Question

A circular disc of radius 𝑅 carries surface charge density $\sigma \left(r\right)={\sigma }_o(1-\frac{r}{R})$,where ${\sigma }_o$ 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

Let us choose an elementary ring of radius $r$ and thickness $dr$ as shown in the above figure .

So , the area of elementary ring $dA$ $=$ $2$$\pi$$rdr$

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