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 ${\varphi }_0.$ Now consider a spherical surface of radius $2R$ and one of the diameter along axis of cylinder and electric flux through spherical surface is $\varphi .$ Find $10\frac{\varphi }{{\varphi }_0}.$

Answer 1

Answer: 18

${\varphi }_0=\frac{\rho \pi R^2\cdot 2R}{{ϵ}_0}$
${\varphi }_0=\frac{2\rho \pi R^3}{{ϵ}_0}$

$V_1=\underset{0}{\overset{\pi /6}{\int }}\pi 4R^2{\sin }^2\theta 2R\sin \theta d\theta$
$=\frac{8\pi R^3}{4}\underset{0}{\overset{\pi /6}{\int }}(3\sin \theta -\sin 3\theta )d\theta$
$=\frac{8\pi R^3}{4}\left(\frac{8}{3}-\frac{3\sqrt{3}}{2}\right)$
Total volume $V=\left(V_1+V_2\right)2$
$=\frac{\pi R^3}{2}\left(\frac{8}{3}-\frac{3\sqrt{3}}{2}\right)+\pi R^{2}2\sqrt{3}R$
$=\pi R^3\left[2\sqrt{3}-\frac{3\sqrt{3}}{4}+\frac{4}{3}\right]$
$\left(\frac{15\sqrt{3}+16}{12}\right)\pi R^3$
$10\frac{\varphi }{{\left(\varphi \right)}_0}=10\left(\frac{15\sqrt{3}+16}{24}\right)\approx 18$