•Electric flux is the amount of electric field going across a surface•It is defined in terms of a direction, or normal unit vector,perpendicular to the surface•For a constant electric field, and a flat surface, it is easy to calculate

•Denoted by E

•Units of Nm2/C•When the surface is flat, and the fields are constant, youcan just use multiplication to get the flux•When the surface is curved, or the fields are not constant,you have to perform an integration

n̂

ˆE AE n

E A E

0E n̂

cosE AE

ˆE dAE n

Gauss’s LawElectric Flux

A point charge q is at the center of a cylinder of radius a and height 2b. What is the electric flux out of (a) each end and (b) the lateral surface?

Electric Flux For a Cylinder

q b

b

a

n̂ E

b

s

r

2 2r s b 2

ek qE

r

2ˆ cosek q

rE n 3

ek qb

r

3/22 2

ek qb

b s

2dA sds

ˆE dAE n

•Consider a ring of radius s and thickness ds

3/22 20

2aek qbsds

b s

2 2

0

2a

ek qb

b s

2 2

22 e

e

k qbk q

b a

lateral surface

E

an̂ z

r

2 2r z a 3ˆ ek qa

r E n 2dA adz

ˆE dAE n 2

3/22 2

2be

b

k qa dz

a z

2 2

2b

e

b

k qz

a z

2 2

4 ek qb

a b

top

Total Flux Out of Various ShapesA point charge q is at the “center” of a (a) sphere (b) joined hemispheres

(c) cylinder (d) cube. What is the total electric flux out of the shape?

q

q

q

E

n̂

a

ˆE dAE n 2

ˆ ek q

aE n

22

4ek qa

a

4E ek q

1 2 3E E E E

En̂

ab

1E EA 22

2ek qa

a

1 2E ek q

E

n̂2 0E E

n̂

3 2E ek q

4E ek q

cap lat2E E E

2 2 2 2

2 42 2 e e

e

k qb k qbk q

b a b a

4E ek q

Gauss’s Law•No matter what shape you use, the total electric flux out of a region containing a point charge q is 4keq = q/0.Why is this true?•Electric flux is just measuring how many fieldlines come out of a given region•No matter how you distort the shape, the field linescome out somewhere

•If you have multiple charges inside the region their effects add•However, charges outside the region do not contribute

q

4E ek q 0E q

q1

q2

q3

1 2 3 0E q q q q4

in

0E

q

Using Gauss’s Law

Gauss’s Law can be used to solve three types of problems:1. Finding the total charge in a region when you know the electric field outside

that region2. Finding the total flux out of a region when the charge is known

a) It can also be used to find the flux out of one side in symmetrical problems

b) In such cases, you must first argue from symmetry that the flux is identical through each side

3. Finding the electrical field in highly symmetrical situationsa) One must first use reason to find the direction of the electric field

everywhereb) Then draw a Gaussian surface over which the electric field is constantc) Use this surface to find the electric field using Gauss’s Lawd) Works generally only for spherical, cylindrical, or planar-type problems

Sample ProblemA very long box has the shape of a regular pentagonal prism. Inscribed in the box is a sphere of radius R with surface charge density . What is the electric flux out of one lateral side of the box?

End view

Perspective view

•The flux out of the end caps is negligible•Because it is a regular pentagon, the flux from the other five sides must be the same

in 0E q ,side 05 E A 2

,side 05 4E R 2

,side 04 5E R

Using Gauss’s Law to find E-fieldA sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere?

a

•Clearly, all directions are created equal in this problem•Certainly the electric field will point away from the sphere at all points•The electric field must depend only on the distance•Draw a sphere of radius r around this charge•Now use Gauss’s Law with this sphere

ˆEE r

r

E

in 0 Eq

0 ˆV AE n 3 2403 4a r E

3

20

ˆ3

a

r

E r

Note: We implicitly assumed r > a when we calculated the enclosed charge.

Using Gauss’s Law to find E-field (2)A sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere?

a

•When computing the flux for a Gaussian surface, only include the electric charges inside the surface

ˆEE r

r

E

in 0 Eq

0 ˆV AE n 3 24

03 4r r E

3 20

0

ˆ 3 for ,

ˆ 3 for .

a r r a

r r a

rE

r

0ˆ 3r E r

Electric Field From a Line ChargeWhat is the electric field from an infinite line with linear charge density ?

ˆEE r

in 0 Eq

E

•Electric field must point away from the line charge, and depends only on distance•Add a cylindrical Gaussian surface with radius r and length L•Use Gauss’s Law

•The ends of the cylinder don’t contribute•On the sides, the electric field and the normal are parallel

r

L

0 ˆL daE n

n̂n̂

0EA 0 2E rL 02

Er

0

ˆ

2 r

rE

Electric Field From a Plane ChargeWhat is the electric field from an infinite plane with surface charge density ?

ˆEE k

in 0 Eq

E

•Electric field must point away from the surface, and depends only on distance d from the surface•Add a box shaped Gaussian surface of size 2d L W •Use Gauss’s Law•The sides don’t contribute•On the top and bottom, the electric field and the normal are parallel

0ˆLW daE n 0EA 0 2E LW

02E

0

ˆ

2

kE

n̂

n̂n̂

Conductors and Gauss’s Law•Conductors are materials where charges are free to flow in response to electric forces•The charges flow until the electric field is neutralized in the conductor

Inside a conductor, E = 0

•Draw any Gaussian surface inside the conductor

in 0 Eq 0 ˆdAE n 0

In the interior of a conductor, there is no charge The charge all flows to the surface

Electric Field at Surface of a Conductor•Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductor•Will be perpendicular to surface

•We can calculate it from Gauss’s Law•Draw a small box that slightly penetrates the surface•The lateral sides are small and have no flux throughthem•The bottom side is inside the conductor and has no electric field•The top side has area A and has flux through it

n̂

ˆE AE n EA•The charge inside the box is due to the surface charge •We can use Gauss’s Law to relate these inq A

0 E 0EA

0

ˆE n

Sample problemAn infinitely long hollow neutral conducting cylinder has inner

radius a and outer radius b. Along its axis is an infinite line charge with linear charge density . Find the electric field everywhere.

a

b perspective view

end-on view

•Use cylindrical Gaussian surfaces when needed in each region•For the innermost region (r < a), the total charge comes entirely from the line charge

•The computation is identical to before

•For the region inside the conductor, the electric field is always zero•For the region outside the conductor (r > b), the electric field can be calculated like before

•The conductor, since it is neutral, doesn’t contribute

0

ˆ

2 r

rE

0E

0

ˆ

2 r

rE

Where does the charge go?How can the electric field appear, then disappear, then reappear?

+

end-on view

•The positive charge at the center attracts negative charges from the conductor, which move towards it•This leaves behind positive charges, which repel each other and migrate to the surface

0

ˆif or

2

0 if

r a r br

a r b

r

E

–

–

––

–

–

––+

+

+

+

+

+

++

•In general, a hollow conductor masks the distribution of the charge inside it, only remembering the total charge•Consider a sphere with an irregular cavity in it cutaway

view q–

– ––

––+

+

+

++

+

out 2ˆek q

rE r