Take a look at the following problem:


I have solved the problem; the question is related to something else. We can calculate the electric flux by calculating the fraction of solid angle subtended by the center of the sphere on the desired surface, which will also be the fraction of flux. i.e. $$\int d\Omega= \int \frac{\hat{n} \cdot d\vec{S}}{r^2}= \int_{\phi=-\pi/2}^{+\pi/2} \int_{\theta=\pi/2 -\alpha}^{\pi/2} \sin \theta\ d \theta\ d\phi=\pi (\sin \alpha)$$ after which the flux comes out to be $$\frac{Q}{4 \epsilon_0}(\sin \alpha)$$ which corresponds to the second option.

But something regarding this result is bugging me. It is that, when we the angle $\alpha$ is $\pi$, the flux should come out to be $Q/2\epsilon_0$ whereas here it says it is $0$. What exactly is it that I am missing?


1 Answer 1


The result is not $\sin\alpha$ but $(1-\cos{\alpha})$. It is always good to check as you did with half sphere or something trivial so you can be sure you didn't get the right answer.

You should check the integration boundaries. Just make it simple like

$$ \int\limits_0^\pi \int\limits_0^\alpha \sin\alpha \mathrm{d}\alpha \mathrm{d}\theta = \pi \int\limits_0^\alpha \sin\alpha \mathrm{d}\alpha = \dots $$

and I let you do the rest.

  • $\begingroup$ Isn't the angle $\theta$ defined from the z axis as the initial line? $\endgroup$
    – Physiker
    Feb 3, 2021 at 10:35
  • $\begingroup$ What $z$ axis? You mean the dashed one? That is a bad choice of frame. Though you can do it with it of course. $\endgroup$
    – kyril
    Feb 3, 2021 at 10:38
  • $\begingroup$ Yes. But if we did it with angle $\theta$ starting from the $xy$ plane, then the $\sin$ in the equations will have to be replaced by a cosine, which will again give the same result I have derived above. What I mean is that the surface element defined above has angle $\theta$ taken from the dashed line $\endgroup$
    – Physiker
    Feb 3, 2021 at 10:41
  • $\begingroup$ Look at my edit and convince yourself what frame I choose. The whole point of spherical coordinates is to make the problem easy. Of course you can do it in your way and if someone wants to check what is wrong he/she is welcome ;) $\endgroup$
    – kyril
    Feb 3, 2021 at 10:46
  • $\begingroup$ Ah. I see your doubt, no if you put the $z$ axis on what you call the $x y$ plane the sine in the spherical coordinates doesn't change it is always a sine and is due to the jacobian of the spherical transformation. It is important in physics not to be too earth to earth. $x y$ or $z$ can be defined as anything and has nothing to do with horizontal or vertical concept. $\endgroup$
    – kyril
    Feb 3, 2021 at 10:56

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