A Hard Kind Of Ring

At first we'll take an infinitesimal part of the toroid, which is a disc with radious $r$ , which center is at a distance $R$ from the origin and a distance $\sqrt{R^2+L^2}$ from the point $P$ . By Gauss' Law we have that: $\displaystyle \oint E_{dE} ds=\frac{Q_{enlosed}}{\epsilon_{0}}$ $E_{dE}\cdot 2\pi \sqrt{R^2+L^2}=\frac{\rho\pi r^2}{\epsilon_{0}}$ $\Rightarrow \vec{E}_{dE}=\frac{\rho r^2}{2\epsilon_{0} \sqrt{R^2+L^2}}\cdot \frac{(-R \cos \theta , -R \sin \theta , L)}{\sqrt{R^2+L^2}}$ $\vec{E}_{dE}=\frac{\rho r^2 (-R\cos\theta,-R\sin\theta,L)}{2\epsilon_{0}(R^2+L^2)}$ Note that this can be done only because $L\leq R$ . As we have an infinitesimal part of the total electric field at $P$ , we have: $\vec{E}=\displaystyle \int_{0}^{2\pi}\vec{E}_{dE}d\theta$ $\vec{E}=\displaystyle \int_{0}^{2\pi}\frac{\rho r^2 (-R\cos\theta,-R\sin\theta,L)d\theta}{2\epsilon_{0}(R^2+L^2)}$ By the geometry of the electric field we know that the components in $x$ and $y$ will be cancelled at summing up them: $\vec{E}=\displaystyle \int_{0}^{2\pi}\frac{\rho r^2 L d\theta}{2\epsilon_{0}(R^2+L^2)}$ $\vec{E}=\frac{\rho r^2 L \pi}{\epsilon_{0}(R^2+L^2)}$ As $R=10[m]$ , $r=2[m]$ , $\rho=125[\frac{C}{m^3}]$ and $L=5$ : $\vec{E}=\frac{20\pi}{\epsilon_{0}}$