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Homework6 - Homework 6 Answers 95.657 Fall 2011...

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Homework 6 Answers, 95.657 Fall 2011, Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Problem 1 Consider a sphere of radius b centered at the origin whose surface is held at some fixed potential V (θ, φ). (a) Find the electrostatic potential everywhere external to the sphere in terms of spherical harmonics and V (θ, φ). (b) If V (θ, φ) = sin 2 θ e 2 i φ , what does the solution reduce down to? SOLUTION: (a) External to the sphere, there are no charges, so the electrostatic potential obeys the Laplace equation. As derived previously in class, the solution to the Laplace equation in spherical coordinates when the region of interest includes the full angular sweep is given by: Φ= l = 0 m =− l l ( A lm r l + B lm r l 1 ) Y lm , ϕ) Because the region of interest includes the full angular sweep, we have essentially already applied the boundary conditions in the polar angle and azimuthal angle. We have two sets of constants left to determine, and two boundary conditions left to apply: r max : Φ( r =∞)= finite r min : Φ( r = b )= V , ϕ) Applying the first leads to the statement that A l = 0 for all l > 0. The solution now becomes: Φ= A 00 + l = 0 m =− l l B lm r l 1 Y lm , ϕ) Note that unless the boundary conditions specifically require otherwise, we can always add overall an overall constant to the potential and still have the same physical fields. We add the constant - A 00 so that we have: Φ= l = 0 m =− l l B lm r l 1 Y lm , ϕ) This is equivalent to assuming that the potential approaches zero at r equals infinity. While the problem did not explicitly require this, it left us the freedom to apply this. If the problem had been different and had required that the potential at infinity approach some non-zero constant C , then we would have to keep the A 00 term and follow it through. Apply the last boundary condition: V , ϕ)= l = 0 m =− l l B lm b l 1 Y lm , ϕ)

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We need to solve this equation for B lm , but it is stuck in the sums. We want to take advantage of the orthogonality of the spherical harmonics, shown below, to collapse the sums.
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