volume x velocity of fish

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 m 3, 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 = 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 m 3, 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, A p = 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. =

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 Each half is bent 60° back

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. 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.