Beauty Of Flux in Physics

The picture is not clear, but the plate $P$ is the region of the $xz$ -plane defined by $0 \le z \le \ell$ , $x \ge z$ . The flux $\Phi$ through the plate is then $\Phi \; = \; \iint_P \frac{q}{4\pi\varepsilon_0} \frac{\mathbf{r}}{r^3}\cdot d\mathbf{S} \; = \; \iint_P \frac{q}{4\pi\varepsilon_0} \frac{\mathbf{r}}{r^3}\cdot (-\mathbf{k})dA \; = \; \frac{q\ell}{4\pi \varepsilon_0}\iint_P \frac{dx\,dz}{(x^2 + z^2 + \ell^2)^{\frac32}} \; = \; \frac{q}{4\pi \varepsilon_0} \int_0^1 \left(\int_Z^\infty \frac{dX}{(X^2 + Z^2 + 1)^{\frac32}}\right)\,dZ$ making the substitution $x = \ell X$ , $z = \ell Z$ . The substitution $X = \sqrt{Z^2 + 1}\sinh u$ now gives $\begin{aligned} \Phi & = \; \frac{q}{4\pi \varepsilon_0}\int_0^1 \left(\int_{\sinh^{-1}\frac{Z}{\sqrt{Z^2+1}}}^\infty \frac{\mathrm{sech}^2u}{Z^2+1}\,du\right)\,dZ \; = \; \frac{q}{4\pi \varepsilon_0} \int_0^1 \frac{dZ}{Z^2+1}\Big[\tanh u\Big]_{\sinh^{-1}\frac{Z}{\sqrt{Z^2+1}}}^\infty \\ & = \; \frac{q}{4\pi \varepsilon_0} \int_0^1 \frac{1}{Z^2+1}\left(1 - \frac{Z}{\sqrt{2Z^2 + 1}}\right)\,dZ \; =\; \frac{q}{4\pi \varepsilon_0} \left(\tfrac14\pi - \int_0^1 \frac{Z\,dZ}{(Z^2+1)\sqrt{2Z^2+1}}\right) \\ & =\; \frac{q}{4\pi \varepsilon_0} \left(\tfrac14\pi - \tfrac12\int_0^1 \frac{dw}{(w+1)\sqrt{2w+1}}\right) \end{aligned}$ Finally the substitution $2w+1=p^2$ gives $\Phi \; = \; \frac{q}{4\pi \varepsilon_0} \left(\tfrac14\pi - \int_1^{\sqrt{3}}\frac{dp}{p^2+1}\right) \; = \; \frac{q}{4\pi \varepsilon_0} \left(\tfrac14\pi - \big(\tfrac13\pi - \tfrac14\pi\big)\right) \; = \; \frac{q}{24\varepsilon_0}$ making the answer $\boxed{24}$ .