How Do I Solve Poisson's Equation for V?

[leoflindall]

Homework Statement

\nabla^{2}V=(\rho)/(\epsilon_{0})

Homework Equations

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The Attempt at a Solution

I have a function of x which i can supstitute into Charge density and boundary conditions, however my problem with this is very simple. How to i manipulate the equation to get a funvtion for V. Obviously i need to get rid of the div of the div of v and i assume this has to be done through some form of intergration?

I assume this isn't as simple as intergrating both sides twice?

I would appreciate any guidance that can be given on this.

Regards.

Leo

 

[CompuChip]

In general, differential equations in physics are hard to solve. From what you posted, I understand that the function \rho = \rho(x) is given, and you are working in one dimension only.

In that case,
\nabla^2 V = \frac{d^2 V}{dx^2}
so you can solve your equation (I'm wiping the constant into rho, for notational convenience)
\frac{d^2 V}{dx^2} = \rho(x)
by integrating twice.

For example, when \rho(x) = \epsilon_0 x^2 you would simply get
V''(x) = x^2
so
V'(x) = \frac13 x^3 + c
and
V(x) = \frac{1}{12} x^4 + c x + k

The only snag might be that the integration is very hard to do: just pick a nasty function like \rho / \epsilon_0 = \sqrt{1 + x e^x} and your scr*wed :)

Note that once you go to two dimensions, things already get far less trivial, you would have to solve
\frac{\partial^2V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = \rho(x, y)
which takes a little more than two integrations (usually, you already need things like symmetry and polar coordinates here to make anything of it).

 

[leoflindall]

That makes makes perfect sense! Thank you for your help!