# Solve Poisson's Equation

1. Jan 20, 2010

### leoflindall

1. The problem statement, all variables and given/known data

$$\nabla$$$$^{2}$$V=($$\rho$$)/($$\epsilon$$$$_{0}$$)

2. Relevant equations

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3. 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

2. Jan 20, 2010

### 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).

3. Jan 20, 2010

### leoflindall

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