Hemisphere E-Field (Quantitative)

Note that $\begin{aligned} \int_0^\pi \frac{\sin\theta(\cos\theta - \frac34)}{(\frac{25}{16} - \frac32\cos\theta)^{\frac32}}\,d\theta & = \; 16\int_0^\pi \frac{\sin\theta(4\cos\theta-3)}{(25-24\cos\theta)^{\frac32}}\,d\theta \; = \; 16\int_{-1}^1 \frac{4x-3}{(25 - 24x)^{\frac32}}\,dx \\ & = \; \tfrac83\int_{-1}^1 \left[ \frac{7}{(25 - 24x)^{\frac32}} - \frac{1}{(25 - 24x)^{\frac12}}\right]\,dx \; = \; \tfrac29\left[ \frac{7}{\sqrt{25-24x}} + \sqrt{25 - 24x}\right]_{-1}^1 \\ & = \; 0 \end{aligned}$ With a suitable system of polar coordinates, we have $\begin{aligned} \mathbf{E}_1 & = \; \int_0^\pi \,d\theta \int_0^\pi d\phi \frac{\sigma R^2 \sin\theta}{4\pi \varepsilon_0} \frac{1}{R^2} \left(\begin{array}{c} \sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta\end{array}\right) \; = \; \frac{\sigma}{4\pi \varepsilon_0} \int_0^\pi \left(\begin{array}{c} 0 \\ 2\sin\theta \\ \pi \cos\theta \end{array} \right) \sin\theta\,d\theta \; = \; \frac{\sigma}{4\pi \varepsilon_0} \left[ \left( \begin{array}{c} 0 \\ \theta - \tfrac12\sin2\theta \\ \tfrac12\pi\sin^2\theta \end{array}\right)\right]_0^\pi \; = \; \frac{\sigma}{4\varepsilon_0}\mathbf{j} \end{aligned}$ On the other hand $\begin{aligned} \mathbf{E}_2 & = \; \int_0^\pi \,d\theta \int_0^\pi d\phi \frac{\sigma R^2 \sin\theta}{4\pi \varepsilon_0} \frac{1}{R^2(\frac{25}{16} - \frac32\cos\theta)^{\frac32}} \left(\begin{array}{c} \sin\theta\cos\phi \\ \sin\theta\sin\phi \\ \cos\theta-\tfrac34\end{array}\right) \; = \; \frac{\sigma}{4\pi \varepsilon_0} \int_0^\pi \frac{\sin\theta}{(\frac{25}{16} - \frac32\cos\theta)^{\frac32}}\left(\begin{array}{c} 0 \\ 2\sin\theta \\ 2\pi(\cos\theta-\tfrac34) \end{array} \right) \,d\theta \\ & = \; \frac{\sigma}{4\pi \varepsilon_0}\int_0^\pi \frac{2\sin^2\theta}{(\frac{25}{16} - \frac32\cos\theta)^{\frac32}}\,d\theta \mathbf{j} \; = \; \frac{128}{\pi}\int_0^\pi \frac{\sin^2\theta\,d\theta}{(25 - 24\cos\theta)^{\frac32}} \mathbf{E}_1 \end{aligned}$ so that $\alpha \; = \; \frac{128}{\pi}\int_0^\pi \frac{\sin^2\theta\,d\theta}{(25 - 24\cos\theta)^{\frac32}}$ This integral can be evaluated in terms of elliptic functions, obtaining $\alpha \; = \; \frac{8}{9\pi}\big(25K(-48) - E(-48)\big) \; = \; \boxed{1.34119}$ It is worth noting that a generalisation of these arguments shows that the electric field is $\mathbf{E}(u) \; = \; \frac{\sigma}{2\pi\varepsilon_0}\left\{ \frac{1+u^2}{u^2(1+u)}K\left(\frac{4u}{(1+u)^2}\right) - \frac{1+u}{u}E\left(\frac{4u}{(1+u)^2}\right)\right\} \mathbf{j}$ at a point on the $xy$ -plane a distance $uR$ from the origin. The singularity in this expression at $u=0$ is removable.