Can gauss save us?

$E=\frac{1}{4\pi \epsilon_{0}}\displaystyle \int_{0}^{2\pi} \int_{0}^{R} \frac{b\cdot r \cdot r \cdot d\theta \cdot dr \cdot x}{(r^1+1)^{\frac{3}{2}}}$ $E=\frac{b}{4\pi \epsilon_{0}}\displaystyle \int_{0}^{2\pi}d\theta \int_{0}^{R} \frac{r^2}{(r^2+1)^{\frac{3}{2}}}dr$ $E=10^{-9}\cdot 9\cdot 10^9 \cdot 2\pi \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{sin^2(\alpha)}{cos(\alpha)}d\alpha$ $E=19\pi \cdot (\displaystyle \int_{0}^{\frac{\pi}{4}} sec(\alpha)d\alpha - \displaystyle \int_{0}^{\frac{\pi}{4}} cos(\alpha)d\alpha)$ $E=18\pi(ln(\sqrt{2}+1)-\frac{1}{\sqrt{2}}) \approx 10$

2 $\times k \times \pi \times \int_{0}^{x} \frac{r \times br}{(r^{2}+x^{2})^{3/2}} dr$ $= 10$ at $x=1$

The electric field due to charged ring is

$\large E = \frac{qz}{4\pi\epsilon_0\cdot (z^2+r^2)^{\frac{3}{2}}}$

So for a disk we have to integrate it over radius R

$\large dE = \frac{z\sigma 2\pi r dr}{4\pi\epsilon_0\cdot (z^2+r^2)^{\frac{3}{2}}}$

$\large E=\int^R_0 dE = \int^R_0\frac{zbr \cdot2\pi r dr}{4\pi\epsilon_0\cdot (z^2+r^2)^{\frac{3}{2}}}$

$\large E=\frac{zb\cdot 2\pi}{ 4\pi\epsilon_0}\int^R_0\frac{r^2 dr}{ (z^2+r^2)^{\frac{3}{2}}}$

Taking $R=1,$ $\dfrac{1}{4 \pi \epsilon_{0}} = 9\times 10^{9}$ , $b = 10^{-9}$ and $z=1$ we get

$\large E=18{\pi}\int^1_0 \frac{r^2dr}{(r^2+1)^{\frac{3}{2}}}$

So the answer is $\boxed {10}$

thumbs up to u bro.. great problem!!!!