A square conducting plate of side length $2 a$, centered about the origin in the xy-plane, is charged with a uniform surface charge density $\sigma$.
(a) Prove that following integral where $z$ is independent of $x$ and $y$ :
$$
\int_0^a \int_0^a \frac{d x d y}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}=\frac{1}{z} \tan ^{-1} \frac{a^2}{z \sqrt{2 a^2+z^2}} .
$$
(b) Determine the electric field $\boldsymbol{E}$ at $(0,0, z)$ due to this surface charge distribution. Find the limits where $z \rightarrow 0^{+}$and $z \rightarrow 0^{-}$and explain why they make sense.
Now, in addition to the previous charged square plate, there is another square plate of the same size, parallel to the xy-plane and centered at $(0,0, d)$. This additional plate is uniformly charged with a surface charge density $-\sigma$.
(c) Determine the electric field at $(0,0, z)$ for all $z$ due to the uniform charge distributions on both plates (we assume, albeit incorrectly, that they stay uniform in the presence of each other).
(d) Assuming $d<<a$, find the asymptotic solution to the previous field for all $z$. As an aside, the charge distributions on both plates indeed remain uniform in this limit as the plates are effectively infinitely large as compared to the separation between them.
(e) Based on your previous answer, determine the potential difference $V$ between the two plates. The capacitance of the two plates is defined as
$$
C=\left|\frac{Q}{V}\right|
$$
where $Q$ is the total charge on either plate. Find $C$ and state the variables it depends on.