Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
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Chapter 3, Problem 3.27P

Consider a slab of material containing a volume charge distribution throughout. The slab is of length d in the z direction, and its dimensions in x and y represent a cross-sectional area of A. Free space permittivity exists throughout. The electric field in the slab is given by E = p 0 : e p a exp ( a z ) a z V / m

where p0 is a positive constant. (a) Find the volume charge density pv in the slab, (b) Find the total charge in the slab. (c) Verify your result for part b by evaluating the net outward flux of D through the slab surfaces.

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Two 1.20 m non-conductive wires form a right angle. A segment has +2.50 µC of charge, distributed evenly along its length; while the other segment has -2.50 µC of charge, distributed uniformly along its length, as illustrated in the figure. Find the magnitude and direction of the electric field produced by these wires at point P, which is 60.0 cm from each wire.
(al:Determine E caused by the spherical cloud of electrons with a volume charge density of - 1.68 x 10 -18 for 0 10mm. Clearly mention the surfaces, there differential components and write the equation properly by doing all the steps. (b): For the dielectric composition shown in the figure find out its total capacitance.
Fill in the blank: A uniform sheet of charge with Ps1 = -0.2 nC/m² occupies the plane z = 17 and a second uniform sheet of charge with ps1 = 0.2 nC/m² occupies the x-y plane. The region between the plates is free space. The potential difference VBC B(0, 0, 4) and C(0, -2, 10) is..... Bc for V.

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