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Solving for electric potential using separation of variables
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[QUOTE="TSny, post: 5451542, member: 229090"] Your second approach should work. However, I noticed a few errors: [U]Page 1[/U]: Your expression for σ is incorrect. You want to write a surface charge density for the surface r = R. Thus, ∫σdA = q for this surface. There is no integration over r here. Think about the expression for dA for this surface. [U]Page 4[/U]: Your powers of r in ##\Phi_I## and ##\Phi_{II}## don't appear to be correct. For example, see eq. 14 here [URL]http://www.luc.edu/faculty/dslavsk/courses/phys301/classnotes/laplacesequation.pdf[/URL] [U]Page 4[/U]: Check the signs in the equation at the bottom of the page for the boundary condition involving R[SUP]+[/SUP] and R[SUP]-[/SUP]. Recall that ##\vec{E} = -\nabla \Phi##. ( I don't understand the equation in the middle of the page, just before "in Volume II".) [U]Page 7[/U]: You forgot the ##\delta_{l,l'}## that occurs when doing the integral on the left at the top of the page. [U]There is another boundary condition[/U] that you can use. What can you say about the continuity of ##\Phi## at r = R when ##\theta \neq 0##? [/QUOTE]
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Solving for electric potential using separation of variables
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