Triangular Electric Field

This is just 1/4 of flux of a tetrahedron (due to guess law). So sigma/4€o

The electric field at a point $P$ with position vector $\mathbf{r}$ is $\mathbf{E} \; = \; \frac{\sigma}{4\pi \varepsilon_0} \iint_T \frac{\mathbf{r} - \mathbf{x}}{|\mathbf{r} - \mathbf{x}|^3}\,dA(\mathbf{x})$ where the integral is over the triangle $T$ . By symmetry, the electrc field at the particular point $P$ will be normal to the triangle, and so we calculate $\mathbf{E} \; = \; \frac{\sigma}{4\pi\varepsilon_0} CP \iint_T \frac{dA(\mathbf{x})}{|\mathbf{r}-\mathbf{x}|^3}\mathbf{k} \; = \; \frac{\sigma}{4\pi\varepsilon_0}\iint_T \frac{\cos\theta\,dA(\mathbf{x})}{|\mathbf{r}-\mathbf{x}|^2}\mathbf{k} \;= \; \frac{\sigma}{4\pi\varepsilon_0}\iint_T d\Omega(\mathbf{x})\mathbf{k} \; = \; \frac{\sigma\Omega}{4\pi\varepsilon_0}\mathbf{k}$ where $C$ is the centroid of the triangle, $\Omega$ is the solid angle subtended by the triangle $T$ at the point $P$ and $\mathbf{k}$ is a unit vector in the direction of $PC$ . Now $P$ is at the centroid of a regular tetrahedron of which $T$ is one of the faces. Thus, by symmetry, $\Omega = \tfrac14 \times 4\pi = \pi$ , and hence $\mathbf{E} \; = \; \frac{\sigma}{4\varepsilon_0}\mathbf{k}$ and, with the given values of $\sigma$ and $\varepsilon_0$ , we have $|\mathbf{E}| = \boxed{28.2486}$ .