# Given the potential between two infinite parallel plates, how to find charge densities on the plates?

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?

$$\mathbf{E} = -\frac{\partial{\phi}}{{\partial z}}\, \hat{z} = -\frac{\phi_0}{a}\, \hat{z}.$$
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, $$|\sigma| = \frac{\epsilon_0 \, \phi_0}{a}.$$
• 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. Commented Feb 15, 2022 at 6:47
• 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. Commented Feb 15, 2022 at 7:55