Finding potential using poisson's equation, not a homework problem.

[yungman]

This is not a homework problem. I just want to have a better understanding of scalar electric potential.

In electrostatic, $\; V=-\int_A^B \vec E \cdot \hat{T} dl \;$ where the solution is:

$$V=\frac{q}{4\pi \epsilon_0} \frac{1}{B}$$

Where we assume $\; A=\infty$.


At the same time, $\vec E = -\nabla V \Rightarrow \nabla \cdot \nabla \vec E = -\nabla^2 V$.

I want to see whether I can solve V using partial differential equation technique by using:

$$V(x,y)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} E_{mn} sin(\frac{m\pi}{a}) sin(\frac{n\pi}{b})$$

Where a and b are the boundary condition. My question is how do I set up the boundary condition a and b?

 

[kuruman]

I want to see whether I can solve V using partial differential equation technique ...

What exactly do you mean by "solve V"? Solve what for V? Your title mentions the Poisson equation which is $$\frac{\partial^2 V(x,y)}{\partial x^2}+\frac{\partial^2 V(x,y)}{\partial y^2}=\rho(x,y)$$To do that, you will need the charge density function $\rho(x,y)$ (I assume this is a 2-D situation) which you do not provide.
If instead you want to solve Laplace's equation,$$\frac{\partial^2 V(x,y)}{\partial x^2}+\frac{\partial^2 V(x,y)}{\partial y^2}=0,$$then you need to do so by separating variables (there are many videos on the internet to show you how), find the most general solution and then apply the boundary conditions of your choice to narrow down the most general solution and figure out the separation constant.

What you have for $V(x,y)$ is not the most general solution. You probably meant to write$$V(x,y)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} E_{mn} \sin(\frac{m\pi x}{a}) \sin(\frac{n\pi y}{b}),$$ but $\sin(\frac{m\pi x}{a}) \sin(\frac{n\pi y}{b})$ is not a solution of Laplace's equation as you can verify by direct substitution and neither is a linear combination of such terms. I don't think that your $V(x,y)$ can be of any use. It is a function that vanishes at the rectangular boundaries because $V(0,y)=V(x,0)=V(a,y)=V(x,b)=0$. Now if the potential vanishes at all four boundaries of the rectangle, the uniqueness theorem says that the solution inside the rectangle is the trivial solution $V(x,y)=0$.