Essential Electromagnetism Solutions

Essential Electromagnetism Solutions tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn về tất cả các...

Essential Electromagnetism: Solutions Raymond John Protheroe Download free books at Raymond John Protheroe Essential Electromagnetism Solutions Download free eBooks at bookboon.com Essential Electromagnetism: Solutions First edition © 2013 Raymond John Protheroe & bookboon.com (Ventus Publishing ApS) ISBN 978-87-403-0404-6 Download free eBooks at bookboon.com Essential Electromagnetism: Solutions Contents Contents Preface Electrostatics Poisson’s and Laplace’s equations 17 Multipole expansion for localised charge distribution 32 Macroscopic and microscopic dielectric theory 39 Magnetic ield and vector potential 54 Magnetism of materials 67 www.sylvania.com We not reinvent the wheel we reinvent light Fascinating lighting offers an ininite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges An environment in which your expertise is in high demand Enjoy the supportive working atmosphere within our global group and beneit from international career paths Implement sustainable ideas in close cooperation with other specialists and contribute to inluencing our future Come and join us in reinventing light every day Light is OSRAM Download free eBooks at bookboon.com Click on the ad to read more Essential Electromagnetism: Solutions Preface Preface This book gives the solutions to the exercises at the end of each chapter of my book Essential Electromagnetism (also published by Ventus) I recommend that you attempt a particular exercise after reading the relevant chapter, and before looking at the solutions published here Often there is more than one way to solve a problem, and obviously one should use any valid method that gets the result with the least effort Usually this means looking for symmetry in the problem for example from the information given can we say that from symmetry arguments the field we need to derive can only be pointing in a certain direction If so, we only need to calculate the component of the field in that direction, or we may be able to use Gauss law or Amp re s law to enable us to write down the result In some of these exercise solutions the simplest route to the solution is deliberately not taken in order to illustrate other methods of solving a problem, but in these cases the simpler method is pointed out The solutions to the exercise problems for Each chapter of Essential Electromagnetism are presented here in the corresponding chapters of Essential Electromagnetism - Solutions I hope you find these exercises useful If you find typos or errors I would appreciate you letting me know Suggestions for improvement are also welcome please email them to me at protheroe.essentialphysics@gmail.com Raymond J Protheroe, January 2013 School of Chemistry & Physics, The University of Adelaide, Australia Download free eBooks at bookboon.com Essential Electromagnetism: Solutions Electrostatics Electrostatics 1 The surface of a non-conducting sphere of radius a centred on the origin has surface charge density σ(a, θ, φ) = σ0 cos θ and is uniformly filled with charge of density ρ0 Find the electric field at the origin Solution z θ dθ dS=r dS O x φ dE dφ At the centre of the sphere the electric field due to the volume charge will be zero because the contribution of a volume element located at r′ will be exactly cancelled by that of an equivalent volume element at −r′ , so we only need to consider the surface charge E(0, θ, φ) = 4πε0 ∫ σ(a, θ, φ) (−� r)dS, a2 ∫ 2π ∫ π σ(a, θ, φ)(−� r)[a2 sin θdθdφ], 4πε0 a2 0 ∫ 2π ∫ 1 d cos θ(σ0 cos θ)(−� r) dφ = 4πε0 −1 = (1.1) (1.2) (1.3) Because of the symmetry of the problem, the electric field at the centre can only be in the ±z direction, and so we only need to find the z-component E(0, θ, φ) · � z= 4πε0 = ∫ 2π dφ 2π 4πε0 ∫ ∫ −1 d cos θ(σ0 cos θ)(−� r) · � z, d cos θ(σ0 cos θ)(− cos θ), −1 Download free eBooks at bookboon.com (1.4) (1.5) Essential Electromagnetism: Solutions Essential Electromagnetism - Solutions ∴ E(0, θ, φ) · � z=− ∴ E(0, θ, φ) = − σ0 2ε0 Electrostatics Electrostatics ∫ d cos θ cos2 θ = − −1 σ0 3ε0 σ0 � z 3ε0 (1.6) (1.7) A spherically symmetric charge distribution has the following charge density profile ρ(r, θ, φ) = { ρ0 (r < a) −β (r ≥ a) ρ0 (r/a) (1.8) where β is a constant (2 < β < 3) Find the electric field and electrostatic potential everywhere Solution The charge density is spherically symmetric, with no dependence on θ or φ, so the electric field must be in the radial direction and depend only on r This is the ideal case to exploit Gauss law in integral form, ∮ E · dS = ε0 ∫ (1.9) ρd3 r For r < a 4πr2 Er = πr ρ0 , ε0 ∴ E(r) = For r > a 4πr2 Er = ∴ 4πr Er ∴ Er πa ρ0 + ε0 ε0 ∫ ρ0 r � r 3ε0 r (1.10) 4π(r′ )2 ρ0 aβ (r′ )−β dr′ , (1.11) a [ ]r 4πρ0 aβ (r′ )3−β πa ρ0 + , = ε0 ε0 3−β a (1.12) 4πρ0 a3 = ε0 (1.13) ( (r/a)3−β − 1 + 3−β ) (( ) ρ0 a r 3−β β − = (3 − β)ε0 r2 a ) Download free eBooks at bookboon.com (1.14) Essential Electromagnetism: Solutions Essential Electromagnetism - Solutions Electrostatics Electrostatics This electric field is due entirely to the charge distribution, and so must be conservative, and we would expect that ∇ × E = as E is directed radially outward and so has no circulation It follows that: ρ0 a3 V (r ≥ a) = − (3 − β)ε0 ∫ r ) ( β(r′ )−2 ′ 1−β β−3 dr′ , a − (r ) ∞ [ ]r ρ0 a3 (r′ )2−β aβ−3 β(r′ )−1 + , = − (3 − β)ε0 2−β ∞ ( ) ρ a3 3r2−β aβ−3 − β(β − 2)r−1 3(3 − β)(β − 2)ε0 ∫ r ρ0 V (r ≤ a) = V (a) − r′ dr′ , 3ε0 a [ ]r (3 + 2β − β )ρ0 a2 ρ0 (r′ )2 = − , 3(3 − β)(β − 2)ε0 3ε0 a = = (3 + 2β − β )ρ0 a2 ρ0 + (a − r2 ) 3(3 − β)(β − 2)ε0 6ε0 (1.15) (1.16) (1.17) (1.18) (1.19) (1.20) 360° thinking Discover the truth at www.deloitte.ca/careers Download free eBooks at bookboon.com © Deloitte & Touche LLP and affiliated entities Click on the ad to read more Essential Electromagnetism: Solutions Essential Electromagnetism - Solutions Electrostatics Electrostatics The electric field is given by E(r) = E0 cos(z/z0 ) exp(−r/r0 ) � r, where z0 and r0 are constants Find the charge density Solution In this problem the electric field is given in terms of z and r We will need to write E in terms of either Cartesian or spherical coordinates, and then use Gauss law in differential form Choosing spherical coordinates because E is in the radial direction, ρ(r) = ε0 ∇ · E, (1.21) ( ) ( )] [ r cos θ r ∂ = ε E0 exp − , (1.22) r cos r ∂r z0 r0 ( ( [ ( ) ( ) ) ) r cos θ cos θ r r r cos θ ε0 E0 exp − exp − − r2 sin + = 2r cos r z0 r0 z0 z0 r0 ( ) ( ) ( )] r cos θ −1 r exp − , (1.23) r cos z0 r0 r0 ( ) ( )[ ( ) ] r cos θ r r r cos θ r cos θ ε0 E0 cos − exp − − tan , (1.24) = r z0 r0 z0 z0 r0 ( ) ( )[ ( ) ] z r z z ε0 E0 r cos − exp − − tan (1.25) = r z0 r0 z0 z0 r0 If we had a point charge q at the origin we might choose the reference point to be some point at an arbitrary distance r0 (usually infinity) from the origin Then if we wish to find V (r, θ, φ) it would be convenient to have the reference point at r0 = (r0 , θ, φ) Although obtaining the potential in this case is trivial, and one would usually just write it down, obtain the potential by carrying out explicitly the line integral for an appropriately parameterised curve Solution Download free eBooks at bookboon.com Essential Electromagnetism: Solutions Essential Electromagnetism - Solutions Electrostatics Electrostatics z path Γ E r’ λ=0 dr’ r r0 λ = r − r’ λ = r 0− r y q x We start by parameterising the path from r0 to r: r′ (λ) = (r0 − λ) � r; Then, V (r) = − = − =− ∫ ∫ r r0 ∫ dr′ = −dλ � r; E(r′ ) · dr′ , r(λ=r0 −r) r(λ=0) ∫ (0 < λ < r0 − r) r(λ=r0 −r) r(λ=0) r0 −r (1.27) E(r′ (λ)) · dr′ , (1.28) q � r · (−dλ � r), 4πε0 (r0 − λ)2 (1.29) q dλ, 4πε0 (r0 − λ)2 [ ]r0 −r q , = 4πε0 (r0 − λ) = (1.26) (1.30) (1.31) = q q − , 4πε0 [r0 − (r0 − r)] 4πε0 (r0 − 0) (1.32) = q q − 4πε0 r 4πε0 r0 (1.33) Hence, if we set r0 = ∞ we get the usual potential for a point charge q at the origin V (r) = q 4πε0 r (1.34) 10 Download free eBooks at bookboon.com ...Raymond John Protheroe Essential Electromagnetism Solutions Download free eBooks at bookboon.com Essential Electromagnetism: Solutions First edition © 2013 Raymond John Protheroe... on the ad to read more Essential Electromagnetism: Solutions Preface Preface This book gives the solutions to the exercises at the end of each chapter of my book Essential Electromagnetism (also... θ)(− cos θ), −1 Download free eBooks at bookboon.com (1.4) (1.5) Essential Electromagnetism: Solutions Essential Electromagnetism - Solutions ∴ E(0, θ, φ) · � z=− ∴ E(0, θ, φ) = − σ0 2ε0 Electrostatics