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Gauss’ Law
Seed Question: Consider an imaginary sphere of radius r centered on +q:
A) How would the flux change if you doubled r?
B) How would the flux change if you doubled +q?
Use tiny sketches!
Exploration: From B) we conclude that the flux  is proportional to ______________.
In other words  = _______________________.
Let’s apply Gauss’ Law to a charge +qenclosed. Start by sketching the lines of force. Then pick
a shape with high symmetry that is centered on +qenc, so all the lines of force are  to the
surface. Sketch several electric field vectors on the surface.
Find the flux through your surface:
∮ 𝐸⃗ ∙ 𝑑𝐴 =
But we also know that the flux is
(see above)
=
Putting it all together:
+qenc
Solving for E:
E=
We need to find what the constant
is…Put a test charge qo on the surface,
and use Coulomb’s Law to find FE at the
position of qo.
FE =
so
E=
Comparing the electric fields, the constant must be:
constant =
Putting this all together:
∮ 𝐸⃗ ∙ 𝑑𝐴 =
Which is Gauss’ Law!
Let’s test this out on a case we are already familiar with.

Use Gauss’ Law to find E near an infinite sheet with
uniform charge density  : Sketch the lines of force first.
Then pick a simple Gaussian surface where the electric
field will be either parallel or perpendicular to each part
of the Gaussian surface.
Sketch the end cap area vectors and indicate the charge
enclosed.
Calculate the flux through each area and set it equal to
the charge enclosed.
Does your result agree with your earlier entry’s brute
force integral?
Which technique do you like better?
Big Idea: Gauss’
∮ 𝐸⃗ ∙ 𝑑𝐴 =
𝑞𝑒𝑛𝑐
𝜖𝑜
Where 𝜖𝑜 =
⃗
Note: ∮ 𝐸
Law
1
4𝜋𝑘
= 8.85 𝑥 10−12
𝐶2
𝑁𝑚2
∙ 𝑑𝐴 is called “the flux.” The circle means you must integrate over a closed area.
Discussion: ?