3209-2009-Solutions7

3209-2009-Solutions7 - ECE 3209 Electromagnetic Fields...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 7 Solutions 1. Average potential over a sphere with point charge q outside. Consider the geometry of the problem shown below: The average potential over the spherical surface is, Φ avg 1 4 πR 2 I sphere Φ da, where Φ( r ) = 1 4 π² 0 q r is the potential at position ~ r from a point charge. From the Law of Cosines, r 2 = s 2 + R 2 - 2 sR cos θ , so Φ avg = 1 4 πR 2 q 4 π² 0 Z 2 π 0 Z π 0 R 2 sin θ dθ s 2 + R 2 - 2 sR cos θ = q 4 π² 0 1 2 sR p s 2 + R 2 - 2 sR cos θ fl fl fl π 0 = q 4 π² 0 1 2 sR ( p ( s + R ) 2 - p ( s - R ) 2 ) Φ avg = 1 4 π² 0 q s which is the potential due to the point charge evaluated at the center of the sphere. 2. Conductive medium between concentric spheres. (a) For a conducting medium, ~ J = σ c ~ E , where σ c is the conductivity. For a uniform conductor and steady current, ~ ∇ · ~ E = 0 , within the medium, and thus, 2 Φ = 0 . The
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
potential between the conductors satisfies Laplace’s equation which, in spherical coordinates, is written, 2 Φ = 1 r 2 d dr r 2 d Φ dr · = 0 Since the problem is spherically symmetric and independent of the θ and φ coordinates. The region of interest does not contain the point r = 0 , so we can recast Laplace’s equation as, ∂r r 2 Φ · = 0 r 2 d Φ dr = - C d Φ dr = - C r 2 where C is an integration constant. Integrating this equation, we obtain Φ( r ) = C r + B where B is the second integration constant. Setting the boundary conditions to be V ( a ) = 0 and V ( b ) = V 0 we get C a + B = 0 B = - C a C b + B = C b - C a = C a - b ab = V 0 C = ab a - b V 0 and thus the potential between the conductors is, Φ( r ) = ab a - b V 0 1 r - 1 a · and the electric field is, ~ E ( r ) = - ~ Φ = ˆ r d Φ dr = ab b - a V 0 r 2 ˆ r The current density is,
Image of page 2
~ J = σ c ~ E = σ c ab b - a · V 0 r 2 ˆ r We find the total current by integrating over the cross-sectional area (a spherical surface) between the conductors: I = Z S ~ J · d~a = σ c ab b - a · V 0 Z 2 π 0 Z π 0 1 r 2 r 2 sin θdθ = 4 π ab b - a · σ c V 0 Notice that as b → ∞ , I
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern