1

However, in some cases, a SYMMETRIC CHARGE DISTRIBUTION allows us to guess the orientation of the corresponding electric field. That is the case, for example, when we consider a INFINITELY-LONG line of uniform charge distribution (λ= charge per unit length.)

Asymmetric charge distribution

Symmetric charge distribution

Andres La Rosa Lecture Notes Portland State University PH-212

THE GAUSS' LAW

Q

+ + + + + + + + + + + + + + + + + + + + +

The electric field at any point away from the line (A and B, for instance) turns out to be perpendicular to the line.

2

It turns out, the Gauss's law will allow calculating the magnitude, as we will see below

+ + + + + + +

+ + + + + + +

Since the points P, Q, R and S are at the same distance from the charged-line, the corresponding electric fields should have the same magnitude:

Vertical line

Hypothetical ring on a horizontal plane

3 The Gauss's law (to be described below) is a tool that allows to calculate (in a very simplified way) the electric field produced by symmetrically distributed charges Note: But keep in mind that Gauss's law is valid for both, symmetric or asymmetric charge distributions.

4

Gauss's Law THE GAUSS' LAW

5

Definition of the solid angle ΔΩ

ΔΩ

ΔΩ

ΔΩ ΔS ΔS

ΔS

CASE: ΔS is parallel to r

ΔΩ

Magnitude of ΔS

Total solid angleenclosed by aspherical surface

Sphere ofradius R

6

ΔΩ

ΔΩ

CASE: ΔS is not parallel to r

ΔS ΔS

ΔS

7

ΔS

ΔΩ

ΔS ΔΩ =

ΔΩ =

8

Definition of the electric flux

9

Examples of electric flux

10

Question: What is the electric flux that crosses the surface ABCD?

72NC

11

Questions

11

E ΔS Cos θ

θ

Electric flux through a closed surface

CASE: The electric field is produced by one point-charge located inside the closed surface

Flux depends only on the solid angle, independent on the distance of ΔS to the point-charge !!

E

ΔΩ

= 4π

13

whereφ is the electrical flux

crossing the mathematicalsurface S ,

andq is the point charge

inside the surface S

q

q1

q

CASE: More than one point-charge are located inside the mathematical closed surface

q2

IN GENERALnet charge insideS

Gauss' Law

14

CONCLUSION: Charges located outside the surface S do not contribute to the electric flux

Exercise

charge inside

q1q4

q2

q3

q2 + q3

q1, q2, q3 , q4 , q5

Exercise: Evaluate the electric flux across the surface S produced by the give discrete charges indicated in the figure

This result is valid for any arbitrary surface S

q5

E

17

Applying Gauss' Law to problems that present cylindrical symmetry

18

Applying Gauss' Law to problems that present planar symmetry

See also textbook,page 617

S cylindrical surface(it has a circular base of area A)

S

19

σ

σ

20

21

22

σ1 σ2E1 E1 E1

E2 E2E2

E2

E1

EA = E1 + E2

Example

EC = E1 + E2 =

σ1 σ2E1 E1 E1

E2 E2 E2

+ E2 = E1 =

Example

This arrangement ofcharges is used todescribe(approximately) theworking principle of aCAPACITOR

E

E

X

X

-

23

Applying Gauss' Law to problems that present spherical symmetry

E (r) =

E

E

E

CASE 1: Spherical shell uniformly charged.

Question:What is the electric field inside the sphere?

Answer

S: Gaussian surfaceA spherical surface of radius r

E

E

Q is the total charge on the sphere

24

Question:What is the electric field outside the sphereof radius R?

S: Gaussian surfaceA spherical surface of radius r

E

25

E

26

ECASE 2: Compact sphere of radius R uniformly charged.

S: Gaussian surfaceA spherical surface of radius r

27

E

28

29

We have just determined the electric field E for positions r<R (that is, fields inside the charged dielectric sphere). Let's procedd calculating E for positions outside the sphere, that is for r>R

30

31

32 Under electrostatic conditions, what is the electric field inside a conductor?

33 A compact conductor has a total charge Q. Under electrostatic conditions, where are those charges located?

There is not net charge inside the conductor. All the charges distribute at the outer surface of the conductor

r

Would there be charge inside the conducting sphere?

R

r < R

Gaussian surface

Let's use Gauss' law

Therefore, charge inside = 0

34 What is the direction of the electric field near the surface of a conductor?