Demonstration that the electric potential is continuous over a surface

[fluidistic]

Homework Statement

I'm asked to demonstrate that the electric potential is continuous over a surface with a charge density $\sigma$.

Homework Equations

$\Phi (\vec x )= \int _ S \frac{\sigma (\vec x' )}{|\vec x - \vec x '|}da'$

The Attempt at a Solution

I'm not sure what I must show mathematically. Is it that $\Phi (\vec x )= \int _ S \frac{\sigma (\vec x' )}{|\vec x - \vec x '|}da'$ remains integrable when x'=x? Or is it that (I believe it is), when I take the limit of when x tends to x', Phi must equate $\Phi (\vec x')$? Either way I haven't figured out how to do it.
Any tip is welcome. Thanks!

 

[mark726]

Hello there,

To demonstrate that the electric potential is continuous over a surface with a charge density $\sigma$, you can use the definition of continuity. In this case, we want to show that the electric potential at a point on the surface, say $\vec x$, is equal to the limit of the electric potential as we approach that point from any direction. In other words, as we move closer and closer to $\vec x$, the electric potential should not change abruptly.

Using the equation you provided, $\Phi (\vec x )= \int _ S \frac{\sigma (\vec x' )}{|\vec x - \vec x '|}da'$, we can see that the electric potential at $\vec x$ depends on the charge density at all points on the surface $S$. As we approach $\vec x$, the distance between the point of interest and the charge density at each point on the surface will decrease, but the charge density itself will remain constant. This means that the integral will remain finite and the electric potential will not change abruptly. Therefore, the electric potential is continuous over the surface with charge density $\sigma$.

I hope this helps. Let me know if you have any further questions. Good luck with your demonstration!