Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Textbook Question
Chapter 3, Problem 3.19P
A spherical surface of radius 3 mm is centered at P(4,1,5) in free space. Let D=xax C/m2. Use the results of Section 3.4 to estimate the net electric flux leaving the spherical surface.
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Chapter 3 Solutions
Engineering Electromagnetics
Ch. 3 - Prob. 3.1PCh. 3 - An electric field in space is E=(5z2/C0)azV/m....Ch. 3 - Consider an electric dipole in free space,...Ch. 3 - An electric field in free space is E=(5z3/0)z V/m....Ch. 3 - A volume charge distribution in free space is...Ch. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Use Gauss, law in integral form to show that an...Ch. 3 - A sphere of radius a free space contains charge of...Ch. 3 - An infinitely long cylindrical dielectric of...
Ch. 3 - Consider a cylindrical charge distribution having...Ch. 3 - The sun radiates a tota1 power of about 3.86...Ch. 3 - Spherical surfaces at r = 2, 4, and 6 m carry...Ch. 3 - Prob. 3.14PCh. 3 - Volume charge density is located as follows; pv=0...Ch. 3 - An electric flux density is given by D=D0aP, where...Ch. 3 - In a region having spherical symmetry, volume...Ch. 3 - State whether the divergence of the following...Ch. 3 - A spherical surface of radius 3 mm is centered at...Ch. 3 - A radial electric field distribution in free space...Ch. 3 - In a region exhibiting spherical symmetry,...Ch. 3 - (a) A flux density field is given as F1 = 5 az....Ch. 3 - (a) A point charge Q lies at the origin. Show that...Ch. 3 - In a region in free space, electric flux density...Ch. 3 - Within the spherical shell, 3D= 5(r-3)3a,C/m2 .(a)...Ch. 3 - If we have a perfect gas of mass density Px...Ch. 3 - Consider a slab of material containing a volume...Ch. 3 - Repeat Problem 3.8, but use .D= pv and take an...Ch. 3 - Prob. 3.29PCh. 3 - (a) Use Maxwells first equation. �. D=Pv, to...Ch. 3 - Prob. 3.31P
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- A circular disk of radius 3.3 m has a non-uniform surface charge density of p, = sin nC/m? is located at the 9p origin as shown in the figure. The net flux in "nC" through a closed surface of a cube centered at the origin that has side length of 8.6 m is Ps y Answer:arrow_forwardFigure shows a Gaussian surface in the form of a closed cylinder (a Gaussian cylinder or G-cylinder) of radius R. It lies in a uniform electric field E with the cylinder’s central axis (along the length of the cylinder) parallel to the field. What is the net flux ϕ of the electric field through the cylinder?arrow_forwardThe electric field present in the region is given by the expression shown in the photo in V/m. Determine the amount of flux, in pC, passing through a disk of radius 5m. The said disk lies on the plane z=2marrow_forward
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- A conducting cylinder of radius 3.5 cm and length 5.6 cm has a total charge of 4.5 x10 ^ -9 C distributed uniformly on its surface area. Find the potential at (a) its surface and (b) 2.5 cm and (c) 5.0 cm from the center of the cylinder. I really need an idea huhu.?arrow_forwardThe distance between parallel conductive discs in the Figure ? ? = 5 mm dir. Accept that the area between the disks is space and the potential at ? ? = 2 mm and the load on the disks find their intensity.arrow_forwardA sphere has radius of R. The sphere also has a uniform charge of 4Q. There is a point charge of -Q at sphere's center. Derive an equation for E at points where the radius is less then R.arrow_forward
- The vector magnetic potential inside a conductive cylindrical conductor of radius b is given by A . H =?arrow_forwardTwo conducting cylinders at p= 2 cm and p = 8 cm in free space are held at potentials of 60 mV and -30 mV, electric field at p= 3 cm 2.166ap- 1.3ap- 1.625ap- 1.0833ap-arrow_forwardProblem An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as DE - dÃ= We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain A = (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc On the other hand, the Gaussian surface inside the wire is given by A = Using Equation 1, the electric field in simplified form is E = Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus denc = On the other hand, the Gaussian surface outside the wire is given by A = Using Equation 1, the electric…arrow_forward
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