Download PES 1120 Spring 2014, Spendier Lecture 8/Page 1 Lecture

PES 1120 Spring 2014, Spendier
Lecture 8/Page 1
Lecture today:
1) More on electric flux
2) Gauss’ Law
3) Gauss’ Law and Coulomb’s Law
Announcement: Prof. Karen Livesey will fill in for me Monday 17th and Wednesday
19th of February. Lectures will be held in the usual room at usual time. HW 3 will be
due Feb. 19th. So please look at your HW 3 in advance and ask questions. You can
always e-mail with a specific question as well. I will come early and stay after class
on Friday to answer questions.
Last time we introduced a new idea called “flux” (Latin - fluxus - for flow)
To understand flux, we returned to the idea of a vector fields, and, in particular, I used the
idea of fluid (water) flow.
We used the analogy of water flow to describe flux in more intuitive terms.
When we are considering the electric field flux, nothing is “flowing”, but since both are
described by vector fields, the mathematics is the same.
How did we calculate flux?
When the area vector and the electric field are at an angle θ to each other, only the


component of E parallel to dA causes flux.


Confusingly, the component of E parallel to dA is called E since it is also along the
normal line. E  E cos q
E causes no flux because it doesn’t pass through the surface.
The small amount of flux from this one location is:
 
d   EdA cos q  E  dA
PES 1120 Spring 2014, Spendier
Lecture 8/Page 2
To find the flux through the entire surface, we visit every point on the surface, find d
and add those fluxes together to obtain the net electric field flux through a closed
surface:


 
E
r

dA



units
[N m2/C]
where Φ is the Greek capital letter for Phi (capital Phi). The loop on the integral sign
indicates that the integration is to be taken over the entire (closed) surface = called
Gaussian surface
The net electric flux through a closed surface is proportional to the net number of field
lines entering/exiting the surface.
Electric field flux  the number of electric field lines crossing the surface.
We then showed that the net electric field flux through a cubic surface in a uniform
electric field is zero!
Gaussian surface
This result turns out to be independent of the shape of the closed surface used. We could
have also used a closed sphere instead of a cube. If the number of electric field lines
INTO the surface equals the number of electric field lines OUT of the surface, the net
flux will always be zero!
But is this always true for flux, that it is zero?
Sources and Sinks:
Consider the vector field we discussed last time, in the r-direction, for a positive point
charge. This is a case of a vector field that has a singularity, in this case a source. Note
that there will be a net flux, since all the vectors at the surface are pointing out of the
surface, in the same direction as the area vectors for this surface.
PES 1120 Spring 2014, Spendier
Lecture 8/Page 3
Gaussian surface
If you reversed the direction of all the vectors, it would be a sink produced by a negative
charge.
Is this always true of the net flux with a source or sink? No! If the source (or sink) is not
contained INSIDE the volume, then there are vectors that point into and out from the
volume. It can be shown that the net flux in this situation is again zero.
Gaussian surface
Summary:
So, using the electric field line way of thinking about electric flux, the net flux
through a closed surface is always zero (ever field line that enters, also exits), unless
there is a source or sink enclosed within the volume bounded by the surface.
PES 1120 Spring 2014, Spendier
WORKSHEET: A) The concept of Flux
Lecture 8/Page 4
PES 1120 Spring 2014, Spendier
Lecture 8/Page 5
So what have we learned?
1) Φ is non-zero only if some charge exists inside the Gaussian surface
2) Φ is proportional to the charge enclosed by the Gaussian surface:
qenclosed  0    0
qenclosed  0    0
qenclosed  0    0
We can put these 2 observations together:


 
E
r

dA
 q enclosed



This is precisely the law Carl Friedrich Gauss (1777-1855), German mathematician and
physicist) came up with! It turns out that the constant of proportionality is ε0.
Gauss’ Law: The total electric flux through any closed surface depends on the total
charge inside that surface.


 
q
E
r

dA
 enc



0
A closed surface is one that has a finite volume and the shape of the surface does not
matter. But we will see soon that the shape
 of the Guassian surface should be chosen
carefully in order to be able to calculate E using Gauss’ Law.
WORKSHEET: A) The concept of Flux
PES 1120 Spring 2014, Spendier
Lecture 8/Page 6
Gauss’ Law and Coulomb’s Law:
Example 1: Calculate the Electric field at a distance r from a positive point charge using
Gauss’ law:

 
E
r

dA




1) First, we note that the electric field is everywhere perpendicular to the surface, so that
E and dA are in the same direction everywhere on the surface.
2) Since the sphere is centered on the charge, then the electric field also has the same
magnitude everywhere on the surface.
Since the magnitude of the field is constant, and the dot product just gives 1 everywhere,
then both can be taken out of the integral over the surface for the flux.


 
 E  r   dA  E  r  rˆ  rˆ   dA  E  r   4r 
2
2
since 4r equals the surface area of a sphere. Hence
  E  r   4r 2 
and
E r 
q

q
 enc 2 
2
4 r
4 0 r
4 0 r 2
Since  
q enc
and q enc  q
0
The above result is the same as Coulombs Law:

1 q
E r  
4pe0 r 2
So Gauss’ Law gives us the same result as Coulomb’s Law for the electric field of a point
charge.
Another example where Gauss’ Law makes calculations simpler than Coulomb’s Law.
Normally this is in situations where the electric field has a certain symmetry:
point charge  radial symmetry
PES 1120 Spring 2014, Spendier
Lecture 8/Page 7
flat sheet with charge density σ  symmetry in x-y plane
Hence, we have to be smart and pick just the right surface for the given distribution. A
good start for picking the right Gaussian Surface is one where at every point on the
boundary:
1) The electric field is constant
2) The angle θ = 0 or θ = 180 or θ = 90 (since E=0 is also a constant!)
Given (1) and (2)  for a Gaussian surface: Φnet = EA
Let’s apply this to our charged disk problem. Remember the difficult integration needed
to compute E of a charged disk using Coulomb’s Law? We will now see that Gauss’ Law
is simple to apply.
Example 2: E-field of an Infinite Sheet of Charge
Step 1: Find a closed surface over which the electric field is constant (zero is also a
constant) and either perpendicular, or parallel to the surface.
Gaussian surface: a cylinder with the flat “caps” parallel to the sheet.
Step 2: If E is constant over a surface, E and the dot product can come out of the integral.
PES 1120 Spring 2014, Spendier
Lecture 8/Page 8
Step 3: Find charge enclosed and solve for E.
NOTE: we could have also picked a cube with two faces parallel to the sheet as Gaussian
surface. You can proof it to yourself that the result will be the same as above.