# Finding the surface from the electric potential [closed]

I am stuck on a request from this problem: I have the electrostatic potential $$V(x, y, z) = -\alpha \ln\dfrac{x^2+z^2}{R^2}$$

for $$x^2+z^2 > R^2$$ (otherwise it's zero), where $$\alpha, R$$ are constants.

It asks me for the Electric field all over the space (done via the gradient), the volumetric charge density all over the space (done via Maxwell $$1$$st), and now it ask me for

the surface over which there is a nonzero surface charge density $$\sigma$$ and it's value.

How can I find the surface?

• @Frobenius There are no efforts, since I did not know where to start... I solved the first two points but I needed a guida for the third one! Jun 7 at 11:54
• @Frobenius So, following the answer below, I first got the gradient of the potential, which reads $$E = -\nabla V = \left\{\frac{2 a x}{x^2+z^2},\frac{2 a z}{x^2+z^2}\right\}$$ Then the divergence is zero... Jun 7 at 11:58
• @Frobenius My bad! $x^2 + z^2 > R^2$. Going to edit. Jun 7 at 12:07

How can I find the surface?

Find where the divergence of the E field, ( or equivalently, the Laplacian of the potential field is non-zero.

$$\nabla^2 V \ne 0$$

By Gauss's Law, this is the only place where there can be charge.

Is there a surface where the Laplacian is discontinuous?

Make sure that you account for the fact that the description of the potential field divides space into two regions, and provides a different formula for each region.

• Following your hint, I calculated the gradient of the Electric Field $$E = -\nabla V = \left\{\frac{2 a x}{x^2+z^2},0, \frac{2 a z}{x^2+z^2}\right\}$$ But then the divergence is zero... OR shall I take the norm, that is $$E = \sqrt{E_x^2 + E_y^2 + E_z^2} = \dfrac{2\alpha^2}{\sqrt{x^2+z^2}}$$ And take the divergence from here? Jun 7 at 12:14
• @selfteachingmaths Aren't there two regions each with a different formula for the electric field? Jun 7 at 12:27

Hint 1 :

(I would prefer to post this as a comment but the site instructs us to avoid extensive use of them so I post as an answer).

To handle the discontinuity a first thought would be to use the Heaviside step or $$\theta$$ function and express the potential by something like that $$\begin{equation} V(x, y, z) \boldsymbol{=-} \alpha \ln\dfrac{x^2+z^2}{R^2}\,\theta\left(\dfrac{x^2+z^2}{R^2}\boldsymbol{-}1\right) \tag{A-01}\label{A-01} \end{equation}$$ Then by differentiation it would appear in our expressions the Dirac $$\delta-$$function. (Sincerely, I don't know if and how this could work.)

Hint 2 :

Consider a right cylinder with axis the $$y-$$axis, height $$h=\Delta y$$ and base radius $$R_{test}\ge R$$. Find the flux of $$\mathbf E$$ (the $$\mathbf E$$ as you found earlier). This flux would be equal to the enclosed charge. How this flux depends on $$R_{test}$$??? Could you interpret the result ??? And what about the flux for $$R_{test}< R$$ ??? After this consider that this cylinder is a $$''$$pipe$$''$$ with internal diameter $$R_{in}=R-\Delta R$$ and external diameter $$R_{ex}=R+\Delta R$$. What is the flux through the wall of this pipe??? How does it depend on $$\Delta R$$??? What in the limit $$\Delta R\rightarrow 0$$ ???

• Cool hint! I will try to go on from here, thanks! Jun 7 at 12:58
• @SelfTeachingMaths : I wish your effort to be successful and if so post it as an answer to your own question. Jun 7 at 13:23