Download Electromagnetic Field Theory: A Collection of Problems by Gerd Mrozynski, Matthias Stallein PDF

By Gerd Mrozynski, Matthias Stallein

After a quick advent into the speculation of electromagnetic fields and the definition of the sphere amounts the publication teaches the analytical answer equipment of Maxwell’s equations via a number of attribute examples. the point of interest is on static and desk bound electrical and magnetic fields, quasi desk bound fields, and electromagnetic waves. For a deeper figuring out, the various depicted box styles are very useful. The publication deals a set of difficulties and strategies which allow the reader to appreciate and to use Maxwell’s thought for a wide category of difficulties together with classical static difficulties correct as much as waveguide eigenvalue problems.

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Additional info for Electromagnetic Field Theory: A Collection of Problems

Sample text

R α, C are positive constants. What is the charge density according to the potential in homogeneous space of permittivity ε ? As the spheres r = const are equipotential surfaces, the direction of the electric field is radial with E(r) = er E(r) = −er The space charge density ΔV dV dr d2 V dr2 − ε (r) = = = = = exp(−αr) dV = er C(1 + αr) . dr r2 satisfies the Laplace–Poisson equation: 2 dV d2 V = − ; + dr2 r dr ε exp(−αr) −C (1 + αr) r2 r>0 −C [αr + (1 + αr) (−αr − 2)] exp(−αr) r3 exp(−αr) −ε C (αr)2 ; r3 exp(−αr) r3 C (αr)2 r > 0.

Its contribution to the potential leads to a complicated integration. A considerably easier approach is to derive this contribution by evaluating the expression for the potential of the initial charge at mirrored observation points. Thus no additional calculation is needed (cp. 9). For example, if a point charge Q is located in front of a grounded sphere of radius a at position rq with |rq | > a, then the resulting potential V (rp ) at points rp with |rp | ≥ a is the sum of the potential Ve (rp ) of the charge Q in homogeneous space and of the potential of the mirrored charge −Qa/|rq | at position rq (a/rq )2 .

As the exciting potential Ve = M Pn (u) 4πε r(n+1) = n=1 M cos ϑ M = 4πε r2 4πεa2 is a solution for n = 1, the overall solution is given ⎧ M ⎨ (a/r)2 + C1 r/a V (r, ϑ) = 4πεa2 ⎩ C2 (a/r)2 a r by ⎫ ⎬ ⎭ cos ϑ ; 2 cos ϑ r≤a r≥a . The constants C1 and C2 result from the boundary conditions for the potential and the normal component of the electric flux density. V (r, ϑ)| r>a = V (r, ϑ)| ra r→a = ε ∂V ∂r r

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