Download Notes on Electric Fields of Continuous Charge Distributions

Notes on Electric Fields
of Continuous Charge Distributions
For discrete point-like electric charges, the net electric field is a vector sum of the fields
due to individual charges. For a continuous charge distribution along 1, 2, or 3 dimensions
of space,
dQ(x) = λ × dx,
or dQ(x, y) = σ × dx dy,
or dQ(x, y, z) = ρ × dx dy dz,
(1)
the sum becomes an integral
E(r) =
Z
k(r − r′ )
|r −
r′ |3
dQ(r′ )
(2)
In my notations, r = (x, y, z) is the point where we measure the electric field E while
r′ = (x′ , y ′ , z ′ ) the location of the charge. Also, k is the Coulomb constant: in MKSA units
k = 1/(4πǫ0 ) ≈ 9 · 109 N m2 /C2 , while in Gaussian units k = 1.
In these notes, I am going to calculate the electric field for a few uniform examples of
continuous charges: (1) Infinite thin rod; (2) finite thin rod; (3) infinite 2D plate; (4) a pair
of parallel plates; (5) a finite thin 2D disk.
Cylindrical Coordinates
All the examples in these notes have axial symmetry — rotating the whole world through
any angle around a fixed axis (which I’ll use as the z axis) will not change the locations
of the charges or their density, so the electric field should also remain unchanged by such
a rotation. To make use of this symmetry, I am going to use the cylindrical coordinates
(ρ, φ, z), where (ρ, φ) act as polar coordinates in the horizontal (x, y) plane while z is the
vertical elevation. In terms of Cartesian (x, y, z) coordinates
ρ =
p
x2 + y 2 ,
x = ρ × cos φ,
y
,
x
z = z.
y = ρ × sin φ,
z = z.
φ = arctan
(3)
The unit vector ~ıρ in the ρ direction point horizontally away from the z axis while the unit
1
vector ~ıφ point along the circle surrounding that axis,
y
~ıφ
~ıρ
~ıρ = cos φ~ıx + sin φ~ıy ,
x
~ıφ = − sin φ~ıx + cos φ~ıy .
(4)
Finally, the unit vector ~ız in the z direction point vertically up, just as it does in the Cartesian
coordinate system. In terms of these 3 unit vectors,
E = Eρ ~ıρ + Eφ ~ıφ + Ez ~ız
(5)
which translate to Cartesian components as
Ex = Eρ × cos φ − Eφ × sin φ,
Ey = Eρ × sin φ + Eφ × cos φ,
(6)
Ez = Ez .
Example 1: Uniformly charged infinite thin rod.
In this example the electric charges are uniformly distributed along an infinitely thin, infinitely long rod. We use coordinates where this rod lies along the z axis, so the charges are
at (x′ , y ′, z ′ ) = (0, 0, any z ′ ). The charge density is uniform, which means
λ =
dQ(z ′ )
= const.
dz ′
This charge system has many of symmetries, including rotations around the z axis, translations along that axis, and flipping the vertical axis upside down around any point we like.
The electric field E(x, y, z) must have similar symmetries, so its vertical component Ez must
2
vanish everywhere, while the horizontal components Ex and Ey do not depend on z. Furthermore, by rotational symmetry, the horizontal components must in the radial direction
~ıρ : directly away from z axis for λ > 0, or directly towards the z axis for λ < 0:
2
2
1
1
0
0
-1
-1
-2
-2
-3
-3
In cylindrical coordinates this means Eφ ≡ 0 as well as Ez ≡ 0. Further more, the radial
component Eρ depends only on ρ — the horizontal distance from the charged rod.
In light of these symmetries, the only component we need to calculate is Eρ as a function
of a single variable ρ. To do that, we need to evaluate the integral
Eρ (ρ) =
Z
ρ
× k dQ =
r3
rod
where r =
q
+∞
Z
−∞
ρ2 + (z − z ′ )2
ρ
× kλ dz ′
r3
(7)
(8)
is the distance from the charge at (ρ′ = 0, z ′ ) to the point (ρ, φ, z) where we measure the
field. To clarify the geometry, here is the picture of the vertical plane through the charged
3
z′
ch
ar
z
ge
rod and the place where the field is measured:
ρ = r × cos θ
r
z ′ − z = r × sin θ
be
pr
o
θ
Eρ
(ρ, z)
r 2 = ρ2 + (z ′ − z)2
z′ − z
tan θ =
ρ
(9)
ρ
To simplify the integral (7), let’s change the integration variable from z ′ to the θ angle on
the diagram (9):
z ′ = z + ρ × tan θ,
dθ
dz ′ = ρ × d tan θ = ρ ×
,
cos2 θ
ρ
,
r =
cos θ
(10)
and hence
ρ
cos3 θ
ρ dθ
cos θ
′
×
dz
=
ρ
×
×
=
× dθ = d
r3
ρ3
cos2 θ
ρ
sin θ
ρ
.
(11)
Also, while z ′ ranges from −∞ to +∞, the angle θ runs from −π/2 to +π/2. Plugging all
these data into the integral (7), we obtain
kλ
Eρ (ρ) =
×
ρ
+π
Z
2
d sin θ
−π/2
kλ × sin(+π/2) − sin(−π/2) = 2
ρ
2kλ
,
=
ρ
=
(12)
In Gaussian units this means
2λ
~ıρ
ρ
E =
4
(13)
while in MKSA units
E =
λ
~ıρ .
2πǫ0 ρ
(14)
For example, at ρ = 1 cm away from a rod of charge density λ = 1 µC/m, the electric field
has a rather large magnitude
E =
λ
1
2
×
≈ 2 × 9 · 109 N m2 /C2 ) × 1 · 10−6 C/m) ×
= 1.8 · 106 N/C
4πǫ0 ρ
0.01 m
or 1.8 million volts per meter.
To complete this example, let me convert the electric field (12) from the cylindrical to
the Cartesian coordinates. In light of eqs. (4),
~ıρ
cos φ
sin φ
=
~ıx +
~ıy
ρ
ρ
ρ
x
y
= 2 ~ıx + 2 iiy
ρ
ρ
y
x
~ıx + 2
iiy ,
= 2
2
x +y
x + y2
(15)
y
x
~ıx + 2
iiy ,
E = 2kλ
x2 + y 2
x + y2
(16)
x
,
+ y2
y
Ey (x, y, z) = 2kλ × 2
,
x + y2
Ez (x, y, z) ≡ 0.
(17)
hence
or in components
Ex (x, y, z) = 2kλ ×
5
x2
: Example 2: Uniform thin rod of finite length.
Similar to the previous example, the charges are uniformly distributed along an infinitely
thin rod, but this time the rod has a finite length L. In other words, the z ′ coordinate along
the rod runs from a finite z1 to a finite z2 = z1 + L instead of from −∞ to +∞.
The finite rod has an axial symmetry of rotations around the z axis. but it no longer has
the translational symmetry along that axis, or the flipping symmetry z → −z. Consequently,
the horizontal components of E must point in the radial direction (away from the z axis or
towards it), but the vertical component Ez does not need to vanish (and generally does not).
In cylindrical coordinates this means Eφ ≡ 0 but both Eρ 6= 0 and Ez 6= 0. Also, the Eρ and
Ez components may depend on both ρ and z coordinates (of the point where we measure the
field), although by axial symmetry they do not depend on the φ. Thus, instead of a single
integral (7), we now need to evaluate two integrals
Eρ (ρ, z) = kλ ×
Zz2
z1
Zz2
Ez (ρ, z) = kλ ×
z1
ρ dz ′
,
r3
(18)
(z − z ′ ) dz ′
,
r3
(19)
Let us start with the first integral (18). Proceeding similarly to the first example, we
change the integration variable from the z ′ to the angle θ, exactly like on the diagram (9).
Consequently, we obtain
1
ρ dz ′
=
× d sin θ
3
r
ρ
(11)
but this time the θ angle runs
from θ1 = arctan
z1 − z
ρ
to θ2 = arctan
z2 − z
.
ρ
Consequently,
Eρ
kλ
×
=
ρ
Zθ2
kλ
d sin θ =
× sin θ2 − sin θ1 ).
ρ
θ1
6
(20)
By trigonometry
z1 − z
r1
sin θ1 =
where r1 =
p
(21)
(z1 − z)2 + ρ2 is the distance from the lower end of the rod to the measuring
point, and likewise
z2 − z
.
r2
sin θ2 =
Therefore, the radial component of the electric field evaluates to
kλ
Eρ (ρ, z) =
×
ρ
z1 − z
z2 − z
−
r2
r1
.
(22)
Now consider the vertical component of the electric field. Inside the integral (19), we
have
z − z ′ ρ dz ′
cos θ dθ
sin θ dθ
1
(z − z ′ ) dz ′
=
× 3 = − tan θ ×
= −
=
× d cos θ .
3
r
ρ
r
ρ
ρ
ρ
(23)
Consequently,
Eρ
kλ
=
×
ρ
Zθ2
kλ
d cos θ =
× cos θ2 − cos θ1 ).
ρ
(24)
θ1
By trigonometry,
cos θ1 =
ρ
,
r1
cos θ2 =
ρ
,
r2
(25)
(26)
and therefore
Ez (ρ, z) = kλ
1
1
−
r2
r1
.
Altogether, the electric field of a finite rod can be described by the following field-line
7
picture:
3
2
1
0
-1
-2
-3
-4
Finally, let’s look at the near-the-rod and far-away-from-the-rod limits of the electric
field. Near-the-rod limit means we are much closer to some point in the middle of the rod
tan to the rod’s ends, thus
z1 < z < z2 ,
ρ ≪ L = z2 − z1 .
In this limit θ1 ≈ −π/2, θ2 ≈ +π/2, hence in eq. (20)
sin θ2 − sin θ1 ≈ 2
and therefore
Eρ ≈
2kλ
= rρ [infinite rod].
ρ
8
(27)
At the same time, in eq. (24)
cos θ2 − cos θ1 ≪ 1
and hence Ez ≪ Eρ . In other words, in the near-the-rod limit, the electric field of a finite
rod is similar to the field of the infinite rod: It’s the nearby part of the rod which dominates
the field, and it does not matter if the far ends of the rod are infinitely far away or just far
enough.
On the other hand, the far-from-the-rod limit means that we measure the field at some
distance form the rod which is much larger than the rod’s length L = z2 = z1 . In this limit
p
ρ2 + z 2 ,
r1 ≈ r2 ≈ r =
θ1 ≈ θ2 ≈ θ = − arctan
z
,
ρ
while to a better approximation
r2 − r1 ≈ −L cos θ,
θ2 − θ1 ≈ +
L
× sin θ.
r
Consequently, in eqs. (20) and (24)
L sin θ
d sin θ
=
× cos θ,
dθ
r
d cos θ
L sin θ
≈ (θ2 − θ1 ) ×
=
× (− cos θ),
dθ
r
sin θ2 − sin θ1 ≈ (θ2 − θ1 ) ×
cos θ2 − cos θ1
(28)
while
k(Q/L)
kλ
=
.
ρ
r sin θ
(29)
Therefore
k(Q/L) L sin θ
×
× cos θ
r sin θ
r
kQ ρ
kQ
= 2 × sin θ = 2 × ,
r
r
r
k(Q/L) L sin θ
≈
×
× (− cos θ)
r sin θ
r
kQ z
kQ
= 2 × (− cos θ) = 2 × ,
r
r
r
Eρ ≈
Ez
9
(30)
which is precisely the cylindrical components of the electric field vector
E = +
kQ ρ~ıρ + z~ız
kQ ~r
×
= + 2 ×
2
r
r
r
r
(31)
of a point charge Q. Physically, this means that far away from the rod its size is too small
to matter so it may be approximated by a charged point of the same net charge Q = λL.
10
Example 3: Uniformly charged infinite 2D plate.
In this example, the charges are spread out in two dimensions, on a plate which we approximate as infinite in x and y directions but infinitely thin in the z direction. The charges have
uniform density in two dimensions, thus
σ =
dQ
dQ(x′ , y ′ )
=
= const.
d Area
dx′ dy ′
The symmetries of this charge system include translations in x and y directions and rotations
around the z axis. The rotational symmetry tells us that the electric field E(x, y, z) does
not have any horizontal Ex or Ey components, while the translational symmetry makes the
field independent on x and y, thus
Ex ≡ 0,
Ey ≡ y,
Ez (6 x, 6 y, z) = Ez (z only).
Thanks to this symmetry, all we need to calculate the Ez along the z axis, the values at any
other point (x, y, z) would be the same as at (0, 0, z). So let’s start with a picture of the
charges and the point where we measure the field:
z
E
probe at (0, 0, z)
r =
r
=
y
p
p
x′2 + y ′2 + z 2
ρ′2 + z 2
charge at (x′ , y ′ , 0)
ρ′
φ′
x
(32)
11
In light of this diagram and the Coulomb Law
dEz =
z
z
× k dQ = 3 × kσ dx′ dy ′
3
r
r
(33)
and therefore
+∞
+∞ Z
Z
z
Ez (z) = kσ × dx′ dy ′ 3 .
r
−∞
(34)
−∞
To evaluate this double integral, let’s change the integration variables from x′ and y ′ to
the polar coordinates (ρ′ , φ′ ) on the charged plate. The area differential in these variables
becomes
dArea = dx′ dy ′ = ρ′ dρ′ dφ′
(35)
while the coordinates ranges are 0 ≤ φ′ ≤ 2π, 0 ≤ ρ′ < ∞. Consequently, the integral (34)
for the electric field of the charged plate becomes
Ez
Z2π
Z∞
z
= kσ × dρ′ ρ′ dφ′ 3 .
r
0
0
Inside the inner integral, r =
p
(36)
ρ′2 + z 2 depends on ρ′ but not on the angle φ′ , and nothing
else depends on the angle φ′ either. Consequently, the inner integral over dφ′ is an integral
over a constant, so the integration amounts to simply multiplying the integrand by 2π, thus
Ez = kσ ×
Z∞
0
2π × zρ′ dρ′
.
r 3 = (ρ′2 + z 2 )3/2
(37)
To evaluate this integral, let’s change the integration variable once again, from ρ′ to r =
p
ρ′2 + z 2 Since z does not change with ρ′ , we have
r 2 = ρ′2 + z 2
=⇒
2r dr = 2ρ′ dρ′
12
=⇒
2πzρ′ dρ′ = 2πzr dr.
(38)
also, while ρ′ ranges from zero to infinity, r ranges from |z| to infinity, thus
Z∞
Z∞
2π × zρ′ dρ′
2πzr dr
=
r3
r3
0
|z|
Z∞
dr
= 2πz ×
r2
|z|
= 2πz ×
(39)
∞
−1 r |z|
+1
|z|
= 2π × sign(z).
= 2πz ×
Consequently,
Ez = 2πkσ × sign(z)
(40)
E = 2πkσ × sign(z)~ız .
(41)
and hence
In particular, the electric field of an infinite charged plate has the same magnitude E =
2πk × σ everywhere in space, but has opposite directions on the two sides of the plate: For
σ > 0 the field point vertically up above the plate but vertically down below the plate; for
σ < 0 the directions are opposite, down above the plate but up below the plate.
(42)
In coordinate-independent terms, the electric field of a charged plate is always ⊥ to the plate
itself; for a positively-charged plate E points away from the plate, while for a negatively
charged plate E points towards the plate.
13
As to the magnitude of the electric field, in MKSA units
|E| =
σ
.
2ǫ0
(43)
For example, a plate of charge density of 1.0 microCoulomb per square meter creates electric
field of magnitude
E =
1.0 · 10−6 C/m2
≈ 56, 000 V/m.
2 × 8.85 · 10−12 C/(V m)
Example 4: Two parallel plates with opposite charges.
In this example, we have two infinite plates parallel to each other. One plate has positive
uniform charge density +σ while the other has negative uniform charge density −σ, so the
net electric charges of the two plates cancel each other.
−σ
(44)
+σ
In the previous example we have obtained the electric field of a single plate, so in this example
we do not face any integrals; all we need to do is to add up the fields of the two plates,
Enet = E(1) + E(2) .
(45)
Moreover, for parallel plates both fields are ⊥ to the two plates — for the horizontal plates
as on figure (44) the fields are vertical — and have equal magnitudes everywhere in space,
(1) E ≡ E(2) ≡ 2πkσ,
(46)
so all we need is to check where each field points up and where it points down. Specifically,
the field E(1) of the positive plate always points away from that plate while the field E(2) of
the negative plate always points towards that plate. So assuming the negative plate is above
the positive plate as in the picture (44), there are three distinct situations:
14
• Above both plates, the field E(1) of the positive plate points up while the field E(2) of
the negative plate points down. Consequently, the two fields cancel each other, and
the net electric field vanishes!
• Below both plates, the field E(1) of the positive plate points down while the field E(2)
of the negative plate points up. Again, the two fields cancel each other, and the net
electric field vanishes!
⋆ Between the two plates — i.e., above the positive plate but below the negative plate
— the field E(1) of the positive plate points up while the field E(2) of the negative
plate also points up. Consequently, the two fields add up, and the net electric field is
Enet = +4πkσ~ız .
(47)
To summarize, outside the two plates the electric field vanishes, while between the plates the
field has magnitude E = 4πkσ and direction from the positive plate towards the negative
plate, as on the following figure:
E=0
−σ
E = 4πkσ
(48)
+σ
E=0
In MKSA units, the field between the plates has magnitude
E =
σ
ǫ0
(49)
For example, if one plate has charge density +1.0 µC/m2 while the other plate has charge
density −1.0 µC/m2 , then the field between the plates is E = 113, 000 V/m.
15
Example 5: Uniformly charged disk.
In this example, the charged 2D surface is a finite disk or radius R. We use a coordinate
system with an origin at the center of the disk, with the z axis ⊥ to the disk, thus the charges
lie at
(x′ , y ′ , z ′ ) with z ′ ≡ 0,
finite x′ , y ′ ,
x′2 + y ′2 ≤ R2 .
For simplicity, we limit this example to calculating the electric field on the z axis only —
alas, the off-axis field is much more complicated — thus
z
E
cha
rg
ea
t(
b
pro
ea
t
y ′,
, z)
r =
r
x ′,
0
(0 ,
=
y
p
p
x′2 + y ′2 + z 2
ρ′2 + z 2
0)
ρ′
φ′
x
R
(50)
Thanks to the rotational symmetry of the disk, on the z axis the electric field is purely
vertical, Ex = Ey = 0, while the z components obtains from a double integral similar two
the one we had for an infinite plate,
Ez = kσ ×
ZZ
dx′ dy ′
z
,
r3
(51)
disk
except that now the integration range is limited to the disk, x′2 + y ′2 ≤ R2 . In Cartesian
coordinates (x′ , y ′ ) this is a rather complicated range of a double integral, but it becomes
16
much simpler in the polar coordinates (ρ′ , φ′ ) where ρ′ runs from zero to R while φ′ runs
from zero to 2π, thus
Ez
ZR
Z2π
z
= kσ × dρ′ ρ′ × dφ′ 3
r
0
(52)
0
Thanks to the rotational symmetry, nothing inside the integral depends on the polar angle
φ′ , so integrating over that angle produces simply 2π × the integrand, thus
Ez = kσ ×
ZR
0
2π × zρ′ dρ′
.
r 3 = (ρ′2 + z 2 )3/2
(53)
At this point, we again proceed as in example#3 and change the remaining integration
variable from ρ′ to r, which now runs from |z| (for ρ − 0) to
rmax =
p
R2 + z 2 .
(54)
Consequently,
Ez = kσ ×
rZmax
2πz × r dr
r3
|z|
= 2πkσz ×
rZmax
dr
= 2πkσz ×
r2
|z|
= 2πkσz ×
1
1
− √
|z|
R2 + z 2
r
−1 max
r |z|
(55)
,
or equivalently
|z|
.
E(0, 0, z) = 2πkσ × sign(z)~ız × 1 − √
R2 + z 2
(56)
Note that the first two factors on the right hand side are exactly as in the field of an infinitely
charged plane, and only the third factor depends on the disk’s radius R.
17
To complete this example, lets consider the near-the-disk and far-away-from-the-disk
limits of the electric field (56). In the near-the-disk limit, we are measuring the electric field
much closer to the disk than its radius, |z| ≪ R, so in the last factor in eq. (56)
1− √
|z|
≈ 1.
(57)
E ≈ 2πkσ × sign(z)~ız
(58)
R2 + z 2
Consequently,
— near the disk, its electric field is similar to the field of an infinite charged plane.
In the opposite far-away-from-the-disk limit we measure the field at distance |z| much
larger than the disk radius, so we expect the disk’s field to be similar to the field of a point
source with the same net charge Q = πR2 × σ. And indeed, for |z| ≫ R, the last factor in
eq. (56) becomes
|z|
1− √
= 1−
R2 + z 2
r
R2
≈ 1−
1− 2
z + R2
1
R2
R2
R2
1− × 2
=
≈
,
2 z + R2
2(z 2 + R2 )
2z 2
(59)
hence
E(0, 0, z) ≈ 2πkσ × sign(z)~ız ×
R2
k × πR2 × σ
=
× sign(z)~ız
2z 2
z2
(60)
Note that we are measuring this field on the z axis, so z 2 in the denominator is the distance2
from the measurement point to the disk, while sign(x)~ız is the unit vector in the direction
of the measurement point. Finally, πR2 × σ is the total electric charge Q of the disk, so
eq. (60) indeed agrees with the Coulomb field of a point charge Q,
E(r) =
kQ
×~ır .
|r|2
18
(61)