Electric Flux (4-4-2021)

Let a point on the rod be:

$\vec{r}_c = (0,0,z)$

At the vicinity of this general point on the rod, consider an elementary rod of length $dz \ \hat{r}_c$ . The charge on this elementary rod is $dQ = dz$ .

A point on the sphere:

$\vec{r}_p = (R\sin{\theta}\cos{\phi},R\sin{\theta}\sin{\phi},R\cos{\theta})$ $R = 2$

The electric field at a point on the sphere due to the elementary rod is:

$d\vec{E} = \frac{dQ}{4 \pi \epsilon_o} \left(\frac{\vec{r}_p - \vec{r}_c}{\lvert \vec{r}_p - \vec{r}_c \rvert ^3} \right)$

A surface area element on the sphere can be defined as follows:

$d\vec{S} = R^2 \sin{\theta} \ d\theta \ d\phi \ \hat{r}_p$

The flux due to the elementary rod through the surface element in the vicinity of the point on interest on the sphere is:

$d\Phi = d\vec{E} \cdot d\vec{S} = f(z,\theta,\phi) \ dz \ d\theta \ d\phi$

Finally, the flux through the top half of the sphere is:

$\Phi_p = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} f(z,\theta,\phi) \ dz \ d\theta \ d\phi \approx 0.6343$

the flux through the bottom half of the sphere is:

$\Phi_n = \int_{0}^{2\pi} \int_{\pi/2}^{\pi} \int_{0}^{1} f(z,\theta,\phi) \ dz \ d\theta \ d\phi \approx 0.3949$

It is possible to derive the expression of the exact integrands by plugging in all expressions and simplifying them. The integrals can then be outsourced to Wolfram Alpha or any similar online tool. I chose to perform these steps using a simple script of code where $dz = 10^{-2}$ , $d\theta = d\phi = \pi/100$ . The result can be verified using Gauss' law, applying which leads to the result:

$\Phi_p + \Phi_n = 1$

The above expression is the ideal result. But since a numerical method is used to evaluate the integrals, some error in the solution naturally exists. The sanity check for me yields the result:

$\Phi_p + \Phi_n \approx 1.0292$