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TRANSCRIPT
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.