10/21/2004
section 4_4 E-field calculations using Coulombs Law blank.doc
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4-4 E-field Calculations
using Coulomb’s Law
Reading Assignment: pp. 93-98
1. Example: The Uniform, Infinite Line Charge
2. Example: The Uniform Disk of Charge
3. Example: An Infinite Charge Plane
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Infinite Line Charge.doc
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The Uniform, Infinite
Line Charge
Consider an infinite line of charge lying along the z-axis. The
charge density along this line is a constant value of ρ A C/m.
Q: What electric field E ( r ) is produced by this charge
distribution?
A: Apply Coulomb’s Law!
We know that for a line charge distribution that:
E (r ) =
∫
C
r
r′
ρ A ( r′ ) r-r′
d A′
4πε 0 r-r′ 3
E (r )
r-r′
ρA
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Infinite Line Charge.doc
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Q: Yikes! How do we evaluate this integral?
A: Don’t panic! You know how to evaluate this
integral. Let’s break up the process into smaller steps.
Step 1: Determine d A′
The differential element d A′ is just the magnitude of the
differential line element we studied in chapter 2 (i.e.,
d A′ = d A′ ). As a result, we can easily integrate over any of the
seven contours we discussed in chapter 2.
The contour in this problem is one of those! It is a line parallel
to the z-axis, defined as x’ =0 and y’ =0. As a result, we use for
d A′ :
d A′ = ˆaz dz ′ = dz ′
Step 2: Determine the limits of integration
This is easy! The line charge is infinite. Therefore, we integrate
from z ′ = −∞ to z ′ = ∞ .
Step 3: Determine the vector r-r′ .
Since for all charge x’ = 0 and y’ =0, we find:
r-r′ =
=
( xˆa
( xˆa
) (
) − z ′ˆa
x
ˆ y + za
ˆ z − x′ˆax + y ′ˆay + z ′ˆaz
+ ya
x
ˆ y + za
ˆz
+ ya
)
z
ˆ y + ( z − z ′ ) ˆaz
= xˆax + ya
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Infinite Line Charge.doc
Step 4: Determine the scalar r-r′
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3
2
Since r-r′ = x 2 + y 2 + ( z − z ′ ) , we find:
3
2
r-r′ = ⎡⎣x + y + ( z − z ′ ) ⎤⎦
2
2
3
2
Step 5: Time to integrate !
E (r ) =
∫
C
=
ρ A ( r′ ) r-r′
d A′
3
4πε 0 r-r′
1
4πε 0
ρA
=
4πε 0
=
∫ρ
−∞
∞
∫
−∞
A
⎡x 2 + y 2 + ( z − z ′ ) ⎤
⎣
⎦
x ˆax + y ˆay + ( z − z ′ ) ˆaz
2
2
⎡x 2 + y 2 + ( z − z ′ ) ⎤
⎣
⎦
ρ A ( x ˆax + y ˆay )
4πε 0
ρ ˆa
+ A z
4πε 0
=
x ˆax + y ˆay + ( z − z ′ ) ˆaz
∞
∫
2
dz ′
dz ′
dz ′
∞
∫
⎡x 2 + y 2 + ( z − z ′ )2 ⎤
⎣
⎦
( z − z ′ ) dz ′
2
⎡
′ ⎤
⎣x + y + ( z − z ) ⎦
2
ρ A ( x ˆax + y ˆay )
4πε 0
2
−∞
∞
−∞
3
3
2
3
3
2
2
2
+0
x + y2
2
ρ A ( x ˆax + y ˆay )
=
x2 + y2
2πε 0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Infinite Line Charge.doc
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This result, however, is best expressed in cylindrical
coordinates:
ˆy
xˆax + ya
ρ cosφ ˆax + ρ sinφ ˆay
ρ2
cosφ ˆax + sinφ ˆay
=
ρ
=
x2 + y2
And with cylindrical base vectors:
cosφ ˆax + sinφ ˆay
ρ
=
+
+
=
+
+
1
cosφ ˆa
(
ρ
1
)
⋅ ˆaρ + sinφ ˆay ⋅ ˆaρ ˆaρ
( cosφ ˆa
⋅ ˆaφ + sinφ ˆay ⋅ ˆaφ ˆaφ
cosφ ˆa
(
ρ
⋅ ˆaz + sinφ ˆay ⋅ ˆaz ˆaz
ρ
x
1
1
x
)
)
( cos φ + sin φ ) ˆa
ρ
1
ρ
1
ρ
ˆaρ
=
ρ
Jim Stiles
x
2
2
ρ
( -cosφ sinφ + sinφ cosφ ) ˆaφ
( cosφ ( 0 ) + sinφ ( 0 ) ) ˆaz
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Infinite Line Charge.doc
5/5
As a result, we can write the electric field produced by an
infinite line charge with constant density ρ A as:
E (r ) =
ρ A âρ
2πε 0 ρ
Note what this means. Recall unit vector âρ is the direction that
points away from the z-axis. In other words, the electric field
produced by the uniform line charge points away from the line
charge, just like the electric field produced by a point charge
likewise points away from the charge.
It is apparent that the electric field in the static case appears
to diverge from the location of the charge. And, this is exactly
what Maxwell’s equations (Gauss’s Law) says will happen ! i.e.,:
∇ ⋅ E (r ) =
ρv ( r )
ε0
Note the magnitude of the electric field is proportional to 1 ρ ,
therefore the electric field diminishes as we get further from
the line charge. Note however, the electric field does not
diminish as quickly as that generated by a point charge. Recall
in that case, the magnitude of the electric field diminishes as
1 r2 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Disk of Charge.doc
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The Uniform Disk
of Charge
Consider a disk radius a, centered at the origin, and lying
entirely on the z =0 plane.
E (r )
r
r-r′
ρs
r′
This disk contains surface charge, with density of ρs C/m2.
This density is uniform across the disk.
Let’s find the electric field generated by this charge disk!
From Coulomb’s Law, we know:
E (r ) =
Jim Stiles
∫∫S
ρs ( r′ ) r-r′
ds ′
4πε 0 r-r′ 3
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Disk of Charge.doc
2/5
Step 1: Determine ds’
This disk can be described by the equation z’ = 0. That is, every
point on the disk has a cordinate value z’ that is equal to zero.
This is one of the surfaces we examined in chapter 2. The
differential surface element for that surface, you recall, is:
ds ′ = dsz = ρ ′ d ρ ′ d φ ′
Step 2: Determine the limits of integration .
Note over the surface of the disk, ρ ′ changes from 0 to radius
a, and φ ′ changes from 0 to 2π . Therefore:
0 < ρ′ < a
Step 3:
0 < φ ′ < 2π
Determine vector r-r′ .
We know that z’ = 0 for all charge, therefore we can write:
r-r′ =
=
( xˆa
( xˆa
) (
) − ( x′ˆa
x
ˆ y + za
ˆ z − x′ˆax + y ′ˆay + z ′ˆaz
+ ya
x
ˆ y + za
ˆz
+ ya
x
+ y ′ˆay
)
)
ˆz
= ( x-x′ ) ˆax + ( y − y ′ ) ˆay + za
Since the primed coordinates in ds’ are expressed in cylindrical
coordinates, we convert the coordinates to get:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Disk of Charge.doc
(
) (
ˆ y + za
ˆ z − x′ˆax + y ′ˆay
r-r′ = xˆax + ya
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)
= ( x − x') ˆax + ( y − y ′ ) ˆay + z ˆaz
= ( x − ρ ′ cosφ ′ ) ˆax + ( y − ρ ′ sinφ ′ ) ˆay + z ˆaz
Step 4: Determine r-r′
3
We find that:
r-r′
3
2
2
= ⎡( x-ρ ′cosφ ′ ) + ( y − ρ ′sinφ ′ ) + z 2 ⎤
⎣
⎦
3
2
Step 5: Time to integrate !
E (r ) =
∫∫S
ρs ( r′ ) r-r′
ds ′
4πε 0 r-r′ 3
ρs
=
4πε 0
2π a
∫∫
0 0
( x-ρ ′cosφ ′ ) ˆax + ( y − ρ ′sinφ ′ ) ˆay + z ˆaz
⎡( x-ρ ′cosφ ′ )2 + ( y − ρ ′sinφ ′ )2 + z 2 ⎤
⎣
⎦
3
2
ρ ′d ρ ′d φ ′
Yikes! What a mess! To simplify our integration let’s determine
the electric field E ( r ) along the z-axis only. In other words,
set x = 0 and y = 0.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
The Uniform Disk of Charge.doc
E ( x=0,y=0,z ) = ∫∫
S
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ρs ( r′ ) r-r′
ds ′
3
4πε 0 r-r′
ρs 2π a ( 0 − ρ ′cosφ ′ ) aˆx + ( 0 − ρ ′sinφ ′ ) aˆy − zaˆz
ρ ′d ρ ′d φ ′
=
3
2
2
2
4πε 0 ∫0 ∫0 ⎡
( 0 − ρ ′cosφ ′) + ( 0 − ρ ′sinφ ′ ) + z 2 ⎤
⎣
⎦
− ρs 2π a ( ρ ′cosφ ′ ) aˆx + ( ρ ′sinφ ′ ) aˆy − zaˆz
ρ ′d ρ ′d φ ′
=
3
∫
∫
4πε 0 0 0
⎡⎣ ρ ′2 + z 2 ⎤⎦ 2
2π a
( ρ ′cosφ ′ ) ρ ′d ρ ′d φ ′
ρs
aˆx ∫ ∫
=
3
4πε 0
0 0
⎡⎣ ρ ′2 + z 2 ⎤⎦ 2
2π a
( ρ ′sinφ ′ ) ρ ′d ρ ′d φ ′
− ρs
aˆy ∫ ∫
+
3
4πε 0 0 0
⎡⎣ ρ ′2 + z 2 ⎤⎦ 2
2π a
− ρs
z ρ ′d ρ ′d φ ′
aˆz ∫ ∫
+
4πε 0 0 0 ⎡ ρ ′2 + z 2 ⎤ 32
⎣
⎦
Note that since:
2π
2π
0
0
∫ sin φ d φ = 0 = ∫ cos φ d φ
The first two terms (Ex and Ey) are equal to zero. Integrating
the last term, we get:
⎧ ρ
⎡
⎤
z
⎪ s ˆaz ⎢1 −
⎥
⎪ 2ε 0
z 2 + a2 ⎦
⎣
⎪
E ( x=0,y=0,z ) = ⎨
⎪
⎤
z
⎪ ρs ˆa ⎡ −1 −
⎥
2
2
⎪ 2ε 0 z ⎣⎢
z
a
+
⎦
⎩
Jim Stiles
The Univ. of Kansas
if z > 0
if z < 0
Dept. of EECS
10/21/2004
The Uniform Disk of Charge.doc
5/5
From this expression, we can conclude two things. The first is
that above the disk (z > 0), the electric field points in the
direction ˆaz , and below the disk (z < 0), it points in the direction
- ˆaz .
What a surprise (not)! The electric field points away from the
charge. It appears to be diverging from the charged disk (as
predicted by Gauss’s Law).
Likewise, it is evident that as we move further and further
from the disk, the electric field will diminish. In fact, as
distance z goes to infinity, the magnitude of the electric field
approaches zero. This of course is similar to the point or line
charge; as we move an infinite distance away, the electric field
diminishes to nothing.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
An Infinite Charge Plane.doc
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An Infinite Charge Plane
Say that we have a very large charge disk. So large, in fact,
that its radius a approaches infinity !
What electric field is created by this infinite
plane?
Q:
A: We already know! Just evaluate the charge disk
solution for the case where the disk radius a is
infinity.
In other words:
⎧
⎤
ρ ⎡
z
⎪ ˆaz s ⎢1 −
⎥
⎪ 2ε 0 ⎣
z 2 + a2 ⎦
⎪
lim E ( x=0,y=0,z ) = ⎨
a →∞
⎪
⎤
z
⎪ˆa ρs ⎡ −1 −
⎥
2
2
⎪ z 2ε 0 ⎢⎣
+
z
a
⎦
⎩
⎧ ρs
⎪ 2ε ˆaz
⎪⎪ 0
=⎨
⎪ −ρ
⎪ s ˆaz
⎪⎩ 2ε 0
if z > 0
if z < 0
if z > 0
if z < 0
Therefore, the electric field produced by an infinite charge
plane, with surface charge density ρs , is:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
An Infinite Charge Plane.doc
⎧ ρs
⎪ 2ε ˆaz
⎪⎪ 0
E (r ) = ⎨
⎪ −ρ
⎪ s ˆaz
⎪⎩ 2ε 0
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if z > 0
if z < 0
Think about what this says!
* First, we note that the electric field points away from the
plane if ρs is positive, and toward the plane if ρs is
negative.
* Second, we notice that the magnitude of the electric field
is a constant—the magnitude is independent of the
distance from the infinite plane!
ρs > 0
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/21/2004
An Infinite Charge Plane.doc
3/3
The reason for this result is, that no matter how far you are
(i.e., |z|) from the infinite charge plane, you remain infinitely
close to plane, when compared to its radius a.
We will find these results are useful when we study the
behavior of a parallel plate capacitor.
Jim Stiles
The Univ. of Kansas
Dept. of EECS