Question

A circular disc of radius R carries surface charge density $$\sigma(r)=\sigma_0\left(1-\dfrac{r}{R}\right)$$ 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 $$R/4$$ and concentric with the disc is $$\Phi$$. Then the ratio $$\dfrac{\Phi_0}{\Phi}$$ is________.

Answer 1

$$\phi_0 = \dfrac{\int dq}{\varepsilon_0} = \dfrac{\displaystyle \int_0^R \sigma_0 \left(1 - \dfrac{r}{R}\right) 2 \pi r \ dr}{\varepsilon_0}$$

$$\phi_0 = \dfrac{\int dq}{\varepsilon_0} = \dfrac{\displaystyle \int_0^{\pi/4} \sigma_0 \left(1 - \dfrac{r}{R}\right) 2 \pi r \ dr}{\varepsilon_0}$$

$$\therefore \dfrac{\phi_0}{\phi} = \dfrac{\sigma_0 2 \pi \displaystyle \int_0^R \left(r - \dfrac{r^2}{R} \right) dr}{\sigma_0 2\pi \int_0^{R/4} \left(r - \dfrac{r^2}{R}\right) dr}$$

$$\therefore \dfrac{\phi_0}{\phi}=\dfrac{\dfrac{R^2}{2} - \dfrac{R^2}{3}}{\dfrac{R^2 }{32} - \dfrac{R^2}{3 \times 64} }$$

$$\therefore \dfrac{\phi_0}{\phi}= \dfrac{32}{5}$$

$$\therefore \dfrac{\phi_0}{\phi}= 6.40$$