Phys102 Lecture 4

Gauss’s Law

Key Points

• Electric Flux

• Gauss’s Law

• Applications of Gauss’s Law

References

SFU Ed: 22-1,2,3.

6th Ed: 16-10,+.

Phys102 Lecture 4 - 1

Electric flux:

Electric flux through an

area is proportional to

the total number of field

lines crossing the area.

Electric Flux

The direction of area if defined as

the normal direction.

Example 22-1: Electric flux.

Calculate the electric flux through the

rectangle shown. The rectangle is 10 cm

by 20 cm, the electric field is uniform at

200 N/C, and the angle θ is 30°.

Flux through a closed surface:

in (-)

out (+)

(Just a new notation)

The total flux through a closed surface is equal to

the charge enclosed divided by e0:

This can be used to find the electric field in situations with

a high degree of symmetry.

To use Gauss’s law to calculate E, we need to choose a

closed surface such that it is easy to calculate the

summation. Specifically, E is constant on the surface.

Sometimes the surface can consists of a few parts, on

each part E should be a constant or 0.

22-2 Gauss’s Law

AE

Total

For a point charge, we choose a spherical surface as our

Gaussian surface

Therefore,

Non-calculus version:

Solving for E gives the result we

expect from Coulomb’s law:

2

0

4 rEAEQ

e

On every point of the GS,

E = constant, and AE

//

If there is a positive charge enclosed, the total outward flux is positive.

This means the positive charge is are the source of the E-field lines.

If there is a negative charge enclosed, the total outward flux is

negative, meaning there is a net inward flux. This means the negative

charge is a sink of the E-field lines.

Gauss’s Law and the source of E-field

0eEncl

Total

QAE

I-clicker question 4-1:

The total flux of magnetic field through a closed surface should be:

A) Always positive.

B) Always negative.

C) Always zero.

D) Can be any value.

E) No idea.

Example 22-3:

Spherical conductor.

A thin spherical shell of

radius r0 possesses a

total net charge Q that

is uniformly distributed

on it. Determine the

electric field at points

(a) outside the shell,

and (b) within the shell.

(c) What if the

conductor were a solid

sphere?

Example 22-4: Solid sphere

of charge.

An electric charge Q is

distributed uniformly

throughout a nonconducting

sphere of radius r0.

Determine the electric field

(a) outside the sphere

(r > r0) and (b) inside the

sphere (r < r0).

I-clicker question 4-2:

This question is the same

as part C of the previous

question.

A) True.

B) False.

Example 22-6: Long uniform line of charge.

A very long straight wire possesses a uniform positive charge

per unit length, λ. Calculate the electric field at points near (but

outside) the wire, far from the ends.

Example 22-7: Infinite

plane of charge.

Charge is distributed

uniformly, with a surface

charge density σ (σ =

charge per unit area =

dQ/dA) over a very large

but very thin

nonconducting flat plane

surface. Determine the

electric field at points

near the plane.

Example 22-8: Electric

field near any

conducting surface.

Show that the electric

field just outside the

surface of any good

conductor of arbitrary

shape is given by

E = σ/ε0

where σ is the surface

charge density on the

conductor’s surface at that point.

Procedure for Gauss’s law problems:

1. Identify the symmetry, and choose a

gaussian surface that takes advantage of it

(with surfaces along surfaces of constant

field).

2. Draw the surface.

3. Use the symmetry to find the direction of E.

4. Evaluate the flux in terms of E.

5. Calculate the enclosed charge.

6. Solve for the field.

E

Gauss’s law only needs the enclosed charge

inside the Gaussian surface. Any charges

outside the GS do not appear in the formula.

Why? (i-clicker question 4-3)

A conceptual question about Gauss’s law

0eEncl

Total

QAE

A) Gauss’s law doesn’t apply when there are charges outside the

Gaussian surface (GS).

B) Charges outside the Gaussian surface do not contribute to the electric

field inside the GS.

C) We choose the GS surface such that the charges outside does not

appear in the formula.

D) Because of the symmetry.

E) Charges outside the GS do not contribute to the total flux through the

GS.