Question 851: Electrostatics & Capacitance

Question

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

Accepted Answer

To determine the discÃ¢ÂÂs total charge, integrate the radial surfaceÃ¢ÂÂcharge distribution across its entire area:
$ \begin{aligned} q_{\text{enc}} &= \int_{0}^{a} \sigma(r)\, 2\pi r\,dr                 = 2\pi \sigma_{0}\int_{0}^{a}\!\left(r - \frac{r^{2}}{a}\right)dr \[4pt]                &= 2\pi \sigma_{0}\!\left[\frac{r^{2}}{2} - \frac{r^{3}}{3a}\right]_{0}^{a}                = \frac{\pi \sigma_{0}a^{2}}{3}. \end{aligned} $
The electric flux through the spherical surface that encloses the entire charged disc is
$$ \phi_0=\frac{q_{\mathrm{enc} 0}}{\epsilon_0}=\frac{\pi \sigma_0 a^2}{3 \epsilon_0} . $$
Next, we find the flux through the smaller spherical surface of radius $a / 4$, which encloses a portion of the disc. The charge on this portion of the disc is
$$ q_{\mathrm{enc}}=2 \pi \sigma_0 \int\limits_0^{a / 4}\left(r-\frac{r^2}{a}\right) \mathrm{d} r=\frac{5}{96} \pi \sigma_0 a^2 $$
The flux through this spherical surface is
$$ \phi=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{5 \pi \sigma_0 a^2}{96 \epsilon_0} $$
The ratio of the two fluxes is
$$ \frac{\phi_0}{\phi}=\frac{32}{5}=6.4 $$

Electrostatics & Capacitance

Integer / Numeric Answer Type