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I know from my class that to calculate the flux of a point charge with Gauss's law, I have to make a surface that intersects with all of the flux lines resulting from the charge, and then make this integration on that surface

$$ \iint \vec{E}\cdot\vec{dA} $$

and that surface will of course be a sphere. But I thought that if I only need to make a surface which intersects with all of the field lines, then I may use the surface bounded by a circle with infinite radius and then multiply the result by two because this surface will only intersect with the lines at the upper side of the charge. This image shows what I mean. Gaussian surface to calculate the flux of a point charge
But when I tried to do this I got an infinite flux. Was this approach WRONG in the first place? If so, why?

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  • $\begingroup$ Assuming you are using a filled-in circle (and not a ring as you have drawn), this sounds like it could work. Can you add some detail in your post as to how you are handling the dot product in the integral? $\endgroup$ – Bryson S. Oct 23 '14 at 19:24
  • $\begingroup$ I think that Gauss' law is specifically about closed surfaces (such that there is a clearly defined volume and charge enclosed), i.e., surfaces with zero boundary. Though your approach can be made to work, keep in mind that there are an infinity of surfaces bounded by a circle. Given that degree of freedom, shouldn't you choose a surface that is normal to the field lines? hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html $\endgroup$ – Alfred Centauri Oct 24 '14 at 2:12
  • $\begingroup$ @Alfred Centauri - all of what you say is completely correct. But you can calculate Gauss's law using any combination of surfaces that sum to a closed surface. It does work (see below) and would work for any of the infinity of surfaces you mentioned. This is the under-used power of Gauss's law. $\endgroup$ – Rob Jeffries Oct 24 '14 at 6:18
  • $\begingroup$ I did some mistake that gave me infinity but finally I got the right result. see Mateus answer below $\endgroup$ – Abdo Saeed Anwar Oct 24 '14 at 17:47
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You probably did some wrong calculation, because your reasoning works. Take a circle of radius $r$ a distance $a$ above the charge. The increase $d\Phi$ of the flux by increasing the radius by $dr$ is given by $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{2\pi r \cos \theta dr}{a^2+r^2}=\frac{q}{2\epsilon_0}\frac{ar dr}{(a^2+r^2)^{3/2}},$$ where $\theta$ is the azimuthal angle. Then $$\Phi=\frac{q}{2\epsilon_0}\int_0^\infty\frac{ar dr}{(a^2+r^2)^{3/2}}=-\frac{q}{2\epsilon_0}\left.\frac{a}{\sqrt{a^2+r^2}}\right|_{r=0}^\infty=\frac{q}{2\epsilon_0},$$ which is the expected result, since the flux over a closed surface must be $\dfrac{q}{\epsilon_0}$.

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  • $\begingroup$ That is exactly what I tried to do but I made a silly mistake that gave me infinity, I used $sin\theta$ insteat of cos Thanks! $\endgroup$ – Abdo Saeed Anwar Oct 24 '14 at 17:45
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Your reasoning is correct, it's just a lot harder to do with the surfaces you've chosen.

Draw a small, elemental ring at some arbitrary height above the charge. A line from any point on the ring to the charge subtends the polar angle $\theta$ with the z-axis and is a radial distance $r$ from it. (i.e. $r$ and $\theta$ are the usual spherical polar coordinates (and $\phi$ will be the azimuthal angle).

The area of the ring is given by $$d{\bf A}= (2\pi r \sin\theta)(r d\theta/\cos \theta) {\bf \hat{z}}$$ The first term is the circumference of the ring, the second term is the thickness of the ring [it was the division by $\cos \theta$ that evaded me for quite a while - it is required because for a given $d\theta$, as $\theta$ gets bigger, so does the thickness of the ring].

From there you say ${\bf E} = Q/(4\pi\epsilon_0 r^2) {\bf \hat{r}}$ (in SI units). Then $$ \int {\bf E} \cdot d{\bf A} = \frac{Q}{4\pi\epsilon_0 r^2} \int^{\pi/2}_{0} 2\pi r^2 \frac{\sin \theta}{\cos \theta} d\theta\ ({\bf \hat{r}}\cdot {\bf \hat{z}})$$ where the limits mean you extend the rings to cover an infinite plane above the charge.

But ${\bf \hat{r}}\cdot {\bf \hat{z}} = \cos \theta$, so the integral simplifies to $$ \int {\bf E} \cdot d{\bf A} = \frac{Q}{2\epsilon_0} \int^{\pi/2}_{0} \sin \theta\ d\theta = \frac{Q}{2\epsilon_0}$$ Two of these planes, above and below the charge sum to give you the flux obtained over a sphere - as required.

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The Gauss Law indicates that the field lines $\vec{E}$ should be normal to the Gaussian Surface taken $dA$. Thus we take the dot-product to take the normal component of field $\vec{E}$ with the area.

$\vec{E}\cdot \hat{n}dA = EdAcos\theta$

The reason to take Gaussian surface as a sphere, with the point charge being its center, is because 1) the field radiates outwards in all directions uniformly, and 2) the Gaussian surface will always be perpendicular to field lines. Hence making $cos\theta =1$

I'm guessing you haven't taken the $EdAcos\theta$. Because intuitively, as the circles become infinitely large, the field lines will also become parallel to the Gaussian surface taken, thus making $cos\theta =0$

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  • $\begingroup$ I did considered that but the problem was that when substituting for cos $\theta$ I substituted for sin instead Thank you $\endgroup$ – Abdo Saeed Anwar Oct 24 '14 at 17:41

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