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 $\phi_{0}.$ Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi.$ Then the ratio is $\frac{\phi_{0}}{\phi}=\frac{x}{5}$ , the value of $x$ is___

Answer 1

$\phi_{0}=\frac{\int dq }{\varepsilon_{0}}=\frac{\int_{0}^{ R } \sigma_{0}\left(1-\frac{ r }{ R }\right) 2 \pi rdr }{\varepsilon_{0}}$
$\phi=\frac{\int dq }{\varepsilon_{0}}=\frac{\int_{0}^{ R / 4} \sigma_{0}\left(1-\frac{ r }{ R }\right) 2 \pi rdr }{\varepsilon_{0}}$
$\therefore \frac{\phi_{0}}{\phi}=\frac{\sigma_{0} 2 \pi \int_{0}^{ R }\left( r -\frac{ r ^{2}}{ R }\right) dr }{\sigma_{0} 2 \pi \int_{0}^{ R / 4}\left( r -\frac{ r ^{2}}{ R }\right) dr }$
$=\frac{\frac{ R ^{2}}{2}-\frac{ R ^{2}}{3}}{\frac{ R ^{2}}{32}-\frac{ R ^{2}}{3 \times 64}}=\frac{32}{5}=6.40$