Study Guide
Gauss’ Law
-
University:
Kentucky State University -
Course:
PHY 212 | General Physics II Academic year:
2021
-
Views:
333
Pages:
42
Author:
Quintin Ware
Gauss’ Law
Mathematically*, we express the idea of the last two slides as
q enclosed
Φ E E dA
ε0
We will find that Gauss’ law gives a very simple way to calculate
electric fields for charge distributions that exhibit a very high
degree of symmetry…
Hmm… Let’s check our the calculation of the electric flux
through the cylinder of the last few slides: the charge enclosed
is zero and thus the total flux is zero too!
Gauss’ Law
Always true, not always useful!
*“Mathematics is the Queen of the Sciences.”—Karl Gauss Example: use Gauss’ Law to calculate the electric field from an
isolated point charge q.
E
Let’s assume the point charge
is +.
The electric field everywhere
points away from the charge.
+q
If you go any distance r away
from +q, the electric field is
always directed “out” and has
the same magnitude as the
electric field at any other r,
i.e. E is constant in
magnitude and in direction.
What is the symmetry of the electric field? Spherical symmetry. Example: use Gauss’ Law to calculate the electric field from an
isolated point charge q.
q enclosed
E dA ε 0
To apply Gauss’ Law, we really
want to pick a surface for which
we can easily evaluate E dA.
So let’s draw a Gaussian
sphere of radius r, enclosing
and centered on +q.
“Centered on” makes it easy to
evaluate E dA.
E
+q
r
dA
Everywhere on the sphere, E and dA are parallel and E is
constant so
You do know the formula
for Asphere, don’t you? If
not, ask your 4th grade
Math teacher; Mrs. Layla! Example: use Gauss’ Law to calculate the electric field from an
isolated point charge q.
q enclosed
E dA ε 0
E
The charge enclosed by my
Gaussian sphere is q, so
+q
q
E 4r
o
2
r
dA
q
E=
4o r 2
The direction of E is shown in the diagram.
Or you can say E is “radially out.” Example: use Gauss’ Law to calculate the electric field from an
isolated point charge q.
E
q
E=
, away from +q
2
4o r
“But wait,” you say, “we
already knew the equation for
the electric field of a point
charge. We haven’t learned
anything new. It was a lot of
work for nothing.”
+q
r
Wrong! You have learned how to apply Gauss’ Law. You will
find this technique extremely useful on the following examples. Strategy for Solving Gauss’ Law Problems
Select a Gaussian surface with symmetry that “matches” the
charge distribution.
You want E to be constant in magnitude and in direction
Evaluate the surface integral (electric flux).
Determine the charge inside the Gaussian surface.
Solve for E. Example: calculate the electric field for 0b. We should do each
region separately.
What is the symmetry of the charge
distribution?
It is obviously spherical!
a
b Example (First Part): calculate the electric field for 00
a
b r
The charge distribution is spherically symmetric (you see the
“same” thing at any given r). Therefore you must see the
same electric field at any given r, so the electric field is also
spherically symmetric.
A spherically symmetric electric field is everywhere radial and
has the same magnitude at any given r. For >0, E is “out.” Example (Second Part): calculate the electric field for a0
a
b r
dA
That was easy so far, wasn’t it.
The “tricky” part is finding qenclosed. We have to determine the
charge inside the dashed-line Gaussian sphere.
The volume charge density is . The amount of charge in a
volume V is simply V. Example (Second Part): calculate the electric field for a0
a
b r
q enclosed q shell of inner radius a and outer radius r
q enclosed Vshell of inner radius a and outer radius r
qenclosed Vsphere of radius r Vsphere of radius a
q enclosed
4 3 4 3 4
r a r 3 a 3
3
3
3
Notice it doesn’t matter how
large b is. It’s only the
charge inside r that counts. Example (Second Part): calculate the electric field for a0
a
b r
3
3
r
a
3
0 r 2
Above is the magnitude E, the direction is radially out.
We still need to calculate the electric field for r>b. Example (Third Part): calculate the electric field for r>b.
q enclosed
E dA ε 0
E
>0
Vshell of inner radius a and outer radius b
E 4r
o
a
2
Hmm.. How about the
volume of the shell of
inner radius b and outer
radius r?
4
b3 a 3
E 3 2
4r o
E
b3 a 3
3o r
2
b r
dA
This is just Q/40r2,
like a point charge.
Why is it so?
, radially out Summary: electric field for 00
R
E E
r
dA
Looking down the axis of the cylinder. dA
>0
r
L
E
R E
dA
E dA 0 because E dA
dA
E dA 0 because E dA
E E
dA
r
L
E dA E dA because E // dA
Also E E must be constant at any given r. E
dA
r
You do know the formula
for Acylinder, don’t you? If
not, ask your 4th grade
Math teacher; Mrs. Layla!
L
E
E dA E dA E dA E dA
tube
tube
tube
E circumference of Gaussian cylinder length of GC
E 2r L dA
E
>0
r
R
L
q enclosed
A E 2r L
EEddA
o
Venclosed A base length
E 2r L
o
o
R 2 L
o dA
E
>0
r
L
E 2r L
R 2 L
o
R 2 R 2
E
2o r 2o r
R dA
E
>0
r
L
For positive :
R 2
E
, radially out
2o r
R2
In general: E
2o r
R Example 3: use Gauss’ Law to calculate the electric field due to
a long line of charge, with linear charge density .
This is very easy using Gauss’ Law (remember what a pain it
was in the previous chapter). Example: use Gauss’ Law to calculate the electric field due to a
long line of charge, with linear charge density .
Line is looooooong.
E
>0 E
r
dA
Looking down the line. dA
E
>0
r
L dA
E
>0
r
L
q enclosed λL
E dA E2π r L ε 0 ε 0
λ
For positive : E
, radially out
2πε0 r
2k λ
λ
E
or:
In general: E
r
2πε0 r Recall the result from
the previous chapter:
Q
E y kd
L
dx
L2
L 2
x
2
d
L
d
2 3/2
P
Evaluation of the integral (with the aid of an integral table or a
software package like Mathematica or Maple) gives us
2 πε 0 d
This is the same result but attained
here (using Gauss’ law) in a much
simpler way! Example 4: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
This is very easy using Gauss’ Law (study example 21.11 in
your text!!) Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
>0
Two views of sheet of charge; side view looking edge on, and
top view looking down. Sheet extends infinitely far in two
dimensions. Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
E
>0
E
For this electric field symmetry, we
usually use a “pillbox” (cylinder or
rectangular shape) for our Gaussian
surface. Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
E
>0
E
H
r
r
q enclosed
2
E dA 2 Eπ r ε 0
q enclosed σ π r 2 Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
E
>0
E
H
r
r
2Eπ r
2
σ
E
2ε 0
σ π r2
ε0
That sure was way easier
than the derivation back
in Example 21.11! Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density .
E
>0
E
H
r
r
For positive :
E
, away from the sheet
2ε 0
E is constant every where
σ
(r independent!).
E
In general:
We will recall this result
2ε 0
when we study capacitors
(Chapter 24). Example 5: an insulating sphere of radius a has a uniform
charge density ρ and a total positive charge Q. Calculate the
electric field at a point inside the sphere.
r
Q
a
qqenclosed
Venclosed
enclosed
A
EEddA
o
ε0o
4 3
r
3
2
E 4r
o
Calculate the electric field at a point outside the sphere.
Hmm… one has to anticipate that the field
is identical to that for a point charge:
Q/40r2
Quiz: check for the continuity of E(r) inside
& outside the sphere at r = a Gauss’ Law works well for three kinds of symmetry:
Charge Symmetry
spherical
cylindrical
planar
Gaussian Surface
concentric sphere
coaxial cylinder
pillbox
You can tackle very difficult (high-symmetry) problems in a very
easy way by using this technique.
Gauss’ Law
Always true, not always useful!
Gauss’ Law
of 42
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document