Question

A circular disc of radius $R$ carries surface charge density $\sigma(r)=\sigma_{0}\left(1-\frac{r}{R}\right)$, where $\sigma_{0}$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_{0}$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_{0}}{\phi}$ is ______

Answer 1

$d Q=\left(\sigma_{0}\right)\left(1-\frac{x}{R}\right)(2 \pi x) d x$
$Q_{\frac{R}{4}}=\left(2 \pi \sigma_{0}\right) \int\limits_{0}^{R / 4}\left(x-\frac{x^{2}}{R}\right) d x$
$=\left(2 \pi \sigma_{0}\right)\left[\frac{5 R^{2}}{192}\right]$
$Q_{0}=\left(2 \pi \sigma_{0}\right) \int\limits_{0}^{R}\left(x-\frac{x^{2}}{R}\right) d x=(2 \pi \Delta) \frac{R^{2}}{6}$
$\phi=\frac{q_{\text {in }}}{\varepsilon_{0}}$
$ \Rightarrow \frac{\phi_{0}}{\phi}=\frac{1 \times 192}{(6) 5}=6.40$