SP212 Lesson 5
Ch. 23 – Electric Flux and Gauss’ Law
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The Electric Field (again)
Whew! Chapter 22 was brutal. Unfortunately we can’t escape the electric field,
it’s absolutely vital. But we can work smarter, not harder. We’ll still use calculus,
but now let’s be elegant. Let’s use Gauss’s Law.
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Flux
Gauss’s Law depends on the concept of electric flux through a surface, which can
be thought of as the amount of electric field piercing the surface.
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Defining our Terms
~ to represent our area.
We define an area vector, A
~ has magnitude equal to the area in question
A
~ has direction perpendicular to the surface, pointing out
A
The flux Φ through the surface is
~
Φ = vA cos θ = ~v · A
That’s for any vector field ~v. It therefore applies to an electric field!
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Flux of an Electric Field
Now that we know how to calculate the flux through a single area (like a window)
it’s easy to define the flux through an object with more than one side: Just add
the flux through each side
X
~E · ∆A
~
Φ=
We can extend that to complicated, curved surfaces by using calculus in the usual
way:
I
~
Φ = ~E · d A
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Example
Example
The square surface shown in the figure measures 3.2 mm on each side. It is
immersed in a uniform electric field with magnitude E = 1800 N/C and with field
lines at an angle of 35◦ with a normal to the surface, as shown. Calculate the
electric flux through the surface.
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Solution
Very straightforward, we apply our definition of flux:
~
Φ = ~E · A
= EA cos θ
= (1800 N/C)(0.0032 m)2 cos(145◦ )
Φ = −1.5 × 10−2 Nm2 /C
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Example The Second
Example
A cylinder is immersed in a uniform electric field ~E as shown. The axis of the
cylinder is parallel to the field. What is the total flux Φ through the entire closed
surface?
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Solution
Do it side by side!
I
Φ=
~E · d A
~
Z
=
Z
E cos θ dA +
a
Z
E cos θ dA +
b
SP212 Lesson 5
E cos θ dA
c
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Side a
On this side, the electric field is anti-parallel to the area vector, so θ = 180◦
Z
E cos(180◦ ) dA
a
Z
−E
dA
− EA
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Side b
On this side, the electric field is perpendicular to the area vector, so θ = 90◦
Z
E cos(90◦ ) dA
a
0
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Side c
On this side, the electric field is parallel to the area vector, so θ = 0◦
Z
E cos(0◦ ) dA
a
Z
E
dA
EA
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Add ’em up!
Now add our results from all three sides to get the total flux:
Φ = −EA + 0 + EA = 0
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Challenge: FLux
A surface as the area vector:
~ = (2 ı̂ + 3 ̂ )m2
A
What is the electric flux through the area if the field is....
(a) ~E = 4 ı̂N/C
(b) ~E = 4 k̂k N/C
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Gauss’s Law
Well, that was fun, but what does this have to do with electrostatics? It turns
out, there’s a simple connection between the flux of an electric field through a
closed surface, and the net charge contained within it. Gauss’s law states:
I
0 Φ = qenc
~E · d A
~ = qenc
0
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Gauss’s Law and Coulomb’s Law
Let’s look at a single positive point charge q, surrounded by a
spherical gaussian surface. We know that all the differential
~ must be pointed radially outward. We also
area vectors d A
know the electric field must be radially directed outward
everywhere, so everywhere they are parallel, and θ = 0.
Gauss’s Law then becomes
I
I
~E · d A
~ = qenc = E dA
0
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Gauss’s Law and Coulomb’s Law
E is constant all over the surface of the sphere, so it comes out of the integral.
The integral is then just the surface area of the sphere:
Z
q
E dA =
0
q
E (4πr 2 ) =
0
1 q
E=
4π0 r 2
Behold! The same expression we obtained from Coulomb’s Law!
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Challenge: Gauss’s Law
The figure shows, in cross section, three solid cylinders, each of length L and
uniform charge Q. Concentric with each cylinder is a cylindrical Gaussian surface
with all three surfaces having the same radius.
Rank the Gaussian surface according to the value of the electric field on any point
on the sides, greatest first.
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