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?

  • 5
    $\begingroup$ This seems to be merely the fact that the gradient is perpendicular to the equipotential surface. $\endgroup$
    – Qmechanic
    Mar 19, 2018 at 18:53
  • $\begingroup$ @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. $\endgroup$
    – Sebastiano
    Mar 19, 2018 at 21:44
  • 1
    $\begingroup$ @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. $\endgroup$
    – Siva
    Mar 25, 2018 at 21:26
  • $\begingroup$ @Qmechanic I have added some details. I have not studied and resolved PDE when I was at university. $\endgroup$
    – Sebastiano
    Mar 26, 2018 at 21:33

1 Answer 1


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 $$V_1\cdot x + V_2 \cdot y + V_3 \cdot z = k, \quad k \in \mathbb{R}$$

  • $\begingroup$ I modified my question again, hoping it would be clearer than before. I would like to understand how to find the general case and not the particular one. I thank you for your answer, which obviously vote positively, both for the effort and for the time you have dedicated to me. $\endgroup$
    – Sebastiano
    Mar 29, 2018 at 19:26
  • $\begingroup$ I edited my answer, I hope I made it better. $\endgroup$ Mar 30, 2018 at 10:35
  • $\begingroup$ 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. $\endgroup$
    – Sebastiano
    Mar 30, 2018 at 20:26
  • $\begingroup$ 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 :) $\endgroup$ Mar 31, 2018 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.