Question

A circular disc of radius $R$ carries surface charge density $\sigma \left(r\right)={\left(\sigma \right)}_0\left(1-\frac{r}{R}\right),$ 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 ${\varphi }_0.$ Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\varphi .$ Then the ratio is $\frac{{\varphi }_0}{\varphi }=\frac{x}{5}$ , the value of $x$ is___

Answer 1

Answer: 32

${\varphi }_0=\frac{\int dq}{{\epsilon }_0}=\frac{{\int }_0^R{\sigma }_0\left(1-\frac{r}{R}\right)2\pi rdr}{{\epsilon }_0}$
$\varphi =\frac{\int dq}{{\epsilon }_0}=\frac{{\int }_0^{R/4}{\sigma }_0\left(1-\frac{r}{R}\right)2\pi rdr}{{\epsilon }_0}$
$\therefore \frac{{\varphi }_0}{\varphi }=\frac{{\sigma }_{0}2\pi {\int }_0^R\left(r-\frac{r^2}{R}\right)dr}{{\sigma }_{0}2\pi {\int }_0^{R/4}\left(r-\frac{r^2}{R}\right)dr}$
$=\frac{\frac{R^2}{2}-\frac{R^2}{3}}{\frac{R^2}{32}-\frac{R^2}{3\times 64}}=\frac{32}{5}=6.40$