1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

11.11 • SOURCES AND SINKS 501

(0, 0, h) _qf2 l!.h

L i"·.

I • • • z

\ .... ;

... ,.. .. .. .. ... ...

...

(0, o. - h) • qt2 l!.h

Figure 11.94 Contributions to E from the elements of charge q~h situated at {O, 0, ± h),

above and below the z plane.

fluid flow problems, except that their corresponding potentials differ by a sign
change.
To establish Equation ( 11 - 41 ), we start with Coulomb's law, which states
that two particles with charges q and Q exert a force on one another with mag-
nitude cr~Q, where r is the distance between particles and C is a constant that
depends on the scientific units. For simplicity, we assume that C = 1 and the

test particle at the point z has charge Q = 1.

The contribution /::;.E 1 induced by the element of charge ~ along the seg-

ment of length /::;.h situated at a height h above the plane has magnitude !!::;.E 1 I

given by

1

1::;.E I = (q/2) tJ..h

I r 2 + h2.

It has the same magnitude as /::;.E2 induced by the element ~ located a

distance - h below the plane. From the vertical symmetry involved, their sum,
li.Ez + tJ..E2, lies parallel to the plane along the ray from the origin, as shown in
Figure 11.94.
By the principle of superposition, we add all contributions from the elements

of charge along L to obtain E = E fJ..Ek. By vertical symmetry, E lies parallel to

the complex plane along the ray from the origin through the point z. Hence the

magnitude of E is the sum of all components 1 1::;.EI cost that are parallel to the

complex plane, where t is the angle between t:;.E and the plane. Letting li.h - 0

in this summation process produces the definite integral

1

00

IE (x,y)I = lti.Elcostdh= 1°" (q/2) cost

• oo -oo r^2 + h^2 dh.