Poisson's equation boundary conditions (electrostatics)

[gnegnegne]

Hi everyone!
I have to solve a problem using Poisson's equation.
There are two parallel infinite conductor planes in vacuum. The distance between them is $d$ and they are both kept at a potential $V=0$. Between them there is a uniform volume density charge $\rho_0>0$ infinite along the directions parallel to the planes and its width is $d/2$ (it's a parallelepiped that starts at x=0 and ends at d/2). I have to find the the potential and the electric field in the region between the planes.
For $0<x<d/2$, solving the Poisson's equation (I integrated $\frac{\partial^2 V}{\partial x^2}=-\frac{\rho_0}{\epsilon_0}$):
$$V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax+B$$
For $d/2<x<d$, integrating $\frac{\partial^2 V}{\partial x^2}=0$:
$$V_2(x)=Cx+D$$
Setting the two boundary conditions at x=0 and at x=d I obtain $D=-Cd$ and $B=0$, therefore:
$$V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax$$
$$V_2(x)=C(x-d)$$

I can get ride of one of the two unknowns using the condition that V(x) has to be continuous at x=d/2, however I'm missing another condition. What other condition should be considered? Intuitively I would impose the condition that the electric field has to be continuous too, but I'm not 100% sure.

 

[kuruman]

In order to solve a second order differential equation, you need to know what the function is doing at the boundaries and also what its derivative is doing at the boundaries. Can you find the value of the derivative of the potential at one of the boundaries? That will give you $C$. Hint: What is the negative of the derivative of the potential otherwise known as?

Intuitively I would impose the condition that the electric field has to be continuous too, but I'm not 100% sure.

The electric field is discontinuous because it is zero inside the conductors at either end. Can you provide an argument why it must be continuous at $x=d/2$?

 

[gnegnegne]

Using the condition that the the potential has to be continuous:
$$V(x)=<br /> \begin{cases}<br /> (\frac{\rho_0 d}{4\epsilon_0}-A)x-d(\frac{\rho_0 d}{4 \epsilon_0} - A) \quad \text{for} \quad \frac{d}{2} \leq x \leq d \\<br /> -\frac{\rho_0 x^2}{2 \epsilon_0}+Ax \quad \text{for} \quad 0 \leq x \leq \frac{d}{2} \\<br /> \end{cases}$$

I can calculate the electric field as $E=-\nabla V$:
$$E(x)=<br /> \begin{cases}<br /> -(\frac{\rho_0 d}{4\epsilon_0}-A) \quad \text{for} \quad \frac{d}{2} < x < leq d \\<br /> \frac{\rho_0 x}{ \epsilon_0}-A \quad \text{for} \quad 0 < x <\frac{d}{2} \\<br /> \end{cases}$$I know that near the surfaces of the conductors the electric field is $E=\frac{\sigma}{\epsilon_0}$ perpendicular to the surface. But I don't know the surface density $\sigma$.

The electric field is discontinuous because it is zero inside the conductors at either end. Can you provide an argument why it must be continuous at x=d/2x=d/2x=d/2?

Yes, but at $x=\frac{d}{2}$ I don't have a conductor. It's just where the region occupied by a density of charge $\rho_0$ ends. For $d>x>\frac{d}{2}$ I have no free charges. To prove that the electric field has to be continuous I could consider the same argument used to demonstrate that the discontinuity of the electric field where there is a surface charge is $E_2-E_1=\frac{\sigma}{\epsilon_0}$: I consider I cylinder of height $2h$ and base surface $S$, placed so that half of the cylinder is in the region with free charge ($x<\frac{d}{2}$) and the other half in the region where there is no charge ($x>\frac{d}{2}$).
Using Gauss' Law: $E_2-E_1=\frac{Q(h)}{\epsilon_0}$. For $h\rightarrow 0$ the charge in the cylinder goes to zero because there is no surface charge, which implies (I think) that $E_2=E_1$