Gauss.9125736373:)#(#-+.kssjk8#-$+$8$ppt

Gauss

1. Gauss’ Law

2. Electric Flux
We have used electric field lines to visualize electric fields and
indicate their strength.
We are now going to count*the
number of electric field lines passing
through a surface, and use this
count to determine the electric field.
E
*There are 3 kinds of people in this world: those who can count, and those who can’t.

3. The electric fluxpassing through a surface is the number of
electric field lines that pass through it.
Because electric field lines are drawn
arbitrarily, we quantifyelectric flux
like this: 
E=EA,
…except that…
If the surface is tilted, fewer lines cut
the surface.
E
A
Later we’ll learn about magnetic flux, which is
why I will use the subscript E on electric flux.
E

The green lines miss!

4. E


A
The “amount of surface” perpendicular
to the electric field is Acos.
A
Effective= A cos so 
E= EA
Effective= EA cos.
We define A to be a vector having a
magnitude equal to the area of the
surface, in a direction normal to the
surface.
Therefore, the amount of surface area effectively “cut through”
by the electric field is Acos.
Remember the dot product from Physics 1135?E
EA  

5. If the electric field is not uniform, or the surface is not flat…
divide the surface into
infinitesimal surface
elements and add the
flux through each…
dA
Ei
E i i
A0
i
lim E A

   E
E dA  
A
Remember, the direction of dA
is normal to the surface.
a surface integral,
therefore a double integral

6. If the surface is closed (completely encloses a volume)…
E
…we count* lines going
out as positive and lines
going in as negative…E
E dA  
dA
a surface integral, therefore a
double integral
*There are 10 kinds of people in this world: those who can count in binary, and those who can’t.
For a closed surface, dAis normal
to the surface and always points
away from the inside.

7. What the is this thing?
Nothing to panic about!
The circle just reminds you
to integrate over a closed
surface.

8. Question: you gave me five different equations for electric flux.
Which one do I need to use? E
E dA   E
E dA   E
EA   E
EAcos   E
EA
Answer: use the simplest (easiest!) one that works.
Flat surface, E A, E constant over surface. Easy!
Flat surface, E not A, E constant over surface.
Flat surface, E not A, E constant over surface.
Surface not flat, E not uniform. Avoid, if possible.
Closed surface.
If the surface is closed, you may be able to “break it up” into
simple segments and still use 
E=E·A for each segment.
This is the definition of electric flux, so it is on your equation sheet.
The circle on the integral just reminds you to integrate over a closed surface.

9. Karl Friedrich Gauss
1777 –1855
•Made contributions in
–Electromagnetism
–Number theory
–Statistics
–Non-Euclidean geometry
–Cometary orbital mechanics
–A founder of the German
Magnetic Union
•Studies the Earth’s
magnetic field

10. Gauss's law states that the net number of
electric field lines leaving out of any closed
surface is proportional to the net electric
chargeqininside that volume.

11. On the other hand, electric field lines are
also defined as electric fluxΦEpassing
through any closed surface. Therefore, we
can arrive at the following formula for
Gauss's law
ΦE​= ∮E⋅dA=qin/ϵ0​​​

12. Gauss's law simplifies the calculation of
theelectric fieldassociated with the
distribution of a highly symmetric charge.

13. Sample Problems
1. A point charge of2μCis located at the center of a
cube with sidesL=5cm. What is the net electric flux
through the surface?

14. ​=226000 C/N⋅m2​​

15. Sample Problems
2. A uniform electric field of magnitude1.1×10
^4N/C isperpendicular to a square sheet with
sides 2.0 m long. What is the electric flux through
the sheet?
3.Calculate the flux through the sheet if the plane
of the sheet is at an angle of60°to the field.

16. 4.Find the electric flux through a rectangular
area3cm×2cmbetween two parallel plates
where there is a constant electric field of 30 N/C
for the following orientations of the area:
(a)parallel to the plates,
(b)perpendicular to the plates, and
(c)the normal to the area making a30°angle
with the direction of the electric field.
(Note that this angle can also be given
as180°+30°).

17. 5.A square surface of area2cm2is in a space of
uniform electric field of magnitude103N/C. The
amount of flux through it depends on how the
square is oriented relative to the direction of the
electric field. Find the electric flux through the
square, when the normal to it makes the following
angles with electric field:
(a)30°,
(b)90°, and
(c)0°.