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(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.The total charg

Vedclass · Accepted Answer

$6.40$

Vedclass · Answer

(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.The total charge $Q$ on the disc is:$Q = \int_0^R \sigma(r) \cdot 2\pi r \, dr = \int_0^R \sigma_0 \left(1 - \frac{r}{R}\right) 2\pi r \, dr = 2\pi \sigma_0 \int_0^R \left(r - \frac{r^2}{R}\right) dr = 2\pi \sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R = 2\pi \sigma_0 \left( \frac{R^2}{2} - \frac{R^2}{3} \right) = 2\pi \sigma_0 \left( \frac{R^2}{6} \right) = \frac{\pi \sigma_0 R^2}{3}$.Thus,$\phi_0 = \frac{Q}{\varepsilon_0} = \frac{\pi \sigma_0 R^2}{3\varepsilon_0}$.The charge $q$ enclosed by a concentric spherical surface of radius $r' = \frac{R}{4}$ is the charge on the disc within that radius:$q = \int_0^{R/4} \sigma(r) \cdot 2\pi r \, dr = 2\pi \sigma_0 \int_0^{R/4} \left(r - \frac{r^2}{R}\right) dr = 2\pi \sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^{R/4} = 2\pi \sigma_0 \left( \frac{R^2}{32} - \frac{R^3}{3R \cdot 64} \right) = 2\pi \sigma_0 \left( \frac{R^2}{32} - \frac{R^2}{192} \right) = 2\pi \sigma_0 \left( \frac{6R^2 - R^2}{192} \right) = 2\pi \sigma_0 \left( \frac{5R^2}{192} \right) = \frac{5\pi \sigma_0 R^2}{96}$.Thus,$\phi = \frac{q}{\varepsilon_0} = \frac{5\pi \sigma_0 R^2}{96\varepsilon_0}$.The ratio $\frac{\phi_0}{\phi}$ is:$\frac{\phi_0}{\phi} = \frac{\pi \sigma_0 R^2 / 3\varepsilon_0}{5\pi \sigma_0 R^2 / 96\varepsilon_0} = \frac{1}{3} \cdot \frac{96}{5} = \frac{32}{5} = 6.40$.

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