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Complex Numbers: An Electric Field Computation

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

We have established that vectors may be used to code complex numbers. Conversely, complex numbers may be used to code or represent the orthogonal components of any two-dimensional vector. This makes them invaluable in electromagnetic field theory, where they are used to represent the components of electric and magnetic fields.

The basic problem in electromagnetic field theory is to determine the electric or magnetic field that is generated by a static or dynamic distribution of charge. The key idea is to isolate an infinitesimal charge, determine the field set up by this charge, and then to sum the fields contributed by all such infinitesimal charges. This idea is illustrated in Figure 1, where the charge λλ, uniformly distributed over a line segment of length dxdx at point -x-x, produces a field dE(x)dE(x) at the test point (0,h)(0,h). The field dE(x)dE(x) is a “vector” field (as opposed to a “scalar” field), with components E1(x)E1(x) and E2(x)E2(x). The intensity or field strength of the field dE(x)dE(x) is

| d E ( x ) | = λ d x 4 π ϵ 0 ( h 2 + x 2 ) | d E ( x ) | = λ d x 4 π ϵ 0 ( h 2 + x 2 )
(1)
Figure 1: Infinitesimal Charge λdxλdx Producing Field dE(x)dE(x)
This Cartesian graph contains a line segment extending from the x axis in quadrant II to the a point labeled h on the positive portion of the y axis in quadrant I. The line continues from that point following the same positive slope into quadrant I to an point labeled dE(x). Another line extends from point h at a right angel to the x axis into quadrant I to a point labeled dE_1(x). Inside quadrant III there is a figure that is shaped like an arrow pointing left towards a pair of parallel lines and on the other side of these line there is another arrow pointing to the right. To the left of this arrow ther is the phrase dx. There is an arrow extending from the x axis in quadrant III to the diagonal line discussed previously. The arrow marks the internal angle and is labeled θ(x).

But the field strength is directed at angle θ(x)θ(x), as illustrated in Figure 1. The field dE(x)dE(x) is real with components dE1(x)dE1(x) and dE2(x)dE2(x), but we code it as a complex field. We say that the “complex” field at test point (0,h)(0,h) is

dE(x)=λdx4πϵ0(h2+x2)ejθ(x)dE(x)=λdx4πϵ0(h2+x2)ejθ(x)
(2)

with components dE1(x)dE1(x) and dE2(x)dE2(x). That is,

d E ( x ) = d E 1 ( x ) + j d E 2 ( x ) d E ( x ) = d E 1 ( x ) + j d E 2 ( x )
(3)
d E 1 ( x ) = λ d x 4 π ϵ 0 ( h 2 + x 2 ) c o s θ ( x ) d E 1 ( x ) = λ d x 4 π ϵ 0 ( h 2 + x 2 ) c o s θ ( x )
(4)
d E 2 ( x ) = λ d x 4 π ϵ 0 ( h 2 + x 2 ) s i n θ ( x ) . d E 2 ( x ) = λ d x 4 π ϵ 0 ( h 2 + x 2 ) s i n θ ( x ) .
(5)

For charge uniformally distributed with density λλ along the x-axis, the total field at the test point (0,h)(0,h) is obtained by integrating dEdE:

- d E ( x ) = - λ 4 π ϵ 0 ( h 2 + x 2 ) [ cos θ ( x ) + j s i n θ ( x ) ] d x . - d E ( x ) = - λ 4 π ϵ 0 ( h 2 + x 2 ) [ cos θ ( x ) + j s i n θ ( x ) ] d x .
(6)

The functions cos θ(x)θ(x) and sinθ(x)sinθ(x) are

cosθ(x)=x(x2+h2)1/2; sinθ(x)=h(x2+h2)1/2 cosθ(x)=x(x2+h2)1/2;sinθ(x)=h(x2+h2)1/2
(7)

We leave it as a problem to show that the real component E 1 E 1 of the field is zero. The imaginary component E 2 E 2 is

E = j E 2 = j - λ h 4 π ϵ 0 d x ( x 2 + h 2 ) 3 / 2 E = j E 2 = j - λ h 4 π ϵ 0 d x ( x 2 + h 2 ) 3 / 2
(8)
= j λ h 4 π ϵ 0 x h 2 ( x 2 + h 2 ) 1 / 2 | - = j λ h 4 π ϵ 0 x h 2 ( x 2 + h 2 ) 1 / 2 | -
(9)
= j λ h 4 π ϵ 0 [ 1 h 2 + 1 h 2 ] = j λ 2 π ϵ 0 h = j λ h 4 π ϵ 0 [ 1 h 2 + 1 h 2 ] = j λ 2 π ϵ 0 h
(10)
E 2 = λ 2 π ϵ 0 h . E 2 = λ 2 π ϵ 0 h .
(11)

We emphasize that the field at (0,h)(0,h) is a real field. Our imaginary answer simply says that the real field is oriented in the vertical direction because we have used the imaginary part of the complex field to code the vertical component of the real field.

Exercise 1

Show that the horizontal component of the field E E is zero. Interpret this finding physically.

From the symmetry of this problem, we conclude that the field around the infinitely long wire of Figure 1 is radially symmetric. So, in polar coordinates, we could say

E ( r , θ ) = λ 2 π ϵ 0 r E ( r , θ ) = λ 2 π ϵ 0 r
(12)

which is independent of θ θ. If we integrated the field along a radial line perpendicular to the wire, we would measure the voltage difference

V ( r 1 ) - V ( r 0 ) = r 0 r 1 λ 2 π ϵ 0 r d r = λ 2 π ϵ 0 [ logr1-log r 0 ] . V ( r 1 ) - V ( r 0 ) = r 0 r 1 λ 2 π ϵ 0 r d r = λ 2 π ϵ 0 [logr1-log r 0 ].
(13)

An electric field has units of volts/meter, a charge density λλ has units of coulombs/meter, and ϵ 0 ϵ 0 has units of coulombs/volt-meter; voltage has units of volts (of course).

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