# Electric potential between two parallel plates

I'm having a kind of "hesitation" for a relatively simple electric potential problem found in Demarest.

Here is it :

I first solved it by finding the electric field between the two plates and multiplying it by the distance d. Recalling an infinite plane gives $E = \rho_s/2\xi_0$ :

$$V_{ab} = \int_{0}^{d} E dz = \frac{d}{2\xi_0}(\rho_{sa}-\rho_{sb})$$

But I'd like also to find it simply by calculating the two potentials $V_a$ and $V_b$ idependently and then calculating $V_{ab} = V_a - V_b$. And this is where I seem to always get the wrong signs in my answer.

1) gives a representation of the problem...

2) Ignoring the upper plate : $V_a = \frac{\rho_{sb}}{2\xi_0} \cdot d$

3) Ignoring the lower plate : $V_b = \frac{\rho_{sa}}{2\xi_0} \cdot d$

4) $V_{ab} = V_a - V_b = \frac{d}{2\xi_0}(\rho_{sb}-\rho_{sa})$

I have difficulty understanding why I get the correct answer, but with the sign inversed. I also tried calculating $V_{ab}$ independently and then adding them up, I get the wrong answer again... Is my approach even correct? i.e. calculating the potential $V_b$ in regards to $E_a$ and vice-versa?

How can you solve this by calculating both potentials and adding them up using superposition? It might be an easy problem, it might be details, but I wish to be sure. If anyone could explain it so I can understand correctly, I would be very grateful.

The electric field diagram relating to this problem looks like this and I have included a unit z-direction vector$\hat z$:

What you did first was find $V_{\rm ab}$, the potential of plate $a$ relative to plate $B$ which you called $V_{\rm ab}$.

$\displaystyle V_{\rm ab} = - \int^a _b \vec E \cdot d\vec z = - \int^d _0 \left (\dfrac {\sigma_{\rm b}}{2\epsilon_0}- \dfrac {\sigma_{\rm a}}{2\epsilon_0} \right )dz = \dfrac {d}{2\epsilon_0}(\sigma_{\rm a} - \sigma_{\rm b})$

where $d$ is the separation of the plates and this is in agreement with your answer.

Now looking at the method of superposition.

The potential of plate $a$ relative to the potential of plate $b$ due to the charge on plate $b$ alone:

$\displaystyle V'_{\rm ab} = - \int^a _b \vec E_{\rm b} \cdot d\vec z = - \int^d _0 \left (\dfrac {+\sigma_{\rm b}}{2\epsilon_0}\right )dz = -\dfrac {d}{2\epsilon_0} \sigma_{\rm b}$

The potential of plate $a$ relative to the potential of plate $b$ due to the charge on plate $a$ alone:

$\displaystyle V''_{\rm ab} = - \int^a _b \vec E_{\rm a} \cdot d\vec z = - \int^d _0 \left (\dfrac {-\sigma_{\rm a}}{2\epsilon_0}\right )dz = +\dfrac {d}{2\epsilon_0} \sigma_{\rm a}$

$V_{\rm ab} = V'_{\rm ab} + V''_{\rm ab} = \dfrac {d}{2\epsilon_0}(\sigma_{\rm a} - \sigma_{\rm b})$ as before.

Your lack of definition of $V_{\rm a}$ and $V_{\rm b}$ means that one does no know what your reference potential was.