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] 3,4,2,1D] 3&4 tie, 2, 1E] 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) horizontal, leftB) Horizontal, rightC) Vertical, downD) Vertical, upE) At an angle, with both horizontal & vertical components

Symmetry

What is the direction of the field at dot?

A) horizontal, leftB) Horizontal, rightC) Vertical, downD) Vertical, upE) 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 ~ constantLet’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

= density of fish per unit volume x velocity of fish

€

vF

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/2

B] FA

C] 60*FA

D] none is correct - the answer is a vector

€

vF

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.

€

vF

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. =

€

vF •

v 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] 0B] FA/4C] FA/2D] FAE] FA√3/4

If you rotate the net as shown, how many f.p.h will you catch?

A] 0B] FA/4C] FA/2D] FAE] 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 fluxWhat 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] sameB]C]D]

€

EA1 cosθ

€

EA1 /cosθ

€

EA1 /sinθ

What would happen to the flux if we were to double the field?

A] sameB] doubleC] halfD] 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 chargeLine 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