Transcript ppt document

Gauss’ Law
Besides adding up the electric fields due to all
the individual electric charges, we can use
something called Gauss’ Law. Its idea is
similar to mass conservation in fluid flow
involving sources and sinks.
To see the idea behind this law and how it
works, let’s look at the flow of water.
Gauss’ Law for velocity
Consider the flow of water from a pipe. The amount
of water can be measured by the volume, V, per
time, t: V/t is a measure of the flow.
Volume can be found by multiplying the area by the
perpendicular distance: dV = A*ds where ds is
always perpendicular to A. We can write this in
terms of a dot product: dV = Ads
(where the vector A has a direction perpendicular to
the surface)
so we have dV/dt = A ds / dt = v dA.
Gauss’ Law for velocity
If we integrate over a complete closed surface, then
we have dV/dt (created or destroyed) = v dA .
However, we know mass (and water) can’t be
created or destroyed, so v dA = 0.
Electric field is a vector just like velocity is a vector.
The sources of electric field are charges. Thus, in
analogy with water, we suspect that E dA
should be proportional to sources or sinks of
electric field. The sources and sinks of electric
field we can identify as charges!
Single Point Charge
Consider a point charge. It is the source of an
electric field that goes radially out from the
point. The Electric Field is not absorbed by
space; instead it can only come out of a
positive charge or end on a negative charge!
This is similar to water - it is not absorbed by
space but can only come out of a source or go
down a drain.
Single Point Charge
If we enclose this point charge inside a closed
area, the amount of electric field coming out
of the area will depend only on the amount
of charge inside the area and not on the
area. The field coming out of the green
sphere is equal to the field coming out of
the blue
sphere and
equal to
the field coming
out of
the purple box!
Gauss’ Law
This is just like the flow of water. The flow of
water depends on how much comes out of
the source (and how much goes into a drain
or sink).
The mathematical statement of this for
Electric fields (Gauss’ Law) is:
closed area E dA = constant * Qenclosed .
Gauss’ Law
closed area E dA = constant * Qenclosed
To find the constant, we look at our point
charge case, and we use a sphere for our
closed area.
In this case, the Electric field for the point
charge is: E = kQ/r2, and it points directly
away from the charge. The area for each
part of the sphere also points directly away
from the center of the sphere. Hence, the
factor of cos(θ) in the dot product gives
cos(0o) = 1 everywhere.
Gauss’ Law
closed area E dA = constant * Qenclosed
Also, since the sphere is equidistant from the
center (where the charge is), the Electric
field due to the point charge is constant (E =
kQ/r2). Hence the integral can be
integrated to give: EA = constant*Qenclosed
or with the area of a sphere being A = 4pr2 :
(kQ/r2)(1)(4pr2) = constant*Q , or:
constant = 4pk.
Gauss’ Law
Thus the final statement for Gauss’ Law is:
closed area EdA = 4pkQenclosed .
In determining the final form for this relation
we put in the known form for the electric
field, E. However, we often use this
equation to solve for E. But in order to do
this, we need to integrate some function that
we don’t know. We do this by employing
symmetry. (We used the spherical
symmetry of a point for the point charge.)
Gauss’ Law and Coulomb’s Law
Actually, Gauss’ Law and Coulomb’s Law say
the same thing and are really different
expressions of the same law. Recall that a
law is something not derived from previous
ideas, but rather something that we find by
experimentation that works.
Sometimes we express the constant k in the
form: k = 1 / 4peo , where the sub naught
indicates we are working in vacuum.
Hollow Sphere
closed area EdA = 4pkQenclosed = Qencl/eo .
Let’s apply Gauss’ Law to a hollow charged
sphere.
First, what is the electric field (E) inside a
hollow charged sphere?
To be definite, consider the point marked with
an x in the figure below.
+Q
By symmetry, Ey= 0 . But
R
x
what about Ex ?
Inside Hollow Sphere
closed area EdA = 4pkQenclosed .
All the charge to the right of the green dotted
line causes an electric field to point to the
left (away from the charge), while all the
charge to the left of the line causes the
electric field to point to the right (again,
away from the charge). There is less charge
+Q
to the right, but those charges
x
are closer; there is more
charge to the left but they
are farther away! Which wins?
Inside Hollow Sphere
closed area EdA = 4pkQenclosed .
Let’s try using Gauss’ Law. We first need to
choose an area. To take advantage of the
symmetry, let’s choose a sphere that goes
through our point (blue sphere). By
symmetry the integral reduces to:
EAblue sphere = 4pkQenclosed in blue sphere .
However, Qenclosed = 0, and so
+Q
Einside = 0 .
x
Outside Hollow Sphere
Now we’ll consider the electric field at a point
outside the charged sphere. We’ll again
choose a definite point marked with the blue
x on the diagram below.
To use Gauss’ Law, we need to
choose a definite closed area,
+Q
which, again using the symmetry,
R
will be a sphere going through
x
r
our point, x.
Outside Hollow Sphere
closed area EdA = 4pkQenclosed .
Again, by symmetry the integral can be
integrated to give: EA = 4pkQenclosed .
Thus we have, with A = 4pr2 ,
E = 4pkQenclosed / 4pr2 = kQ/r2 .
This is just the same formula
we had for a point charge!
+Q
R
r
x
Charged Sphere
Hollow Sphere
Results using Gauss’ Law:
Inside, E = 0 .
Outside, magnitude: E = kQ/r2 and
direction points away from the center of
the sphere (just like a point charge).
What about the case of a SOLID sphere of
charge? (In this case, the charges must be
locked into place, because they would normally
try to get as far away from each other as possible and end up on the surface - the same situation as a
hollow sphere of charge!)
Solid Sphere of Charge
We can consider that a solid sphere of charge
is made up of a bunch of hollow spheres of
charge that have a common center. Thus
the charges on the outside of any sphere do
not contribute to the electric field, whereas
the charges inside a particular hollow sphere
would contribute to the field at the hollow
sphere.
Solid sphere of mass
Do gravitational fields act just like electric
fields? YES, only the charges are replaced
by masses and the k is replaced by the G.
Is there a Gauss’ Law for gravitational fields?
YES! closed area gdA = 4pGMenclosed
What about the gravitational field due to a
solid sphere? We have just shown that the
gravitational field should be exactly like a
point mass at the center of the sphere.
(This is what we assumed when we considered the gravity of
the earth in PHYS 150 !)
Gauss’ Law: other symmetries
What other symmetries might we be able to
use Gauss’ Law for?
• We’ve already considered spherical
symmetry which works for points and
spheres.
• How about cylindrical symmetry which
should work for charged lines and
cylinders?
• How about plane symmetry which should
work for plates of charge.
Line of Charge
Consider a long line of charge that has a
uniform charge density, l = Q/L . What is
the electric field strength at a point (marked
x) a distance a away from the wire?
By considering symmetry, we know that the
Electric Field, E, must point directly away
from the wire. (We already did this problem the
straightforward way, but let’s do it using Gauss’
Law for practice.)
x
E
a
l=Q/L
Line of Charge
closed area EdA = 4pkQenclosed .
To use Gauss’ Law, we need to choose an area
based on the symmetry of the situation.
Let’s choose a cylinder of length L and radius
a that goes through our point. The cylinder
has three separate areas: the left side, the
right side, and the “can” part.
x
a
l
L
Line of Charge
For a long wire, we know by symmetry that
the Electric field must point away from
the wire and so must be perpendicular to the
area on both the left and right sides.
Because of the dot product between E and
dA, these two parts of the cylinder do not
contribute to the integral.
dA
E
x
E
dA
a
L
l
Line of Charge
closed area EdA = 4pkQenclosed .
For the “can” part, however, E and dA are parallel,
and so the cos(θ) factor in the dot product in
Gauss’ Law gives 1. Also, by symmetry, the
magnitude of the electric field is constant along
the can part. Thus Gauss’ Law gives:
EAcan = 4pkQenclosed .
With Acan = 2paL and
Qenclosed = lL, we have
dA E
E = 4pk lL / 2paL = 2kl/a .
x
a
l
L
Line of Charge
E = 2kl/a directed away from the
positively charged wire.
Note that this is the same formula for electric
field that we obtained using the
straightforward way. However, the
straightforward way involved a hard
integral, whereas the Gauss’ Law way
involved a simple integral (once we chose
the right area based on the symmetry).
Cylinder of Charge
If we have a long cylinder of charge, we can
again use Gauss’ Law to determine the
electric field inside and outside the cylinder.
l= DQ/DL
Inside a Cylinder of Charge
For a point inside the cylinder, we can
choose an area in the shape of a cylinder
that goes through our chosen point, x.
Because on the left and right sides E and dA
are ^, they do not contribute. And by
symmetry, the “can” part is constant to give
EAcan = 4pkQencl. But since Qencl = 0,
Einside= 0 !
l= DQ/DL
E
dA
x
E
dA
R
Outside a Cylinder of Charge
For a point outside the cylinder, we can
choose an area in the shape of a cylinder
that goes through our chosen point, x.
The left and right ends of the cylinder do not
contribute since E and dA are
perpendicular.
x
E
dA
l= DQ/DL
Outside a Cylinder of Charge
The “can” part has E and dA parallel, and the
E is constant along the “can”. Hence
Gauss’ Law gives: EAcan = 4pkQenclosed .
With Acan = 2paL and Qenclosed = lL, we
have Eoutside = 4pk lL / 2paL = 2kl/a , the
same as for a long line of charge!
dA
x
E
l= DQ/DL
Charge on Plates
We have considered a point (spherical) and a line
(cylindrical). Now we extend our analysis to a
plane (plate) of charge.
We are looking for the E field at a point, x, at a
distance a above a large plate of charge with
charge density s= Q/A.
Again by symmetry, we know
the E field points away
E
from the plate.
x
a
s=Q/A
Above a Charged Plate
closed area EdA = 4pkQenclosed .
To use Gauss’ Law, we need to choose an area
based on the symmetry of the situation.
Let’s choose a box of length L that has a side
going through our point (so a = L/2).
E
x
a=L/2
L
L
s=Q/A
Above a Charged Plate
closed area EdA = 4pkQenclosed .
The four sides have dA’s that are horizontal, while
the E field is vertical. Thus the dot product in
Gauss’ Law says these do not contribute.
However, both the top and bottom sides have
areas that are parallel
to the field. Hence we
E
dA
have EAtop + EAbottom =
x
a=L/2
2EA = 4pkQencl, or
s=Q/A
dA
dA
E = 2pks.
L
L
Review
For a point or outside a sphere, E = kQ/r2 .
For a line or outside a cylinder, E = 2kl/a .
For a plate of charge, E = 2pks=s/2eo .
There are several things to note:
1. All the formulas have a k in them.
2. a) The field due to a point falls off as r-2,
b) the field due to a line falls off as a-1,
c) whereas the field due to a plate is
constant!
How come? And what about units?
Units
As to units, note that the line and cylinder
formula has a l in it, and l has an inverse
distance in it (l=Q/L) to go with the a-1.
The plate formula has a s in it, and recall
s=Q/A, and A has distance squared in it.
So the units do work out!
Physical Behavior
But why the difference in strength with distance for
the three cases?
Note that, for the line of charge, as the field point gets farther
away from the line, the charges further along the line do
not get that much farther away (green line versus blue line)
but the angle gets steeper, so the vertical component
actually gets bigger. This somewhat larger vertical
component somewhat diminishes the lesser effect the
charges right underneath the point have on the field as the
point gets farther away.
x
x
Physical Behavior
The same thing happens for the plate, except the plate is
in two dimensions instead of just one. In other
words, at any particular distance from the point on the
line there are only two charges (one on the left and
one on the right); for the plate there is a whole ring at
any particular distance, and as the distance gets
bigger, the size of the ring, and hence the amount
of charge on the ring, gets bigger! This increase in
charge and increase in vertical component of the
further charges cancels the decrease in strength of the
nearer charges exactly.
Special Case: Coaxial Cables
What is the electric field around a coaxial
cable that has a positive charge on the
(inner) wire and an equal but opposite
charge on the (outer) cylinder? All we have
to do is add the fields due to each.
(I leave it to you to investigate this. What is the
electric field between the inner wire and the
cylinder? What is it outside the cylinder?)
Special Case: Parallel Plates
Consider the situation of having two plates
that are parallel with one plate having a
positive charge and the other plate having
an equal but opposite charge. What is the
electric field outside the two plates? What
is the electric field between the two plates?
Computer Homework
for Gauss’ Law
The 2nd computer homework program on
Vol. 3 is on Gauss’ Law and will give you
practice in applying the results to specific
problems.