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How do I use law to determine total flux at Gaussian surface at null point?

It seems , when I consider a Gaussian surface around null point, two axial electric field lines enter it but they don't exit. This leads to supposition that net flux is not zero. This creates a negative charge present in Gaussian surface when it is not.

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  • $\begingroup$ A Gaussian surface should enclose the charge. You cannot take a Gaussian surface just enclosing a null point only $\endgroup$
    – UKH
    Commented May 12, 2016 at 4:15
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    $\begingroup$ @Anubhav First of all "total flux at null point" is meaningless. Flux is associated with "surface" and not "point". $\endgroup$
    – atom
    Commented May 12, 2016 at 4:32
  • $\begingroup$ @atom Edited question. Don't go for words. Go for feelings. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 5:01
  • $\begingroup$ @Unnikrishnan Gaussian surface can be considered anywhere in space. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 5:03
  • $\begingroup$ @AnubhavGoel: Your words (i.e. language) are important than your feelings in science. You will experience this when you study quantum mechanics. Heisenberg has given us a cautious warning regarding this; and by the way "total flux at null point" is not your feeling but your wrong and un-clear way of thinking. $\endgroup$
    – atom
    Commented May 12, 2016 at 5:07

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You forgot to draw two exitting field lines. It's a hyperbolic singular point. It looks like this:

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Two in along the horizontal line, two out along the vertical line.

You seem to have a mathematical problem with singular points. Sure... if you follow along the EXACT horizontal line, sure... you hit the point and don't know how to continue. But that's true for every saddle point in any physical system! It even happens to you when you go to the mountain pass! Imagine rolling a marble down the ridge of this from A to X:

enter image description here

Sure, you may say that you just keep through going to B and back down again... both A→X and B→X are "inbound" field lines. But notice there are X→D and X→C that point outwards. Only if you start precisely on the ABX line, you "won't bend and stay there". But that's a special case, and it's unstable. Deviate for a hairline, and you'll roll downhill. That's exactly how your electrostatic system works. The electric potential is a saddle at that point. There is zero field at X (no slope) but it's unstable you get pulled in from two directions and expelled out from the other two, but you must shift slightly to feel that. Staying exactly at the center, you have no information about the surroundings.

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  • $\begingroup$ Exactly AT the point, the field is zero anyway, so both the ingoing and outgoing field lines actually vanis, get to zero intensity 0. But when you explore the surroundings, you see the picture above: steer slightly to the either side, and you'll make a sharp turn and circle back. $\endgroup$
    – orion
    Commented May 12, 2016 at 7:37
  • $\begingroup$ Sorry! I do not get your answer. How can they exit?How did they bent vertically? There was no vertical component. I had considered field horizontal line with y = 0. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 7:39
  • $\begingroup$ Consider surface with non zero radius. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 7:41
  • $\begingroup$ "But that's a special case, and it's unstable". Exactly, this special case leads to violation of gauss law, two extra field lines entering. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 8:11
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    $\begingroup$ Again... there isn't really such a thing as a field line. It's way of drawing the field, which has intersections of four lines at saddle points, just like elevation lines on a map have crosses there. You're trying to give a physical meaning to a graphing tool. You're forgetting that field lines skip the information about the field intensity, which goes to zero there (if you wanted, you could fadeout the lines to white there). The real thing here is $\vec{E}(x,y,z)$. And the Gauss law says $$e=\epsilon_0\oint \vec{E}\cdot d\vec{S}$$ forget about drawing field lines without lifting the pen. $\endgroup$
    – orion
    Commented May 12, 2016 at 8:33
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Imagine that the electric field in the region of the neutral (null) point looks something like this with a Gaussian surface which is a cylinder shown in red. The exact shape does not matter.

enter image description here

As drawn all looks well in that grey electric field lines enter at faces $A$ and $B$ and leave out of curved area $C$. One can imagine that all the sums work out as the total electric flux entering the Gaussian surface is equal to the total electric flux leaving the Gaussian surface resulting in a net flux through the Gaussian surface of zero ie no net electric charge within the cylinder.

Your contention is that there are two electric field lines which enter the Gaussian surface but do not leave the cylinder and this indicates a net negative charge within the cylinder.

The first thing to note is that those two field lines shown in green are unique in that all other field lines enter and leave the Gaussian surface no matter how close those field lines are to the green field lines.
Now a line is a one dimensional entity as it has no width (cross sectional area) and so when one computes the contribution to the electric flux $\vec E \cdot d\vec A$ of those two electric field lines shown in green the result is zero as $d\vec A$ (the cross-sectional area of the lines) is zero.

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  • $\begingroup$ Just one more thing. What if electric field line was thick? Or, how so ever thin it is , it should contribute something? $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 11:44
  • $\begingroup$ An electric field line is a useful visual aid but you must not push it too far in terms of it being real. For example where do you draw electric field lines? I suggest you draw a few electric field lines to illustrate something about the electric field. The electric field line has no width and only gets a width when you have to draw it. $\endgroup$
    – Farcher
    Commented May 12, 2016 at 11:58
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Gauss' Law in differential form is:

$ \nabla.E = 4\pi\rho $

where $E$ and $\rho$ are electric field and charge density respectively.

i.e.

${\partial E(x,y,z) \over\partial x} +{\partial E(x,y,z) \over\partial y} +{\partial E(x,y,z) \over\partial z} =4\pi\rho(x,y,z) $

So you first differentiate electric field w.r.t. space co-ordinates. Then put particular values of $E(x_o,y_o,z_o)$ at the null point i.e. at $(x_o,y_o,z_o)$. But you know that $E(x_o,y_o,z_o) = 0$. So left side of above equation becomes 0. Also you know that $4\pi\rho(x_o,y_o,z_o) = 0$ since there is no charge at the null point. So both sides are identically zero; hence Gauss' law is verified. Problem does not arise when we think of fields and charge densities as continuous functions of space co-ordinates, and real meaning of differential operators.

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  • $\begingroup$ Look at image. Two electric field lines entering, but not leaving. This means net negative flux. $\endgroup$
    – Anubhav Goel
    Commented May 12, 2016 at 7:37

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