Download Lecture 9 - The Curious Case of Discontinuities

Lecture 9 - The Curious Case of
Discontinuities
A Puzzle...
Electron Jelly
Imagine a sphere of radius a filled with negative charge of uniform density, the total charge being equivalent to
that of two electrons. Assume that two protons are embedded in this jelly, and further suppose that, in spite of their
presence, the negative charge distribution remains uniform. Where must the protons be located so that the force on
each of them is zero?
(This is a surprisingly realistic caricature of a hydrogen molecule; the magic that keeps the electron cloud in the
molecule from collapsing around the protons is explained by quantum mechanics!)
Solution
Recalling Gauss’s Law, the electric field at a distance r ≤ a from the center of the jelly sphere (which has charge
density ρ = - 4 2 e 3 ) equals E 4 π r2  = ϵ1  43 π r3 ρ or equivalently
3
πa
0
E=
A proton at a distance r would feel the force Fjelly = E e =
rρ
3 ϵ0
(1)
r ρ e
3 ϵ0
pulling it towards the center of the jelly. To have
stable equilibrium, the other proton must also be at a distance r on the same diameter as the first proton, but in the
2
opposite direction. The repelling force between the two protons equals Fproton = (2k er)2 . Equilibrium occurs when
Fjelly = Fproton , which allows us to solve for r,
re
3 ϵ0
r ρ e
3 ϵ0
=
2e
π a3
=
r3
a3
=
4
3
r=
In retrospect, this factor of
1
2
k e2
(2 r)2
k e2
(2 r)2
1
8
a
2
(2)
is clear. If all of the -2 e electron charge were located in a point charge at the center,
it would provide a force on one of the protons that is 8 times the force due to the other proton (because the other
proton is twice as far away, and half as big). So the forces will balance if we reduce the effective electron charge
by a factor of 8. This is accomplished by reducing the effective radius of the jelly by a factor of 2. □
Advanced Electrostatics Problems
Building a Sheet from Rods
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Lecture 9 - 02-06-2017.nb
Example
An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly
charged rods. Using the fact that the electric field from an infinite rod is 2 πλϵ r pointing away from the rod,
0
integrate over these rods to show that the field from an infinite sheet with charge density σ is
σ
2 ϵ0
pointing away
from the sheet.
Solution
Assume the infinite sheet lies at z = 0 in the x-y plane. What is the electric field at the point (0, 0, z)? By symmetry, it must point in the z-direction. If we split up the infinite sheet into thin rods of thickness ⅆ x running parallel
to the y-axis, their charge density would equal λ = σ ⅆ x. The contribution of the electric field at our point in the zdirection would equal ⅆ Ez = 2 πλr ϵ zr where the last term picks out the z-component. Using r2 = x2 + z2 , the total
0
electric field equals
∞ σ ⅆx
∫-∞ 2 π ϵ0
z
x2 +z2
x=∞
=  2 πσϵ0 ArcTan xz x=-∞ =
σ
2 ϵ0
(3)
as desired. □
This is one of the most astounding results you find in physics! No matter how far away you are from the shell, the
electric field has the value 2σϵ pointing away from the plane. Additionally, no matter how close you come to the
0
plane, you still feel the electric field
σ
2 ϵ0
pointing away from the plane. However, if you pass through the plane, the
electric field will discontinuously change from
σ
2 ϵ0
pointing in one direction to
σ
2 ϵ0
pointing in the opposite direc-
tion! Furthermore, if you are a piece of charge embedded within the sheet, you will feel an electric field of 0 (by
symmetry), but if you move an infinitesimal distance off the sheet you will feel an electric field of 2σϵ .
0
This type of behavior is only exhibited by infinitely thin sheets of charge. Volume charge distributions create
continuous electric fields (since the integration in E = ∫ k ρ ⅆx'r2ⅆy' ⅆz' r smooths out discontinuities in ρ). And lines of
charges don’t seem as alarming because the electric field goes to ∞ near the line of charge (just like for a point
charge). So only 2D charge configurations have this bizarre behavior where the electric field is discontinuous but
stays finite everywhere.
As stated in the "Field from a Cylindrical Shell, Right and Wrong" section in the previous lecture, a great rule to
remember about discontinuities in nature is:
With all discontinuities in nature, the actual value at the discontinuity equals
the average of the discontinuous values!
(4)
This statement is worth its weight in gold; print it on bumper stickers and share it with the world! You will see it
come up again and again in physics (from charge distributions) and mathematics (Fourier series).
Extra Problem: Intersecting Sheets
Hole in a Shell
Consider a spherical shell of charge, of radius R and surface charge density σ, from which a small circular piece of
radius b ≪ R has been removed.
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Lecture 9 - 02-06-2017.nb
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What is the direction and magnitude of the field at the midpoint of the aperture?
Solution
Method 1: Imagine that we have a complete spherical shell overlaid with a thin hole of charge density -σ. By the
Principle of Superposition, this is identical to the above setup. For the spherical shell, Gauss’s Law easily yields
the electric field E = 0 inside the sphere and E = ϵσ pointing radially outwards. Now we add the effect of the thin
0
hole, which at the midpoint looks like an infinite plane of charge with an electric field
plane. Thus, inside the shell the electric field will be 0 +
electric field will be
σ
ϵ0
-
σ
2 ϵ0
=
σ
2 ϵ0
σ
2 ϵ0
=
σ
2 ϵ0
σ
2 ϵ0
pointing towards the
pointing outwards and outside the sphere the
pointing outwards. Hence, the electric field has magnitude
σ
2 ϵ0
and points
radially outwards at the midpoint.
Note that the electric field is not discontinuous! This makes sense, because in the problem there is no sheet of
charge at the point we are considering (and only sheets of charges can cause discontinuities). In our alternate
setup, there are two sheets of charges, but they have charge densities σ and -σ and therefore cancel each other’s
discontinuities.
Method 2: We could also find the solution by straight up integration. Instead of integrating over the entire sphere.
Note that if the aperture has radius b, then θ spans from Rb to π.
θ
R
Note that the distance from the point at angle θ to the middle of the aperture equals 2 R Sin 2θ  (which can either be
found through the Law of Cosines or simple geometry). Therefore, the electric field in the middle of the aperture
(which only has a component in the vertical direction) has magnitude
2 π π k R2 Sin[θ] σ
θ
2 Sin 2  ⅆθ ⅆϕ
4 R2 Sin 2θ 
π k σ π Sin[θ]
ⅆθ
2 ∫b/R Sin 2θ 
π
π k σ ∫b/R Cos 2θ  ⅆθ
E = ∫0 ∫b/R
=
=
θ=π
= 2 π k σ Sin 2θ θ=b/R
= 2 π k σ 1 - Sin 2bR 
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(7)
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Lecture 9 - 02-06-2017.nb
In the limit b ≪ R, Sin 2bR  ≈
b
2R
≈ 0 so that E ≈ 2 π k σ =
σ
,
2 ϵ0
as desired. □
A Plane and a Slab
Example
An infinite plane has uniform surface charge density σ. Adjacent to it is an infinite parallel layer of charge of
thickness d and uniform volume charge density ρ. All charges are fixed. Find E everywhere.
Solution
We will analyze the sheet and the slab separately, and then combine them using the Principle of Superposition.
We already know that the sheet has an electric field
σ
2 ϵ0
pointing away at all points in space. How can we deter-
mine the electric field from the slab?
There are many ways to compute the slab’s electric field: considering it as a conglomeration of thin sheets, straight
up integration, and Gauss’s Law name three distinct methods. We will use Gauss’s Law (but you are invited to
check the other methods).
Assume that the slab stretches infinitely far in the y- and z-directions, and that it spans from x = - d2 to x = d2 . We
use a Gaussian cylinder with length 2 x centered around x = 0 and cross sectional area A in the y-z plane. (Note that
the Gaussian cylinder must be centered around the middle of the slab, although it does not need to by a cylinder
and can instead be any prism shape.) Gauss’s Law yields 2 E A = ϵ1 (2 x A ρ) for x ≤ d2 and 2 E A = ϵ1 (d A ρ) for
0
x ≥
d
.
2
0
In other words,
E=
x ρ
ϵ0
dρ
2 ϵ0
x ≤
x >
d
2
d
2
(8)
with the direction +x for points x ≥ 0 and -x for points x < 0.
All that remains is to add together the contribution of the electric field from the sheet and the slab (noting the
directions of the fields!) to yield the final answer. The following Manipulate shows E as a function of x for various
values of ρσd .
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Lecture 9 - 02-06-2017.nb
ρd
σ
2
E
Out[4]=
ρd
ϵ0
- d2
d
2
x
σ
ϵ0
Note that there is only a discontinuity in the electric field at x = - d2 where the sheet of charge resides. There is no
discontinuity for the slab of charge. □
Advanced Section: Maximum Field from a Blob
Recommended Problems
This is a list of excellent problems (with solutions) in David Morin’s book.
◼ 1.6 Zero potential energy for equilibrium
Mathematica Initialization
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