Laplace operator to find a bundle of parallel planes (equipotential surfaces) to two plates We known that the potential generated by a charge pointwise $q$ is $V(r) = kq/r $ and the equipotential surfaces (in 3D) are  spheres centered in the charge with $r\geq 0$ where $r=d(O,P)$, i.e. the distance between the origin and a generic point $P$.
In fact if we are in space, where an orthonormal system of cartesian coordinates has been fixed, $ r = r(x,y,z) = \sqrt {x^2 + y^2 + z^2}$ indicates the module of the radius vector, i.e. is the distance from the origin. 
If the potential is given by $V = C/r$, where $C$ is a constant. The equipotential surfaces are the place of the points of the space to a fixed potential, that is all the points $ (x, y, z) \in \mathbb{R}^3$ such that
$$V(x,y,z) = \dfrac{C}{r(x,y,z)} = \dfrac{C}{\sqrt {x^2 + y^2 + z^2}} = V_0$$
equivalently
$$\sqrt{x^2 + y^2 + z^2} = \dfrac C{V_0} \iff x^2 + y^2 + z^2 = \left(\dfrac C {V_0} \right)^2. $$
If I have, instead in 3D (general and not particular case) which is the most concrete situation, 2-planes $\pi$ and $\pi'$ parallels (for example, the flat plates of a plane capacitor) between them at a distance $\ell$, is it possible to find a mathematical relationship of the electric potential $V=V(x,y,z)$ that give me a bundle $\mathcal F$ of planes parallel of the type
$$\mathcal F:\quad  V(x,y,z)=ax+by+cz+k=0,\quad k\in\mathbb{R},$$
orthogonal to the uniform electric field $\overline E=(E_x,E_y,E_z)$?
If I trasform this the problem into a differential equation to partial derivatives (PDE), using the Laplace operator
$$\overline E = -\overline \nabla V \Longleftrightarrow E_x\mathbf{\hat x}+E_y\mathbf{\hat y}+E_z\mathbf{\hat z}=-\left(\dfrac{\partial V_x}{\partial x} \mathbf{\hat x}+\dfrac{\partial V_y}{\partial y} \mathbf{\hat y}+\dfrac{\partial V_z}{\partial z} \mathbf{\hat z}\right)$$
how can I find the bundle $\mathcal F:\, V(x,y,z)=ax+by+yz+k=0$ parallel (equipotential surface) to the two plates?
 A: If I understand the question correctly you are interested in finding the surface of constant potential in the case of infinite planes with uniform charge density.
The potential can be chosen to take the form:
$$V(x, y, z) = \left(V_{1}x, V_{2}y, V_{3}z \right),\quad \text{where $V_1, V_2, V_3$ are some constant which depend on the specific problem}$$
Now you need the family of planes perpendicular to the electric field, these are simply all planes with the normal vector pointing in the direction of the electric field:
$$\hat{n} = \frac{\left(E_x, E_y, E_z\right)}{\sqrt{\left(E_x^2+E_y^2+E_z^2\right)}}$$
As you pointed out $$\vec{E} = -\vec{\nabla}V$$
So what is left to do is to find $\hat{n}$:
$$ \hat{n} = \frac{\left(\frac{\partial V_x}{\partial x}, \frac{\partial V_y}{\partial y}, \frac{\partial V_z}{\partial z}\right)}{\sqrt{\left(\left( \frac{\partial V_x}{\partial x} \right)^2+\left( \frac{\partial V_y}{\partial y} \right)^2+\left( \frac{\partial V_z}{\partial z} \right)^2\right)}} = \frac{\left(V_1, V_2, V_3 \right)}{\sqrt{V_1^2+V_2^2+V_3^2}}$$
And the family of planes is given by:
$$\hat{n}\cdot\left(\vec{r}-\vec{r}_0\right)=0 , \quad\text{where $\vec{r}_0$ is generic.}$$
Finally we get
$$V_1\cdot x + V_2 \cdot y + V_3 \cdot z = k, \quad k \in \mathbb{R}$$
