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Can some one please explain why a charge located outside a box shaped surface area produces zero net flux? I know this question has been asked many times, but I can't seem to find one that answers my problem with it.

I cannot seem to make sense of this because if the electric field decreases in magnitude with increasing distance from the source, won't the electric field lines leaving the surface have less magnitude than the ones entering it since it is farther from the source due to the surface area having a certain width, which leads to a greater flux inwards than outwards? Thanks in advance.

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    $\begingroup$ Don't forget about the other surfaces in the calculation, if you trace any one field line that enters the cube, you will see that it also exits the cube even if not by the face directly opposite the side where it entered. $\endgroup$ – Triatticus Jan 29 '19 at 20:58
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Remember that for the flux calculation, there is a scalar product and the angle between the surface normal and the field at each point must be involved. The larger the angle, the smaller the flux.

If the surface is farther, the field is weaker but the angle is also lower. On the whole cube, the flux compensate each other.

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  • $\begingroup$ Thanks for the reply! But since it is a cube won't the angle between the surface and the field be zero for the surface entered by the field and the surface exited? Why would the angle change the further the field travels when the exited surface is also flat? $\endgroup$ – Jerry Hsieh Jan 29 '19 at 9:20
  • $\begingroup$ I think I understand what you mean. For say a point charge outside the surface, the field vectors projected increases in angle relative to each other and thus on a flat surface further away, the field lines perpendicular to the surface have less magnitude. $\endgroup$ – Jerry Hsieh Jan 29 '19 at 9:23
  • $\begingroup$ But what about say if the cubic surface area is lying on a uniform electric field, not a point charge? $\endgroup$ – Jerry Hsieh Jan 29 '19 at 9:25
  • $\begingroup$ In this last case, the two flux are equal, with opposite signs and the flux is zero. $\endgroup$ – Vincent Fraticelli Jan 29 '19 at 9:27
  • $\begingroup$ How are they equal if the vectors leaving the box has lower magnitude than the ones entering? And since it is a uniform field, the angle shouldn't change the farther the field travels. $\endgroup$ – Jerry Hsieh Jan 29 '19 at 9:33
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I think I understand your confusion. The field lines that you see in illustrations of flux do not get weaker over time, counterintuitively. The 1/r^2 dropoff of an electric field is not due to the lines getting weaker; it is due to the fact that the electric field is being spread over a wider area as it gets further from the source charge, and thus it is less dense. In other words, that 1/r^2 isn't a property of the field- it's a property of geometry. Each line you see entering a closed surface, like a box, is the same strength when entering as it is when exiting.

Hope that helps!

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  • $\begingroup$ I suggest you reconsider the statement: "The 1/r^2 dropoff of an electric field is not due to the lines getting weaker; it is due to the fact that the electric field is being spread over a wider area", which is definitely wrong. It is the nature of the electric field, you can see that in its point-like values without considering any area. The correct argument explaining the 0 flux, is the increase in surface area with distance, at a fixed solid angle. $\endgroup$ – ohneVal Jan 29 '19 at 13:57
  • $\begingroup$ @ohneVal Thanks for your reply. I am rather confused, though. For any "stuff" distributed over some spherical shell with radius R, its density over its surface will drop by 1/r^2 as the total surface area grows exponentially. Is this not a fundamental property of a constant value distributed over an increasing spherical surface? And if this geometrical contribution causes a 1/r^2 decrease in density (i.e. field strength in this case), any inherent weakening of the field itself over some distance will result in a decrease faster than 1/r^2. $\endgroup$ – Leo Jan 29 '19 at 14:24
  • $\begingroup$ Appreciate your input Leo. This taught me something about the inverse square law that I didn't know. Basically what you said makes total sense to me! From what I understand from ohneVal's input, to me, he's basically saying the same thing as you, just in different words. $\endgroup$ – Jerry Hsieh Jan 30 '19 at 8:51
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If you placed any net charge outside the box (for a uniform electric field), the net flux through it is zero. Why? Because what we mean by a uniform electric field is a series of parallel field lines passing in and out of the box.

Uniform electric fields are generated in capacitor plates. The inverse square law would say that the field lines would diverge and you might think that this creates a non-zero flux, but at appropriate distances (in millimetres probably) that divergence would be minimum and we can imagine those field lines to be uniform as they are called.

What I mean by divergence of field lines is literally the spreading of theirs.

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  • $\begingroup$ Thanks! This was very insightful and made me realize that since a uniform electric field is generated by capacitor plates, then the field lines passing through the surface area inside will not decrease with further distance! Also made me realize that the inverse square law only applies to point charges. $\endgroup$ – Jerry Hsieh Jan 30 '19 at 8:45
  • $\begingroup$ In fact, it is also true for every other electric field as well. Not just uniform field lines. This is because electric field is only the electric flux density (i.e. the number of field lines $\phi$ passing through an area $A$). It does not matter how they the field lines diverge indicating that the field strength is not the same at both the entry and the exit boundaries. $\endgroup$ – KV18 Jan 30 '19 at 8:52
  • $\begingroup$ physics.stackexchange.com/q/457691 (I have written an answer to a very similar question yesterday). $\endgroup$ – KV18 Jan 30 '19 at 8:55
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Just visualize the flux as an incompressible fluid flow (that's what the word means after all) and where the charge is a magical source of fluid. Now imagine a box-shaped region in the fluid. Because the fluid is incompressible the total amount of fluid in the box must always stay the same. If you have a charge (source) inside of then box then whatever fluid is appearing at the source must flow out of the box: there is a net flux. If the source is outside the box then whatever flows in through some sides of the box must flow out somewhere else. In this case the net flux is zero.

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  • $\begingroup$ Appreciate the help! $\endgroup$ – Jerry Hsieh Jan 30 '19 at 8:40

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