Page 141 - Engineering Electromagnetics, 8th Edition

CHAPTER 5 Conductors and Dielectrics 123

Because D = 0 E,wehave
D P = 8.854 × 10 −12 E P =−3.54a x − 1.771a y nC/m 2
The field is directed downward and to the left at P;itis normal to the equipotential
surface. Therefore,
D N =|D P |= 3.96 nC/m 2

Thus, the surface charge density at P is
ρ S,P = D N = 3.96 nC/m 2
Note that if we had taken the region to the left of the equipotential surface as the
conductor, the E field would terminate on the surface charge and we would let
2
ρ S =−3.96 nC/m .

EXAMPLE 5.3
Finally, let us determine the equation of the streamline passing through P.
Solution. We see that
E y 200y y dy
= =− =
E x −200x x dx
Thus,
dy dx
+ = 0
y x
and

ln y + ln x = C 1
Therefore,

xy = C 2
The line (or surface) through P is obtained when C 2 = (2)(−1) =−2. Thus, the
streamline is the trace of another hyperbolic cylinder,

xy =−2
This is also shown on Figure 5.5.

D5.5. Given the potential field in free space, V = 100 sinh 5x sin 5yV , and
a point P(0.1, 0.2, 0.3), find at P:(a) V ;(b) E;(c) |E|;(d) |ρ S | if it is known
that P lies on a conductor surface.
Ans. 43.8 V; −474a x − 140.8a y V/m; 495 V/m; 4.38 nC/m 2