Question

Consider a uniformly charged solid cylinder of large length and radius $R.$ Now consider a cylindrical surface of radius $2R$ and length $2R$ coaxial with cylinder and electric flux through cylindrical surface is $\phi_{0}.$ Now consider a spherical surface of radius $2R$ and one of the diameter along axis of cylinder and electric flux through spherical surface is $\phi.$ Find $10\frac{\phi}{\phi_{0}}.$

Answer 1

$\phi_{0}=\frac{\rho \pi R^{2} \cdot 2 R}{\epsilon _{0}}$
$\phi_{0}=\frac{2 \rho \pi R^{3}}{\epsilon _{0}}$

$V_{1}=\int\limits _{0}^{\pi / 6}\pi 4R^{2}\sin^{2}\theta 2R\sin\theta d\theta $
$=\frac{8 \pi R^{3}}{4}\int\limits _{0}^{\pi / 6}\left(\right.3\sin\theta -\sin 3\theta \left.\right)d\theta $
$=\frac{8 \pi R^{3}}{4}\left(\frac{8}{3} - \frac{3 \sqrt{3}}{2}\right)$
Total volume $V=\left(V_{1} + V_{2}\right)2$
$=\frac{\pi R^{3}}{2}\left(\frac{8}{3} - \frac{3 \sqrt{3}}{2}\right)+\pi R^{2}2\sqrt{3}R$
$=\pi R^{3}\left[2 \sqrt{3} - \frac{3 \sqrt{3}}{4} + \frac{4}{3}\right]$
$\left(\frac{15 \sqrt{3} + 16}{12}\right)\pi R^{3}$
$10\frac{\phi}{\left(\phi\right)_{0}}=10\left(\frac{15 \sqrt{3} + 16}{24}\right)\approx18$