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Imran Gujjar
Kaleem Ullah
Dilawar Gull
Sufyan Maqsood
Muhammad Umair Ghufran
Gauss’s Law
Karl Friedrich Gauss
(1777-1855) – German mathematician
Gauss’s Law
Already can calculate the E-field of an
arbitrary charge distribution using
Coulomb’s Law.
Gauss’s Law allows the same thing, but
much more easily…
… so long as the charge distribution is
highly symmetrical.
Karl Friedrich Gauss
(1777-1855) – German mathematician
Gauss’s Law
We found the E-fields in the vicinity of continuous charge
distributions by integration… booooo:
R
r
x
dE

E-field of charged disc (R>>x): E 
2 o
Now, we’ll learn an easier way.
Electric Flux
• Sounds fancy, but it’s not hard
• Electric Flux measures how much an electric
field wants to “push through” or “flow
through” some arbitrary surface area
• We care about flux because it makes certain
calculations easier.
Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is perpendicular to
the field
E  E A
Electric Flux – Case 1
Easiest case:
• The E-field is uniform
• The plane is perpendicular to
the field
E  E A
Electric Flux
Flux depends on how strong
the E-field is and how big the
area is.
Electric Flux – Case 2
Junior Varsity case:
• The field is uniform
• The plane is not
perpendicular to the field
 E  E A   E A cos 
Electric Flux – Case 2
Junior Varsity case:
• The field is uniform
• The plane is not
perpendicular to the field
 E  E A   E A cos 
Flux depends on how strong the E-field is, how big
the area is, and the orientation of the area with
respect to the field’s direction.
Electric Flux – Case 2
 E  E A   E A cos 
And, we can write this better
using the definition of the
“dot” product.
where:

A  A  nˆ
Electric Flux – Case 2
Quick Quiz: What would happen to the E-flux if
we change the orientation of the plane?
Electric Flux – Flux through a Closed Surface
 The vectors dAi point in
different directions
◦ At each point, they are
perpendicular to the surface
◦ By convention, they point
outward
Electric Flux: General Definition
E 



E  dA
surface
E-Flux through a surface depends on three
things:
1.How strong the E-field is at each infinitesimal
area.
2.How big the overall area A is after
integration.
3.The orientation between the E-field and each
infinitesimal area.
Electric Flux: General Definition
 
 E   E  dA
surface
Flux can be negative, positive or zero!
-The sign of the flux depends on the convention you
assign. It’s up to you, but once you choose, stick
with it.
Electric Flux – Calculating E-Flux
• The surface integral means the integral must
be evaluated over the surface in question…
more in a moment.
• The value of the flux will depend both on the
field pattern and on the surface
• The units of electric flux are N.m2/C
• The net electric flux through a surface is
directly proportional to the number of electric
field lines passing through the surface.
EG 24.1 – Flux through a Cube
Assume a uniform E-field
pointing only in +x direction
Find the net electric flux
through the surface of a cube
of edge-length l, as shown in
the diagram.
Gauss’s Law
Gauss’s Law
 NET
  qenclosed
  E  dA 
surface
0
The net E-flux through a closed
surface
Charge inside the surface
Gauss’s Law
In other words…
1. Draw a closed surface around a some
charge.
2.
Set up Gauss’s Law for the surface
you’ve drawn.
3.
Use Gauss’s Law to find the E-field.
 NET
  qenclosed
  E  dA 
surface
0
You get to choose the surface –
it’s a purely imaginary thing.
Gauss’s Law – confirming Gauss’s Law
Let’s calculate the net flux through a Gaussian
surface.
Assume a single positive point charge of
magnitude q sits at the center of our imaginary
Gaussian surface, which we choose to be a
sphere of radius r.
Gauss’s Law – confirming Gauss’s Law
• At every point on the sphere’s surface, the
electric field from the charge points normal to
the sphere… why?
• This helps make our calculation easy.
Gauss’s Law – confirming Gauss’s Law
At every point on the sphere’s surface, the
electric field from the charge points normal to
the sphere… why?
This helps make our calculation easy.
Gauss’s Law – confirming Gauss’s Law
Now we have:
But, because of our choice for the Gaussian
surface, symmetry works in our favor.
The electric field due to the point charge is
constant all over the sphere’s surface. So…
Gauss’s Law – confirming Gauss’s Law
This, we can work with.
 NET  E
 dA
surface
We know how to find the magnitude of the
electric field at the sphere’s surface.
Just use Coulomb’s law to calculate the E-field
due to a point charge a distance r away from the
charge.
Gauss’s Law – confirming Gauss’s Law
Thus:
 NET
 dA  4  r
sphere
ke q
 2  dA
r surface
2
And, this surface integral is easy.
Gauss’s Law – confirming Gauss’s Law
Therefore:
 NET
ke q
 2  (4  r 2 )
r
But, we can rewrite Coulomb’s constant.
  NET
Thus, we have confirmed Gauss’s law:
q

0
Flux due to a Point Charge
• A spherical surface surrounds a point charge.
•
What happens to the total flux through the surface if:
(A) the charge is tripled,
(B) the radius of the sphere is doubled,
(C) the surface is changed to a cube, and
(D) the charge moves to another location inside the surface?
Applying Gauss’s Law
Gauss’s Law can be used to
(1) find the E-field at some position relative to a known charge
distribution, or
(2) to find the charge distribution caused by a known E-field.
In either case, you must choose a Gaussian surface to use.
Applying Gauss’s Law
• Choose a surface such that…
1. Symmetry helps: the E-field is constant over the surface (or some
part of the surface)
2. The E-field is zero over the surface (or some portion of the surface)
3. The dot product reduces to EdA (the E-field and the dA vectors are
parallel)
• 4. The dot product reduces to zero (the E-field and the dA
vectors are perpendicular)