As an example, we take for a source the cylinder from the previous page (reproduced below), and calculate the electric field everywhere on the right hand axis of the cylinder outside the cylinder. We choose a coordinate system with the origin at the center of the cylinder and the zz axis along the axis of the cylinder. The known quantities are the radius RR of the cylinder, its length
L
=
2
R
L
=
2
R
, and the total charge on the cylinder QQ. In addition, we know that the charge is uniformly distributed throughout the volume
V
′
V
′
of the cylinder.
Then the problem we address is to find the electric field
E
⇀
[
z
]
E
⇀
[
z
]
for
z
>
R
z>R. The tool we have for solving this problem is Equation 3:
E
⇀
[
r
⇀
]
=
1
4
π
ϵ
0
∫
V
′
ρ
η
^
η
2
ⅆ
V
′
E
⇀
[
r
⇀
]
=
1
4
π
ϵ
0
∫
V
′
ρ
η
^
η
2
ⅆ
V
′
. How do we begin?
First, we have an integration to do, and the most important part of any integral is the bound variable. So a good place to start is to decide how we represent
ⅆ
V
′
ⅆ
V
′
for this example.
ⅆ
V
′
ⅆ
V
′
is an element of volume, that is, an infinitesimal piece of threedimensional (3D) space, so it actually contains 3 bound variables. We could use
ⅆ
V
′
=
ⅆ
x
′
ⅆ
y
′
ⅆ
z
′
ⅆ
V
′
= ⅆ
x
′
ⅆ
y
′
ⅆ
z
′
, but a moment's reflection and the realization that the cylinder has cylindrical symmetry might indicate that cylindrical coordinates would be better used than Cartesian coordinates. (Indeed, use of
ⅆ
V
′
=
ⅆ
x
′
ⅆ
y
′
ⅆ
z
′
ⅆ
V
′
= ⅆ
x
′
ⅆ
y
′
ⅆ
z
′
in this problem would make the integration very much messier, but of course lead to the same result.)
So we choose the cylindrical coordinates
{
s
′
,
φ
′
,
z
′
}
{
s
′
,
φ
′
,
z
′
}, where
s
′
=
x
′
2
+
y
′
2
s
′
=
x
′
2
+
y
′
2
and
φ
′
=
tan

1
[
y
′
x
′
]
φ
′
=
tan

1
[
y
′
x
′
]
are polar coordinates in the xyxy plane. In terms of these coordinates, the volume element is
ⅆ
V
′
=
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
ⅆ
V
′
=
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
, and we have taken care of the bound variables. Again, as a reminder, the primes indicate variables that are restricted to the source charge distribution.
The second most important parts of a definite integral are the limits, so we tackle those next. In sweeping out the volume of the cylinder, the bound variables vary through the domains
φ
′
∈
[
0
,
2
π
]
φ
′
∈[0,2π],
s
′
∈
[
0
,
R
]
s
′
∈[0,R] and
z
′
∈
[

R
,
R
]
z
′
∈[R,R]. These establish the limits of integration we need, stated in terms of known quantities.
Substituting these results into Equation 3, we have partially specified the basic equation to this example, with the statement
E
⇀
[
r
⇀
]
=
1
4
π
ϵ
0
∫

R
R
∫
0
R
∫
0
2
π
ρ
η
^
η
2
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
E
⇀
[
r
⇀
]
=
1
4
π
ϵ
0
∫

R
R
∫
0
R
∫
0
2
π
ρ
η
^
η
2
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
.
Next, what about ρρ? How is it related to the known quantities? From the definition
ρ
:=
ⅆ
q
ⅆ
V
′
ρ :=
ⅆ
q
ⅆ
V
′
and the fundamental theorem of calculus, we can write
Q
=
∫
Q
ⅆ
q
=
∫
V
′
ρ
ⅆ
V
′
Q =
∫
Q
ⅆq =
∫
V
′
ρⅆ
V
′
. Because we know that ρρ is uniform in this example, it does not change during the integration and can be removed from the integral, leaving
Q
=
ρ
∫
V
′
ⅆ
V
′
Q = ρ
∫
V
′
ⅆ
V
′
. The remaining integral is simply the total volume of the cylinder, so
Q
=
ρ
V
′
=
2
π
ρ
R
3
Q = ρ
V
′
= 2πρ
R
3
. (You should verify for yourself that
∫

R
R
∫
0
R
∫
0
2
π
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
=
2
π
R
3
∫

R
R
∫
0
R
∫
0
2
π
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
= 2π
R
3
.) Therefore,
ρ
=
Q
2
π
R
3
ρ =
Q
2
π
R
3
, which is a constant that comes outside the integral in the Coulomb law.
To this point, our application of Equation 3 to this example has lead to the equation
E
⇀
[
r
⇀
]
=
Q
8
π
2
ϵ
0
R
3
∫

R
R
∫
0
R
∫
0
2
π
η
^
η
2
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
E
⇀
[
r
⇀
]
=
Q
8
π
2
ϵ
0
R
3
∫

R
R
∫
0
R
∫
0
2
π
η
^
η
2
s
′
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
. Evidently, our next task is to write the ratio
η
^
η
2
η
^
η
2
in terms of the bound variables
{
s
′
,
φ
′
,
z
′
}
{
s
′
,
φ
′
,
z
′
}. The geometry of the example tells us how to do this.
A supplemental diagram is helpful now. Consider an arbitrary source element
ⅆ
q
ⅆq in
V
′
V
′
. In the
s
z
sz plane, we have the following situation.
Once we have
η
⇀
η
⇀
, we can generate
η
^
η
2
η
^
η
2
. And, from the diagram,
η
⇀
=
r
⇀

r
⇀
′
=
(
z
z
^
)

(
s
′
s
^
+
z
′
z
^
)
η
⇀
=
r
⇀

r
⇀
′
= (z
z
^
)(
s
′
s
^
+
z
′
z
^
). Therefore
η
⇀
=
(
z

z
′
)
z
^
+
s
′
(

s
^
)
η
⇀
= (z
z
′
)
z
^
+
s
′
(
s
^
). Then
η
2
=
η
⇀
·
η
⇀
=
(
z

z
′
)
2
+
s
′
2
η
2
=
η
⇀
·
η
⇀
=
(
z

z
′
)
2
+
s
′
2
and
η
^
=
η
⇀
η
=
(
z

z
′
)
z
^
+
s
′
(

s
^
)
(
z

z
′
)
2
+
s
′
2
η
^
=
η
⇀
η
=
(
z

z
′
)
z
^
+
s
′
(

s
^
)
(
z

z
′
)
2
+
s
′
2
. So the ratio we need for the Coulomb equation is
η
^
η
2
=
(
z

z
′
)
z
^
+
s
′
(

s
^
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
η
^
η
2
=
(
z

z
′
)
z
^
+
s
′
(

s
^
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
. Substituting this value into the Coulomb law equation gives us
E
⇀
[
r
⇀
]
=
Q
8
π
2
ϵ
0
R
3
∫

R
R
∫
0
R
∫
0
2
π
s
′
(
z

z
′
)
z
^
+
s
′
2
(

s
^
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
E
⇀
[
r
⇀
]
=
Q
8
π
2
ϵ
0
R
3
∫

R
R
∫
0
R
∫
0
2
π
s
′
(
z

z
′
)
z
^
+
s
′
2
(

s
^
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
φ
′
ⅆ
s
′
ⅆ
z
′
.
At first sight, you may find this integral imposing. But don't despair, it turns out to be a lot simpler than it first appears. First consider the inner integral over
φ
′
φ
′
; the only quantity in the integrand that depends on this angle is
s
^
s
^
, and
∫
0
2
π
s
^
ⅆ
φ
′
=
0
∫
0
2
π
s
^
ⅆ
φ
′
= 0. (If this seems mysterious, write out
s
^
=
cos
[
φ
′
]
x
^
+
sin
[
φ
′
]
y
^
s
^
= cos[
φ
′
]
x
^
+sin[
φ
′
]
y
^
and perform the integral. Any variable unit vector integrated around a full circle yields zero.) This disappearance of any dependence of our result on
s
^
s
^
is a consequence of the cylindrical symmetry around the zz axis.
So performing the inner integral simplifies the rest of the work to
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
3
∫

R
R
∫
0
R
s
′
(
z

z
′
)
z
^
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
s
′
ⅆ
z
′
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
3
∫

R
R
∫
0
R
s
′
(
z

z
′
)
z
^
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
s
′
ⅆ
z
′
. (Why do we now write
E
⇀
[
z
>
R
]
E
⇀
[
z
>
R
]
for what was previously written as
E
⇀
[
r
⇀
]
E
⇀
[
r
⇀
]
?) Both of the remaining integrations have straightforward antiderivatives; first,
∫
s
′
(
z

z
′
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
s
′
=

(
z

z
′
)
(
s
′
)
2
+
(
z

z
′
)
2
∫
s
′
(
z

z
′
)
(
(
z

z
′
)
2
+
s
′
2
)
3
/
2
ⅆ
s
′
= 
(
z

z
′
)
(
s
′
)
2
+
(
z

z
′
)
2
. After evaluating the antiderivative at the integration limits one encounters the antiderivative
∫
(
z

z
′
)
(
(
z

z
′
)
2
+
R
2
)
1
/
2
ⅆ
z
′
=
R
2
+
(
z

z
′
)
2
∫
(
z

z
′
)
(
(
z

z
′
)
2
+
R
2
)
1
/
2
ⅆ
z
′
=
R
2
+
(
z

z
′
)
2
. The student needn't feel required to laboriously derive antiderivatives; they have been known and tabulated for generations, and there is no sense in reinventing old wheels. What is important is that the student understand the process of solving a problem starting with one or more basic relationships; in our case here, the Coulomb law.
Straightforwardly evaluating the triple integral then gives the result
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
3
(
2
R

R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
)
z
^
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
3
(2R
R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
)
z
^
. One might wish to factor RR from the brackets, leaving
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
2
(
2

1
+
(
z
R
+
1
)
2
+
1
+
(
z
R

1
)
2
)
z
^
E
⇀
[
z
>
R
]
=
Q
4
π
ϵ
0
R
2
(2
1
+
(
z
R
+
1
)
2
+
1
+
(
z
R

1
)
2
)
z
^
. Now the quantity inside the brackets has no units, and we can see that the units are correct for an electric field. We also see that the electric field points directly along the zz axis. Although this form for the solution doesn't provide a lot of insight into the "shape" of the field, it is fine for numerical calculation of the field, but only in the near neighborhood of the source. If
z
R
≫
1
z
R
≫1, (in other words, for field points very far away from the cylinder of source charge) this form is sensitive to the small difference of two large quantities. For far field points, an accurate numerical calculation of the field requires a great deal of numerical precision, keeping many significant digits in the intermediate terms. In other words, one would have to be very careful not to have roundoff errors make the numerical value incorrect. Is there a better way to write this solution for field points that are far from the source?
For one thing, we know that the electric field must go to zero at an infinite distance away from the source. (Can one show this for the form given above?). For another, from a field point very far away from the source, the cylinder of source charge should "look like" a point charge, with
E
⇀
[
z
]
≈
Q
4
π
ϵ
0
z
2
z
^
E
⇀
[
z
]
≈
Q
4
π
ϵ
0
z
2
z
^
in this case. These limits must be examined for any supposed solution for the electric field of a finite source distribution. (Now we have taken
z
>
R
z>R as understood, and replaced it with simply zz.)
So we need to examine the behavior of the quantity
(
2
R

R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
)
(2R
R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
) in the region where
R
z
≪
1
R
z
≪1. We factor out zz, giving
(
2
R

R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
)
=
z
(
2
R
z

(
R
z
)
2
+
(
1
+
R
z
)
2
+
(
R
z
)
2
+
(
1

R
z
)
2
)
(2R
R
2
+
(
z
+
R
)
2
+
R
2
+
(
z

R
)
2
) = z(2
R
z

(
R
z
)
2
+
(
1
+
R
z
)
2
+
(
R
z
)
2
+
(
1

R
z
)
2
). To save writing, let
δ
=
R
z
δ =
R
z
. Next we do a Taylor series expansion of
(
2
δ

δ
2
+
(
1
+
δ
)
2
+
δ
2
+
(
1

δ
)
2
)
(2δ
δ
2
+
(
1
+
δ
)
2
+
δ
2
+
(
1

δ
)
2
) around
δ
=
0
δ = 0. (Perhaps you'll need to pull out your calculus textbook to review Taylor series.) The result involves a lot of cancellation of terms with the (perhaps surprising) simple result that
(
2
δ

δ
2
+
(
1
+
δ
)
2
+
δ
2
+
(
1

δ
)
2
)
≈
δ
3
(2δ
δ
2
+
(
1
+
δ
)
2
+
δ
2
+
(
1

δ
)
2
) ≈
δ
3
! (Go ahead, try it for yourself. The next term that contributes goes as
δ
5
δ
5
.) This result might also be stated in the form
Limit
z
≫
R
[
z
(
2
R
z

(
R
z
)
2
+
(
1
+
R
z
)
2
+
(
R
z
)
2
+
(
1

R
z
)
2
)
]
=
z
(
R
z
)
3
=
R
3
z
2
Limit
z
≫
R
[z(2
R
z

(
R
z
)
2
+
(
1
+
R
z
)
2
+
(
R
z
)
2
+
(
1

R
z
)
2
)] =
z
(
R
z
)
3
=
R
3
z
2
.
Substituting the Taylor series expansion result into our exact solution, we find
E
⇀
[
z
]
≈
Q
z
δ
3
4
π
ϵ
0
R
3
z
^
=
Q
4
π
ϵ
0
z
2
z
^
E
⇀
[
z
]
≈
Q
z
δ
3
4
π
ϵ
0
R
3
z
^
=
Q
4
π
ϵ
0
z
2
z
^
when
z
≫
R
z≫R. This is precisely what we must expect intuitively. One simply must get this result for any correct electric field solution around a finite charge source. Confirming this limit is an important test of the correctness of a calculated exact electrostatic field.
You may have found this an ambitious example to work through. Presumably you have already worked through simpler examples in your course. This is actually more like a review example, designed to display some considerations that often arise in practical Coulomb calculations of electric fields.
Often the charge distributions are 2D (a charged surface) or 1D (a charged wire) rather than a 3D volume. These are treated in the same way as this example, with fewer integrations to perform. In 2D sources, one defines a surface charge density
σ
:=
ⅆ
q
ⅆ
S
′
σ :=
ⅆ
q
ⅆ
S
′
, where
ⅆ
S
′
ⅆ
S
′
is an element of the charged surface. In 1D sources, one defines a linear charge density
λ
:=
ⅆ
q
ⅆ
L
′
λ :=
ⅆ
q
ⅆ
L
′
, where
ⅆ
L
′
ⅆ
L
′
is an element of the charged line or curve.