# Integrating Poisson Equation over two different regions knowing only two boundary conditions for the potentials

Poisson Equation for electric potential is: $$\nabla^2 V=-\frac{\rho}{\epsilon}$$ Solving the equation require two boundary condition. I'm confused about the use of these boundary conditions in some situations, like for example the following.

Consider two conductor planes, both at zero potential ($V=0$). Between the plates, there is a region with a charge density $\rho$ (the grey one) and another one with no charge.

Suppose that I need to solve Poisson Equation in order to find out the electric potential in all the region. The problem is about the boundary conditions: what are the boundary conditions in this case?

Surely I have that $V(0)=0$ and $V(2d)=0$ but this is not enough since I need to split potential in two and integrate the equation in the two different regions. $$\begin{cases} V(0<x<d)=-\frac{\rho}{\epsilon} x^2+c_1 x+c_2 \\ V(d<x<2d)=c_3x+c_4 \end{cases}$$

Another condition to impose could be that the potential must be continous in $x=d$. The three conditions gives $$\begin{cases} c_2=0 \\ c_4=-2d \,c_3\\c_1+c_3=\frac{\rho}{\epsilon}d \end{cases}$$

But I need one more condition in order to solve the equation, and I do not see where to get it.

In general, if Poisson Equation has to be solved over two different regions, how can I manage with situations like this one where I have only two potentials known but one of the potential $V(d)$ is "in common" between the two regions but it is not known in principle?

• You already tried with the derivative of $V$? remember it is discontinuous when there is net charge in the interface between regions. – Saavestro Nov 13 '16 at 19:28

The missing relation is the continuity of electric displacement at $x=d$ which, due to the same $\epsilon$, is the continuity of electric field, i.e., the continuity of the derivatives of the potentials $V_{left}$ and $V_{right}$ at $x=d$. This yields $$\frac{\partial V_{left}}{\partial x}|_{x=d}=\frac{\partial V_{right}}{\partial x}|_{x=d}$$ which gives for your constants the missing 4th equation $$-2\frac{\rho}{\epsilon} d+c_1= c_3$$
• Thanks for the answer! But, since there is a charge distribution for $0<x<d$ shouldn't $E=\frac{\partial V}{\partial x}$ be discontinous in $x=d$? – Sørën Nov 15 '16 at 17:18
• @Soren - no, for constant $\epsilon$ the electric field should be continuous also when a space charge is present as long as you don't have a surface charge (areal charge $\eta$) concentrated at $x=d$, which would cause a discontinuity $\epsilon (E_{right}-E_{left}) =\eta$. But this is obviously not the case. – freecharly Nov 15 '16 at 17:49
• Thanks again! By "constant $\epsilon$" do you mean that $\epsilon$ has the same value both for $0<x<d$ and $d<x<2d$ or that $\epsilon$ is a constant for $0<x<d$ but it can be different for $d<x<2d$? And if in $0<x<d$ there is a charged dielectric with constant $\epsilon$ and for $d<x<2d$ there is vacuum with constant $\epsilon_0$, and $\epsilon \neq \epsilon_0$, the electric field would be discontinous in $x=d$? – Sørën Nov 15 '16 at 18:45
• @Soren - The electric field continuity holds when $\epsilon$ is the same everywhere. In general, you have a continuity of electric displacement $\epsilon E$. If you have a charged dielectric with absolute permittivity $\epsilon$ left and vacuum with $\epsilon_0$ right, this would give you at $x=d$ the condition $\epsilon E_{left}=\epsilon_0 E_{right}$ . – freecharly Nov 15 '16 at 18:55