Solution to Laplace's Equation in Cylindrical Coordinates 1 ...

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L z V = Vo V = 0 x V = 0 a y Figure 2: The geometry of a cylinder with one eddcap held at potential V = V 0 (ρ) and the other sides grounded V = ∑ n A n J 0 (k n ρ)sinh(k n z) Now V = 0 for ρ = a. This means that; J 0 (k n a) = 0 The values of k n a are the zeros of the bessel function J 0 (k n a). The first few are, α 0n = 2.4048, 5.5201, 8.6537, · · ·. Then at Z = L we find A n using the orthogonality of the Bessel functions. A n = 1 a 2 [J 1 (k n a)] 2 sinh(k n L) ∫ a 0 ρdρ J 0 (k n ρ) V 0 (ρ) The graphic form of the solution is shown in figure 3. As another example we find the potential inside a cylinder when the potential is specified on the end caps and the cylindrical wall is at zero potential, figure 4. The boundry conditions are that; V = V 0 sin(φ) z = L V = −V 0 sin(φ) z = -L V = 0 ρ = a The solution must have the form; 4
Figure 3: A graphical representation of the above solution V = Vo sin( φ ) z a L y x V = 0 −L V = −Vo sin( φ ) Figure 4: The geometry of the problem with endcaps held at potential V = V 0 ± sin(φ) 5
Page 1 and 2: Solution to Laplace’s Equation in
Page 3: J ν (x) = ∞ ∑ s=0 (−1) s s!(
Page 7 and 8: G 0 = [ J 0(k n ρ) J 0 (k n c) −
Page 9 and 10: potential over the enclosing surfac

Solution to Laplace's Equation in Cylindrical Coordinates 1 ...

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