a

b

c

•Field lines are a way of representing an electric field•Lines leave (+) charges and return to (-) charges•Number of lines leaving/entering charge charge•Tangent of line = direction of E•Local density of field lines local magnitude of E •Lines from an isolated charge go to infinity•Field lines cannot cross

Last Week Rules for Field Lines

+ -

Electric Flux• Flux:

Let’s quantify previous discussion about numbers of field-lines

Define: electric flux sthrough the closed surface S

“S” is surfaceof the box

Note: in Fishbane they use dA to refer to an element of surface area. In these slides I use dS for the same thing

Flux

• How much of something is passing through some surface

Ex: How many hairs passing through your scalp.

• Two ways to define1.Number per unit area (e.g., 10 hairs/mm2)

This is NOT what we use here.

2.Number passing through an area of interest

e.g., 48,788 hairs passing through my scalp.

This is what we are using here.

Electric Flux

•What does this new quantity mean?•The integral is over a CLOSED

SURFACE•Since is a SCALAR product,

the electric flux is a SCALAR quantity•The integration vector is normal

to the surface and points OUT of the surface.

• is interpreted as the component of E which is NORMAL to the SURFACE

dS

Electric Flux

•Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface.

•The sign matters!! Pay attention to the direction of the normal

component as it penetrates the surface… is it “out of” or “into” the surface?

•“Out of” is “+” “into” is “-”

How to think about flux

• Consider the flux through this surface and define the surface area vector

• Let E-field point in y-direction– then and are parallel

and

• Look at this from on top

E

S

2wESE

S

surface area vector:

yw

yAreaS

ˆ

ˆ2

ww

How to think about flux• Consider flux through two

surfaces that “intercept different numbers of field lines”– first surface is the surface

from the previous slide– Second surface rotated by an

angle

case 1

case 2

Case 2 is smaller!

case 1 yEE o ˆ 2w

E-field surface area E S

2wEo

Flux:

case 2 yEE o ˆ cos2 wEo

2w

The Sign Problem• For an open surface we can

choose the direction of S-vector two different ways– to the left or to the right– what we call flux would be

different these two ways– different by a minus sign

rightleft

A differential surfaceelement, with its vector

•For a closed surface we can choose the direction of S-vector two different ways

–pointing “in” or “out”–Choose “out” –Integral of EdS over a

closed surface gives net flux “out,” but can be + or -

E

1

2

Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces.

a) Ф1 > Ф2

b) Ф1 = Ф2

c) Ф1 < Ф2

Question 1

Question 1

a b c

9% 10%

81%

1. a

2. b

3. c

E

1

2

Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces.

a) Ф1 > Ф2

b) Ф1 = Ф2

c) Ф1 < Ф2

Question 1

A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube.

a) Фbottom < 0

b) Фbottom = 0

c) Фbottom > 0

Question 2

Question 2

a b c

86%

13%

1%

1. a

2. b

3. c

A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube.

a) Фbottom < 0

b) Фbottom = 0

c) Фbottom > 0

Question 2

Question 3

•Imagine a cube of side a positioned in a region of constant electric field as shown

•Which of the following statements about the net electric flux E through the surface of this cube is true?

(a) E = (b) E 2a (c) E 6a

a

a

Question 3

a b c

85%

3%12%

1. a

2. b

3. c

Question 3

•Imagine a cube of side a positioned in a region of constant electric field as shown

•Which of the following statements about the net electric flux E through the surface of this cube is true?

(a) s = (b) s 2a (c) s 6a

a

a

• The electric flux through the surface is defined by:

• Therefore, the total flux through the cube is:

• on the bottom face is negative. (dS is out; E is in)

• on the top face is positive. (dS is out; E is out)

• is ZERO on the four sides that are parallel to the electric field.

The Fundamental Law of Electrostatics?

• Coulomb’s Law

Force between two point charges

OR

• Gauss’ Law

Relationship between Electric Fields

and charges

dS dS

1 2

• A positive charge is contained inside a spherical shell.

• When the charge moves from position 1 to position 2;

•The electric flux dФs through the surface element dS increases

•BUT the electric flux through the entire surface stays constant

Gauss’ Law

• Consider the dipole flux passing through the ‘Gaussian’ surfaces – Surface 1 has +ve flux proportional to charge +q– Surface 2 has –ve flux proportional to charge –q– Surface 3 has zero flux, as many flux lines enter as leave– Surface 4 also has zero flux, all leaving lines return

Gaussian surface 4

Question 4

R

2R

• Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.

– Which of the following statements about the net

electric flux through the 2 surfaces (2R and R) is true?

(a) R < 2R (b) R = 2R(c) R > 2R

Question 4

a b c

5% 9%

87%1. a

2. b

3. c

R

2R

• Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.

– Which of the following statements about the net

electric flux through the 2 surfaces (2R and R) is true?

(a) R < 2R (b) R = 2R (c) R > 2R

Question 4

•Look at the lines going out through each circle -- each circle has the same number of lines.

•The electric field is different at the two surfaces, because E is proportional to 1 / r 2, but the surface areas are also different. The surface area of a sphere is proportional to r 2.

•Since flux = , the r 2 and 1/r 2 terms will cancel, and the two

circles have the same flux!•There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases.

Gauss’ Law

• Gauss’ Law (a FUNDAMENTAL LAW):

The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

• How do we use this equation??•The above equation is ALWAYS TRUE but it isn’t always easy to use.•It is very useful in finding E when the physical situation exhibits strong SYMMETRY.

Gauss’ Law…made easy

•To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL.

(1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface;

If then

If then

(2) Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface.

•With these two conditions we can bring E outside of the integral…and:

Note that is just the area of the Gaussian surface over which we are integrating. Gauss’ Law now takes the form:

This equation can now be solved for E (at the surface) if we know qenclosed (or for qenclosed if we know E).

Gauss’ Law…made easy

Summary

• Electric Flux

• Gauss’ Law

• Read Chapter 23