Homework9 - Homework 9 Answers, 95.657 Electromagnetic...

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Homework 9 Answers, 95.657 Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Jackson 4.10 Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ± Q . The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ε / ε 0 ), as shown in the figure. (a) Find the electric field everywhere between the spheres. SOLUTION We can split the region where we want to know the electric field into two regions, the left hemisphere and the right hemisphere, solve for the field in each region separately and then apply the boundary conditions to get the final solution. In the left region, there are no dielectrics and no charges. So the electric potential obeys the Laplace equation: 2 L = 0 If we align the z -axis pointing to the right, the problem has azimuthal symmetry. The general solution to the Laplace equation in spherical coordinates for azimuthal symmetry is: L r , , = l A l r l B l r l 1 P l cos  The potential on the surface of a conductor is always constant: L r = a = C C = l A l a l B l a l 1 P l cos  Because this equation must hold for all values of the independent polar variable and because the Legendre polynomials are orthogonal, all coefficients must equate independently: C = A 0 B 0 a and B l =− A l a 2 l 1 for l > 0 a b + Q -Q
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The solution now becomes: L r , , = C B 0 a B 0 r l = 1 A l r a l a r l 1 P l cos  The other surface is also a conductor and must have a constant potential as well: L r = b = D D = C B 0 a B 0 b l = 1 A l b a l a b l 1 P l cos  Because this equation must hold for all values of the independent polar variable and because the Legendre polynomials are orthogonal, all coefficients must equate independently: B 0 = ab D C a b and A l = 0 for l > 1 The solution now becomes: L r , , = C ab D C a b 1 a 1 r At this point C and D are undetermined, so we can redefine them to simplify the equation: L r , , = D C r This leads to a total electric field of: E L = C r 2 r We could have guessed this form of the solution based on the symmetry of the problem, but it is often safer and more instructive to go through all the steps. The right hemisphere is a separate region and can be now solved separately. In the region between the
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This note was uploaded on 03/21/2012 for the course PHY 2030 taught by Professor Avery during the Spring '11 term at University of Florida.

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Homework9 - Homework 9 Answers, 95.657 Electromagnetic...

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