0
$\begingroup$

Suppose I am given a charge density $\rho(x)$. Poisson's equation states that

$$\frac{d^2\phi}{dx^2} = -\frac{\rho}{\epsilon}$$

Is there a simple way to see what the characteristic strength of the electrostatic potential should be from a dimensional analysis?

$\endgroup$
  • 1
    $\begingroup$ I think first you'll have to define what you mean by the "characteristic strength" of the potential. AFAIK that isn't a term that's widely used in electrostatics. $\endgroup$ – The Photon Feb 18 '17 at 16:56
  • $\begingroup$ By this I simply mean: What will its order of magnitude be. $\endgroup$ – user13514 Feb 18 '17 at 23:37
0
$\begingroup$

No, you can't.

Consider as one example a parallel plate capacitor. The capacitors constitutive relationship can be rearranged to get

$$V=\frac{Q}{C}$$

The capacitance is, of course, given by the geometry. Ignoring fringing effects, $C=\varepsilon A/d$. So, we have

$$V=\frac{Qd}{\varepsilon A}$$

That is, if charges $+Q$ and $-Q$ are spread in sheets of area $A$ and separated by distance $d$, there must be a potential difference between the two charged regions that depends not just on the amount of charge, but also on the distance between them, the size of the areas they're spread over, and the properties of the material between them.

$\endgroup$

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.