Ratio of Electric Flux for Concentric Spherical Surfaces Enclosing Charged Disc

According to Gauss's Law, the total electric flux passing through any closed surface is equal to the total electric charge enclosed by that surface divided by the permittivity of free space ($\epsilon_0$):

$\phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}$

Step 1: Calculate the Total Charge on the Disc ($Q_0$)

The large spherical surface completely encloses the disc of radius $R$. The charge on an elemental ring of radius $r$ and width $dr$ is given by:

$dq = \sigma(r) \cdot dA = \sigma_0 \left(1 - \frac{r}{R}\right) (2\pi r \, dr)$

Integrating from $r = 0$ to $r = R$ gives the total charge $Q_0$ on the disc:

$Q_0 = \int_0^R 2\pi \sigma_0 \left(r - \frac{r^2}{R}\right) dr$

$Q_0 = 2\pi \sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R$

$Q_0 = 2\pi \sigma_0 \left( \frac{R^2}{2} - \frac{R^2}{3} \right) = 2\pi \sigma_0 \left(\frac{R^2}{6}\right) = \frac{\pi \sigma_0 R^2}{3}$

Therefore, the electric flux through the large spherical surface is:

$\phi_0 = \frac{Q_0}{\epsilon_0} = \frac{\pi \sigma_0 R^2}{3\epsilon_0}$

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Step 2: Calculate the Charge Enclosed by the Smaller Sphere ($Q$)

The smaller concentric spherical surface has a radius of $r_s = \frac{R}{4}$. Since it is concentric with the disc, it encloses the portion of the disc lying within $r = 0$ to $r = \frac{R}{4}$.

The charge $Q$ enclosed by this sphere is:

$Q = \int_0^{R/4} 2\pi \sigma_0 \left(r - \frac{r^2}{R}\right) dr$

$Q = 2\pi \sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^{R/4}$

Substituting the upper limit $r = \frac{R}{4}$:

$Q = 2\pi \sigma_0 \left( \frac{(R/4)^2}{2} - \frac{(R/4)^3}{3R} \right)$

$Q = 2\pi \sigma_0 \left( \frac{R^2}{32} - \frac{R^2}{192} \right)$

$Q = 2\pi \sigma_0 R^2 \left( \frac{6 - 1}{192} \right) = 2\pi \sigma_0 R^2 \left( \frac{5}{192} \right) = \frac{5\pi \sigma_0 R^2}{96}$

Therefore, the electric flux through the smaller spherical surface is:

$\phi = \frac{Q}{\epsilon_0} = \frac{5\pi \sigma_0 R^2}{96\epsilon_0}$

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Step 3: Calculate the Ratio $\frac{\phi_0}{\phi}$

$\frac{\phi_0}{\phi} = \frac{Q_0}{Q} = \frac{\frac{\pi \sigma_0 R^2}{3}}{\frac{5\pi \sigma_0 R^2}{96}} = \frac{1}{3} \times \frac{96}{5} = \frac{32}{5} = 6.4$