Download Applied Electromagnetics by J. E. Parton, S. J. T. Owen, M. S. Raven (auth.) PDF

By J. E. Parton, S. J. T. Owen, M. S. Raven (auth.)

Show description

Read or Download Applied Electromagnetics PDF

Similar physics books

The Kerr Spacetime: Rotating Black Holes in General Relativity

Rotating black holes, as defined via the Kerr space-time, are the major to knowing the main violent and full of life phenomena within the Universe, from the middle cave in of huge supernova explosions generating strong bursts of gamma rays, to supermassive black gap engines that energy quasars and different lively galactic nuclei.

Extra info for Applied Electromagnetics

Example text

Dl round the two closed paths (1, 1, 0)(-1, 1, 0)(-I, -I, O)(I, -I, O)(I, I, 0) and (0, 0, O)(I, 0, 0) (I, I, 0)(0, I, 0)(0, 0, 0). Is Fa conservative field? Check by curl F. 6 Find the electric field strength E and volume charge density Pv at the point (a, a, a) for each of the fields in which scalar Vis given as follows. 7 Find the energy stored in the region r >a for a point charge Qat the origin. 8 Given an electric field strength£= (p/4ne 0 ilX2 cos Oa, +sin Oa6 ) V/m for a dipole p in the z direction at the origin, find the energy stored in the region r >a.

For the rectangular column of charge shown, the normal electric field strength En= Ex at a symmetrical point P can be found by considering the elementary wide plate, dx thick, having 'surface' charge density Pv dx. From the previous case we have Pv dx dEx = - - (rt- 2'Y)Ox :bteo and using x =c - b (tan 'Y) the integral can be evaluated between the limits 'YF and 'YB· The result is 'Yn)] ax Ex =Pv - [ a('YF + 'Ya) + c('Ya - 'YF) +bIn (cos -1t€o COS'YF Of course in all these cases PL, p 8 , Pv have been taken as constants; they could all be functions of position and the integrals would be correspondingly more difficult.

I1 S = '\72S = 0 a2 s a2 s a2 s ax 2 ay 2 az 2 -+-+-=0 1. 2 laplace equation div Fi= 0, curl F = 0 The field is not solenoidal but it is irrotational or lamellar and "i/2 S a2 s a2 s a2 s ax 2 ay 2 az 2 -+-+-i=-0 *0 Poisson equation 19 VECTOR ANALYSIS 1. 3 div F= 0, curl F=F 0 This is solenoidal but it is also rotational and cannot have been derived from a scalar potentialS. But suppose it has been obtained from a vector potential A so chosen that F = curl A = V' x A then we have curl F =curl (curl A)= grad divA - V' 2A We can also stipulate that the vector potential A is solenoidal so that divA = 0 giving finally grad divA = 0 and curl F=- V' 2A =I= 0 This is Poisson's equation using vector potential A instead of scalarS.

Download PDF sample

Rated 4.94 of 5 – based on 41 votes