0
$\begingroup$

Consider two parallel infinite metal plates, one at $z=0$ and the other at $z=a$, maintained at potentials $\phi=0$ and $\phi=\phi_0$ respectively. The potential in between the plates and also on the plates is given by the expression $$\phi(z)=\phi_0\frac{z}{a}.$$ The electric field between the plates is then given by $$\vec{E}=-\frac{d\phi}{dz}\hat{z}=-\frac{\phi_0}{a}\hat{z}.$$ If we use Poisson's equation in the next step $d^2\phi/dz^2=-\rho/\epsilon_0$, the expression written above gives $\rho=0$ (which is charge density in between the plates). But I have a feeling that this formula also contains the information that the charges on the plates are nonzero. How can we find the surface charge densities $\sigma$ on the plates?

$\endgroup$

1 Answer 1

1
$\begingroup$

Not sure if you need to solve Poisson's equation for this. Couldn't you simply take the gradient of the potential to give you the field?

$$\mathbf{E} = -\frac{\partial{\phi}}{{\partial z}}\, \hat{z} = -\frac{\phi_0}{a}\, \hat{z}.$$

Then, you can use the expression for the field between the plates to solve for $\sigma$, the charge density on the positive plate:

$$\mathbf{E} = \frac{\sigma}{\epsilon_0}\, \hat{z} = -\frac{\phi_0}{a} \, \hat{z};$$ so, $$|\sigma| = \frac{\epsilon_0 \, \phi_0}{a}.$$

$\endgroup$
2
  • $\begingroup$ nice, Thanks! But can we get this using Poisson's equation? I think we can do this by converting the surface charge $\sigma$ to a corresponding volume charge $\rho$ density using an appropriate delta function. It may be tedious but should in principle be doable. $\endgroup$ Commented Feb 15, 2022 at 6:47
  • $\begingroup$ Hmm, I don't think so. You already have the equation for $\phi$, which is the whole point of solving Poisson's equation. To solve for the charge density on the plates, you must use an independent relation. $\endgroup$
    – Yejus
    Commented Feb 15, 2022 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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