# 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?

• This seems to be merely the fact that the gradient is perpendicular to the equipotential surface. Mar 19, 2018 at 18:53
• @Qmechanic But Is there a mathematical proof of the your explanation? I would like to know if there is a mathematical demonstration like the one I reported. Namely, that for two parallel plates (in 3D) the equipotential surfaces turn out to be a bundle of parallel planes when the potential is changed from a higher to a lower one. I hope I have not asked a bad question because of the lack of attention I have had. Best regards. Mar 19, 2018 at 21:44
• @Sebastiano The perpendicularity follows almost by definition of "equipotential". Since equipotential surfaces have the same value of the potential, the gradient of the potential cannot have a nonzero component along the surface. That means the gradient must necessarily be normal to the equipotential surface.
– Siva
Mar 25, 2018 at 21:26
• @Qmechanic I have added some details. I have not studied and resolved PDE when I was at university. Mar 26, 2018 at 21:33

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 $$V_1\cdot x + V_2 \cdot y + V_3 \cdot z = k, \quad k \in \mathbb{R}$$
• I give you my bounty also because you have had attention at my problem. 1) I have not understand because $V(x,y,z)=(V_1x,V_2y,V_3z)$; 2) If $\overline E=|E|\hat n$ why $\hat n = \ldots$? 3) Who is $\overline r$ (position vector) and $\overline r_0$? Can you put a figure please thus I understand better? After $\hat{n}\cdot\left(\vec{r}-\vec{r}_0\right)=0$, $\cdot$ is a scalar product? Can you explain better these details? Buona Pasqua e grazie. Mar 30, 2018 at 20:26
• Sure, let's go one by one: 1) The potential is calculated simply integrating the electric field, you just need to be careful where you put the zero of your potential $V = -\int_0^{\infty}\vec{E}\cdot d\vec{r}$, in my case I choose the potential to be $0$ in the origin. 2) There was actually a mistake in my formula, I corrected it i hope it's clear now. 3)Check out this link tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx Yes it is a scalar product. Grazie e buona pasqua anche a te :) Mar 31, 2018 at 9:03