- #1

sukmeov

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Any hints on how to approach this?

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- Thread starter sukmeov
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- #1

sukmeov

- 10

- 2

Any hints on how to approach this?

- #2

BvU

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Google Poisson equation. Your equation describes the electric potential of a non conducting uniformly charged sphere.

Are you aware of the equation for electric potential ##\Delta\phi = C## ?

- #3

sukmeov

- 10

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My surname's funny? Many thanks.

- #4

BvU

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Ah ... Sorry Ukan !Also it's not a nickname it's my surname.

- #5

HallsofIvy

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- #6

phyzguy

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- #7

wrobel

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sinse ##C## is a constant it is just an ODE in the spherical coordinates

- #8

peterwang

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- #9

PhDeezNutz

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- #10

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First solve for electric field ##\mathbf{E}=\nabla U## and ##\nabla\cdot\mathbf{E}=C## by using Gauss' law in integral form and taking advantage of the symmetry.

Then solve for U by using $$U=\int_{\mathcal{C}} \mathbf{E}\cdot d\mathbf{l}$$ where ##\mathcal{C}## is a suitable path of integration, depending of what the form of E is. In this example I expect E to be in the radial direction, hence a suitable path of integration is along the radial line from position r to infinity.

- #11

pasmith

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\frac1{r^2} \frac{d}{dr}\left(r^2\frac{du}{dr}\right) = C[/tex] subject to [itex]u(1) = 0[/itex] doesn't really require resort to Green's funcions or Sturm-Liouville theory.

Most of the work here is in reducing the problem to that: knowing to use spherical polars and understanding that if the boundary condition does not depend on [itex](\theta,\phi)[/itex] then the solution probably doesn't, that the boundary condition at the origin is that [itex]u[/itex] is finite, and that the uniqueness theorem guarantees that if we find a solution then it must be the solution.

- #12

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Not only the boundary condition but the source term also doesn't depend on ##\theta,\phi## and hence the Laplacian operator ##\nabla^2## reduces to ##\frac{1}{r^2}\frac{d}{dr}\left (r^2\frac{d}{dr}\right )##, something that wasn't obvious from the start , because the problem was expressed with the Laplacian operator in cartesian coordinates.understanding that if the boundary condition does not depend on (θ,ϕ) then the solution probably doesn't

- #13

BvU

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Did we all notice OP wasn't seen since June 28 (this year) ?

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