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 $E = \rho_s/2\xi_0$ :
$$ V_{ab} = \int_{0}^{d} E dz = \frac{d}{2\xi_0}(\rho_{sa}-\rho_{sb}) $$
which gives the correct answer.
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.
Best Answer
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.