4. Gauss’s law4.1 Electric flux

Definition:

•Uniform electric field

N

N

AN

EN

~

cos~

~

~

cosEAE

•Not uniform electric field

Perpendicular to the area

E

E

AA

cosAEE

cosAEEE

- number of electric field linesN

Aarea

VmCNmE 2Units:

Example: Flux through open surface

Compare the electric flux through two squares of areas A1 and A2 (A2>A1) placed in a region with a uniform electric field E:

A. Φ1 < Φ2

B. Φ1 = Φ2

C. Φ1 > Φ2

A1

(Side view)

The number of lines going through them is the same.

The effective area for the large square is A2cosθ = A1

A2

θ E

θ

A2 cosθ

•If you reverse the direction of A, you reverse the sign of the flux•For an open surface, choose any direction• For a closed surface it is conventional to take the area vector pointing outwards Ф > 0 lines going out Ф < 0 lines coming in Ф = 0 no lines or a balance between incoming and outgoing lines

A2

E E

The net flux is zero in both cases!

Every line that comes in, goes out.

Example 1: Flux through closed surface

A cubic box is placed in a region of uniform electric field as shown in figure 1. If the cube is tilted, as shown in figure 2, the net electric flux through it:

A. Increases

B. Decreases

C. Stays the same

Example 2: Two spheres have radii R and 2R.

Their centers lie on a positive charge. Compare the electric flux through the two surfaces:

A. ΦR < Φ2R B. ΦR = Φ2R C. ΦR > Φ2R +

R2R

The number of lines going through both surfaces is the same.

The area at 2R is 4 times larger, but

the electric field is 4 times smaller!

4.2 Gauss’s law - generalization of Coulomb’s law

224cos r

r

QkEAAEAEE 2r

QkE

1cos

4 2

rA

0/4 QkQE 04

1

kGauss’s law:

Example:

4.3 Applications of Gauss’s law

rhEEAAEQ E 2cos/ 0

•Cylindrical symmetry

1cos

2

rhA

E

h

Q

rrh

QE

00 22

•Shell theorem:

q

2r

QqkF r

q~

Q

0~ Fr~

R

Rr 2r

QkE

0~ E Rr ~

Examples (applications of Gauss’s law):Electric field created by

Point charge, charged sphere or spherical shell

(Spherical symmetry)

Long cylinder or tube

(Cylindrical symmetry)

Conducting plane surface, and

two conducting parallel plates

(Planar symmetry)

One conducting plate

(Planar symmetry)

rE

02

204 r

QE

0

E

02

E

L

Q

A

Q

EQF

Planar symmetry

a) Conducting surface b) Conducted plate c) Two conducting plates

+++++

E=0

E

0

0

0

1

/

E

A

QE

QEA

+++++

+++++ EE

0

0

0

2

2

1

/2

E

A

QE

QAE

+++++

-----

0

21

E

EEE

EE=0 E=0

Q Q/2+Q/2=Q +Q -Q

Example: A solid metal sphere of radius 3.00 m carries a total charge of 3.5μC. What is the magnitude of the electric field at a distance from the sphere’s center of (a) 0.15 m, (b) 2.90 m, (c) 3.10 m, and (d) 6.00 m? (e) How would the answers differ if the sphere were a thin shell?

(a) and (b) Inside a solid metal sphere the electric field is 0

(c) Outside a solid metal sphere the electric field is the same as if all the charge were concentrated at the center as a point charge:

6

9 2 2 3

22

3.50 10 C8.988 10 N m C 3.27 10 N C

3.10m

QE k

r

(d) Same reasoning as in part (c):

(e) The answers would be no different for a thin metal shell

6

9 2 2 2

22

3.50 10 C8.988 10 N m C 8.74 10 N C

6.00m

QE k

r

Example: Find the magnitude of the electric field produced by a uniformly charged sphere with radius R and total charge Q .

For r > R: 0QE 2

04 r

QE

3

3

00

3

3

R

rQq

R

rQ

V

VQq

E

encl

304 R

QrE

For r < R:

rR

E

Rr

2

1~r

24cos rEEAAEAEE

For r > R: 0QE

2

2

00

2

2

R

rQq

R

rQ

V

VQq

E

encl

For r < R:

rR

E

Rr

r

1~

rLEEAAEAEE 2cos

rrL

QE

00 22

20

20 22 R

r

LR

QrE

Example: Find the magnitude of the electric field produced by a uniformly charged cylinder with radius R length L and total charge Q .

Spherical conducting shell:

A charged isolated conductor

0E 0E

•The static electric field inside a conductor is zero – if it were not, the charges would move.•The net charge on a conductor in equilibrium is on its surface, because the electric field inside the conductor is zero.

||

0

000

cos

dEV

QE

QAE

inside conductor (+Q) + (-Q)=0

otside shell (+Q) + (-Q) + (+Q) = +Q

inside shell (+Q)

Gauss’s surface Charge inside the surface

Example: Compare the magnitude of the electric field at point P before and after the shell is removed.

A. Ebefore < Eafter

B. Ebefore = Eafter

C. Ebefore > Eafter

Hint: Use Gauss’s law to find EP.σoutσin

Q

P

σin

QIn both cases, the symmetry is the same. We will use the same Gaussian surface: a sphere that contains point P.

So the flux will look just the same:

And the enclosed charge is also the same: Q. Therefore,

22 RE

02

02

4

4

R

QE

QRE