Question

A circular disc of radius R carries surface charge density $\sigma \left(r\right)={\sigma }_0\left(1-\frac{r}{R}\right)$, 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 $R/4$ and concentric with the disc is $\varphi$. Then the ratio $\frac{{\varphi }_0}{\varphi }$ is _________.

Answer 1

Step 1: Given data

Radius of the circular disc,$=R$

Surface charge density, $\sigma \left(r\right)={\sigma }_0\left(1-\frac{r}{R}\right)$

Electric flux$={\varphi }_0$

Radius of another spherical surface $=R/4$

Step 2: To find the ratio of $\frac{{\varphi }_0}{\varphi }$

$$\begin{gathered} {\varphi }_0=\frac{\int dq}{{\epsilon }_0} \\ =\frac{{\int }_0^R{\sigma }_0\left(1-{\displaystyle \frac{r}{R}}\right)2\pi rdr}{{\epsilon }_0} \\ R=\frac{\pi }{4} \\ {\varphi }_0=\frac{{\int }_0^{{\displaystyle \frac{\pi }{4}}}{\sigma }_0\left(1-{\displaystyle \frac{r}{R}}\right)2\pi rdr}{{\epsilon }_0} \end{gathered}$$

Therefore

$$\begin{gathered} \frac{{\varphi }_0}{\varphi }=\frac{{\sigma }_{0}2\pi {\int }_0^R\left(r-{\displaystyle \frac{r^2}{R}}\right)dr}{{\sigma }_{0}2\pi {\int }_0^{{\displaystyle \frac{R}{4}}}\left(r-{\displaystyle \frac{r^2}{R}}\right)dr} \\ \frac{{\varphi }_0}{\varphi }=\frac{{\displaystyle \frac{R^2}{2}}-{\displaystyle \frac{R^2}{3}}}{{\displaystyle \frac{R^2}{32}}-{\displaystyle \frac{R^2}{3\times 64}}} \\ \frac{{\varphi }_0}{\varphi }=\frac{32}{5}=6.40 \end{gathered}$$

The ratio of $\frac{{\varphi }_0}{\varphi }=6.4$

Hence the correct answer is $6.4$