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 ${\varphi }_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\varphi$. Then the ratio $\frac{{\varphi }_0}{\varphi }$ is ______

Answer 1

Answer: 6.40

$dQ=\left({\sigma }_0\right)\left(1-\frac{x}{R}\right)(2\pi x)dx$
$Q_{\frac{R}{4}}=\left(2\pi {\sigma }_0\right)\underset{0}{\overset{R/4}{\int }}\left(x-\frac{x^2}{R}\right)dx$
$=\left(2\pi {\sigma }_0\right)\left[\frac{5R^2}{192}\right]$
$Q_0=\left(2\pi {\sigma }_0\right)\underset{0}{\overset{R}{\int }}\left(x-\frac{x^2}{R}\right)dx=(2\pi \Delta )\frac{R^2}{6}$
$\varphi =\frac{q_{\text{in }}}{{\epsilon }_0}$
$\Rightarrow \frac{{\varphi }_0}{\varphi }=\frac{1\times 192}{(6)5}=6.40$