Poisson's equation to find V

[E&M]

Homework Statement

Two infinite parallel plates separated by a distance s are at potential 0 and V_0
a) Use Poisson's eqn to find potential V in the region between the plates where the space charge density is rho = rho_0(x/s). The distance x is measured from the plate at 0 potential.
b)What are the charge densities in the plate?

Homework Equations

$$\nabla$$$$^2{}$$ V = - rho/ epsilon_0

The Attempt at a Solution

For this problem, I started with number of ways but none of them seem to be working. What will be V in this case?

 

[gabbagabbahey]

Why don't you show me what you've tried?

 

[E&M]

V will be in only x direction and it won't be in y and z direction. double partial derivative of V along x will be equal to -rho/epsilon_0. I was thinking of using differential equation but I am not sure.

 

[gabbagabbahey]

You will need to use a differential equation. Luckily if V is only a function of x, then

$$\frac{\partial ^2V}{\partial x^2}=\frac{d^2V}{dx^2}$$

And so you will have an ordinary differential equation instead of a partial differential equation.

Give integrating it a shot and show me what you get.

 

[E&M]

i got V = -(rho_0 * x^3)/(s* epsilon_0*6) and for charge densities I got rho = 0 at the plate with V = 0 and rho = rho_0 for the plate with V = V_0. does this look right?

 

[gabbagabbahey]

Your V looks close, but there is a small error. On your first integration, you must include an integration constant:

$$\frac{d^2V(x)}{dx^2}=\frac{-\rho_0x}{s \epsilon _0} \Rightarrow \int_{x'=0}^{x'=x} \frac{d^2V(x')}{dx'^2} dx'= \int_0^x \frac{-\rho_0x'}{s \epsilon _0}dx'$$

$$\Rightarrow V'(x)-V'(0)=\frac{-\rho_0x^2}{2s \epsilon _0} \Rightarrow V'(x)=\frac{dV(x)}{dx}=\frac{-\rho_0x^2}{2s \epsilon _0}+C$$

where $C=V'(0)$ is the constant of integration

What do you get for V(x) when you don't forget the constant?