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

$$\mathcal F:\quad V(x,y,z)=x+y+z+k=0,\quad k\in\mathbb{R},$$ 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 $\mathcal F:\, V(x,y,z)=x+y+z+k=0$ parallel to the 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?

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 which is the most concrete situation) two 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 that give me a beam of planes parallel orthogonal to the uniform electric field?

If we translate the problem into a differential equation (to partial derivatives (PDE), using the Laplace operator) how can I get what is required?

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

$$\mathcal F:\quad V(x,y,z)=x+y+z+k=0,\quad k\in\mathbb{R},$$ 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 $\mathcal F:\, V(x,y,z)=x+y+z+k=0$ parallel to the 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 which is the most concrete situation) two 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 that give me a beam of planes parallel orthogonal to the uniform electric field?

If we translate the problem into a differential equation (to partial derivatives, using the Laplace operator) how can I get what is required?

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 which is the most concrete situation) two 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 that give me a beam of planes parallel orthogonal to the uniform electric field?

If we translate the problem into a differential equation (to partial derivatives (PDE), using the Laplace operator) how can I get what is required?