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LaplaceEq-spherical-coordinates

LaplaceEq-spherical-coordinates - LAPLACES EQUATION IN...

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LAPLACE’S EQUATION IN SPHERICAL COORDINATES With Applications to Electrodynamics We have seen that Laplace’s equation is one of the most significant equations in physics. It is the solution to problems in a wide variety of fields including thermodynamics and electrodynamics. In your careers as physics students and scientists, you will encounter this equation in a variety of contexts. It is important to know how to solve Laplace’s equation in various coordinate systems. The coordinate systems you will encounter most frequently are Cartesian, cylindrical and spherical polar. We investigated Laplace’s equation in Cartesian coordinates in class and just began investigating its solution in spherical coordinates. Let’s expand that discussion here. We begin with Laplace’s equation: 0 2 = V (1) We can write the Laplacian in spherical coordinates as: ) ( sin 1 ) (sin sin 1 ) ( 1 2 2 2 2 2 2 2 2 φ θ θ θ θ θ + + = V r V r r V r r r V (2) where θ is the polar angle measured down from the north pole, and φ is the azimuthal angle, analogous to longitude in earth measuring coordinates. (In terms of earth measuring coordinates, the polar angle is 90 minus the latitude, often termed the co- latitude .) To make our initial calculations a little simpler, let’s assume azimuthal symmetry ; that means that our parameter V does not vary in the φ direction. In other words, 0 / = φ V , so we can write the Laplacian in (2) a bit more simply. Assuming azimuthal symmetry, eq. (2) becomes: ) (sin sin 1 ) ( 1 2 2 2 2 θ θ θ θ + = V r r V r r r V (3) This is the form of Laplace’s equation we have to solve if we want to find the electric potential in spherical coordinates. First, let’s apply the method of separable variables to this equation to obtain a general solution of Laplace’s equation, and then we will use our general solution to solve a few different problems. To solve Laplace’s equation in spherical coordinates, we write:
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0 ) (sin sin 1 ) ( 1 2 2 2 2 = + = θ θ θ θ V r r V r r r V (4) First Step: The Trial Solution The first step in solving partial differential equations using separable variables is to assume a solution of the form: ) ( ) ( θ Θ = r R V (5) where R(r) is a function only of r , and Θ ( θ ) is a function only of θ . This means that we can set: ) ( ) ( ); ( ) ( θ θ θ Θ′ = Θ = r R V r R r V (6) Substituting the relationships in (6) into (4) produces: 0 )) ( (sin sin ) ( )) ( ( ) ( 2 2 2 2 = Θ′ + Θ = θ θ θ θ θ r r R r R r r r V (7) If we multiply each term in (7) by r 2 and then divide each term by V = R(r) Θ ( θ ) , we obtain: 0 )) ( (sin sin ) ( 1 )) ( ( ) ( 1 2 2 = Θ′ Θ + = θ θ θ θ θ d d r R r dr d r R V (8) Notice that the derivates in (8) are no longer partial derivatives. This is because the method of separable variables has produced two terms; one is solely a function of r and the other is solely a function of θ .
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