1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gaussean Surfaces [Can they pass through charge distributions?]

  1. Nov 11, 2005 #1
    Hello.

    My textbook says that a Gaussean surface must be carefully chosen so that a point charge (or point charges constituting a discrete charge distribution) does not lie ON it, as otherwise the electric field at the location of the charge would be infinite and hence, it would not be possible to compute the flux through such a surface. In contrast, it says, that a Gaussean surface can well pass through a continuous charge distribution (a volume charge distribution) as the field everywhere is defined for such a distribution.

    The simplest example I can think of which justifies this proposition is that of a uniform solid nonconducting sphere with uniform volume charge: the field inside grows linearly with radial distance and falls of as the inverse square of radial distance outside. So, clearly, any spherical Gaussean surface drawn in the interior of the solid sphere passes through a continuous charge distribution yet the field and flux are well defined there.

    But now suppose we build our continuous charge distribution by somehow piling discrete charges on the solid material so that when this process has been carried out, the charge distribution has become continuous. At any intermediate stage, we could identify the distribution as discrete and so point charges may well reside on our Gaussean surface but at the end of the process, there is no problem because now, the field is well defined everywhere!

    Yet, we do regard a continuous charge distribution as made up of infinitely many point charges somehow stacked together though they would--when distinguishable--fly off due to mutual repulsion. Then, what is the fundamental explanation for this apparent inconsistency: that the Gaussean surface does not pose problems though infinitely many point charges lie on it..but it does pose a problem when countably finite charges lie on it?

    According to the book, the reason is that the field for a volume charge is continuous and defined everywhere as opposed to that of a discrete distribution. Perhaps this is the reason but I am not convinced so if you have a better fundamental explanation to give for this, please do post your views here.

    Cheers
    Vivek
     
    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Here's an attempt at an explanation: Charge distributions (be it a surface charge density or a volume charge density) implies "amount of charge per unit surface/volume".

    i) If your Gaussian surface crosses (not lie on) a surface distribution, there is no charge on the Gaussian surface per say since the region of intersection is a line, not a surface. But there is only charge when we consider a surface ==> no charge there on the gaussian surface.

    ii) If your Gaussian surface bathe in a volume density, again, the Gaussian surface is 2-D, while the volume distribution says there is only charge in a volume ==> no charge on the gaussian surface.

    iii) If the gaussian surface lies on a charge distribution, now there is a problem since there is actually charges on the gaussian surfaces and the field is not defined where there is charges.
     
  4. Nov 11, 2005 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Others will disagree on this and I encourage them to do so, but I assert that volume charge distributions are not physical. You cannot do something such as piling up discrete charges and make it continuous. The notion of a volume charge density is that the volume of interest is made of some sort of jelly of charge. There are no charges making up the jelly, the charges ARE the jelly. The jelly has the property that when you integrate its density over a volume, it gives the amount of charge inside the volume. We chose to approximate a physical (i.e. discrete) collection of charges by this jelly, and we define said "density of the jelly" at points (x,y,z) as the number of point charges in a not too small, not too big volume, divided by that volume.
     
  5. Nov 11, 2005 #4
    I don't believe that your textbook has worded it the way you have. Flux through a small area on a gaussian surface is:

    ElectricFieldperpendicular x Area

    Since a point charge has a symmetric electric field, the net field of the area where the charge is located is zero, not infinite (remember that Electric field is a vector not a scalar).

    Hopes this helps. Regards,
    Sam
     
  6. Nov 11, 2005 #5
    Thanks for all your explanations.

    BerryBoy, my textbook has indeed worded it that way.

    The electric field at the location of the charge is infinite. It has got nothing to do with the flux. Please read my post again: it clearly says: "the electric field at the location of the charge would be infinite".
     
    Last edited: Nov 11, 2005
  7. Nov 11, 2005 #6
    The chosen Gaussian surface is a spherical. So its 3D.

    Also, a point charge is a point...it can lie on a line. What do you mean by "crosses"?
     
  8. Nov 11, 2005 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No it's 2D. It's a sheet.

    Exemple of when a gaussian surface crosses a charge distribution: consider a plane which passes through the origin on which lies a surface charge density, and a spherical gaussian surface centered on the origin. Then the sphere crosses (intersect!) the plane, and the line of intersection is a circle in the plane of the charged plane.

    And your last objection is: "a point charge is a point...it can lie on a line". This issue I have handled in my discussion about the jelly. you mustn't consider your surface density as made up of point charges, but as a kind of sheet of material in which charges are diluted. Yes, diluted! That's the word I was looking for!

    Recipe courtesy of quasar987, Canada: Take any amount of point charges you like, put them in boiling water, stir until they are diluted. Put in freezer until it solidify into a jelly and you got yourself a bucket of uniform density jelly! You can also create jelly of non-uniform density by using clever tricks such as sticking charges in the solution before it solidifies.

    But the important point is that there is no point charges in surface and volume density. In a surface density , there is only charge on an area of that surface, not on a line.
     
  9. Nov 11, 2005 #8
    Sorry about my last post, I didn't make it clear what I was trying to say:
    The definition of Electric field forbids this statement:
    [​IMG]
    The reason I used flux was to prove that at exactly the location of the point charge, the net Electric Field is zero. By choosing a surface where the area directly passes through the point charge you can prove that the electric field at the location of the point charge is zero (because flux = charge/E0 and there is no charge enclosed).
    You can also show that there is no field at the location of a point charge from colombs law, because it has the restriction of (mod(r) > 0). Therefore, E is undefined and no existant at r = 0 (where r is the displacement vector from the point charge). Electric field MUST, by definition, have a direction, if if your displacement was zero, which direction would the field point?
    Hope this is clearer,
    Sam
     
  10. Nov 11, 2005 #9

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First time I hear something like that! I always read that electric field of a point charge is not defined at r=0.
     
  11. Nov 13, 2005 #10
    First of all, [itex]|r|^2[/itex] is just [itex]r^2[/itex].

    The field at a point charge is indeed infinite, no matter how you calculate it! It has got nothing to do with the flux or with Gauss's Law. As r tends to zero, the field asymptotically rises. This is in fact a FUNDAMENTAL truth. To make the picture clearer, if we were to draw a graph of E versus radial distance r, it would be seen to fall off as [itex]\frac{1}{r^2}[/itex] becoming infinite at r = 0.

    I don't see how you came to the conclusion that the field is zero at the location of the point charge. :surprised
     
    Last edited: Nov 13, 2005
  12. Nov 13, 2005 #11

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    For all practical purposes, I think that it's okay to have a Gaussian surface pass through a point charge. Locally, at the position of the point charge, the surface looks like a plane passing through the charge. Since the electric field due to the point charge points radially away it will contribute zero flux through that small surface patch. All the other surface elements will then be far away enough so that the contribution will not be infinite.
    In fact, you may use Gauss' law with a surface passing through a point charge and treat it as if the charge is inside the surface, like with a cube with a charge sitting at the corner.
     
  13. Nov 13, 2005 #12
    field theory

    i think the fundamental point here is that the field at the location of a point charge is ill-defined. Surely the limit of the field as r goes to zero is infinite but in classical electromagnetics there is no well defined field exactly at r = 0. another thing that the previous posters have mentioned is that the idea of a linear or surface or volume charge is an idealization of reality. You have to dilute (i like that metaphor) the charges and smear them out to get the continuous charge distributions. As far as people know charge is quantized and only comes in packets of e (leaving out quarks for now) when we speak of the electric field in a material body we are speaking of the average field that is created by all the point charges. The actual field would be oscillating and changing direction very rapidly. In order to make sense of it we choose a small volume and describe a density of chage to smooth this very spikey thing. Another way to approach this stuff is to give some sort of 'radius' to charged particles. Griffiths 'Introduction to Electrodynamics' like to use exercise where we postulate some very tiny radius and see what the results would be. Hope this helps

    Gabe
     
  14. Nov 14, 2005 #13
    Well for a cube sitting at the corner, the flux contribution is 1/8th the flux out of a cube when a charge would be completely inside the cube. And flux out of the three faces which share their vertices with the cube is zero so the net flux out of the cube when the charge is at the corner is [itex]\frac{q}{24\epsilon_{0}}[/itex].

    As for the field being ill-defined, I think a field which is arbitrarily large is just as ill-defined as an indeterminate field. The problem is that at one point in the domain of the surface integral [itex]\int \vec{E}.d\vec{S}[/itex] the field abruptly increases to an infinite value. As a result, the integral does not converge to a finite value.
     
  15. Nov 14, 2005 #14

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    No, you can treat the charge to be inside the cube. So the total flux is [itex]q/\epsilon_0[/itex], for the reasons I gave earlier. In this particular example with its simple geometry it's very easy to show. The flux through the 3 faces adjacent to the charge is zero and the flux through the other three is the same and equal to [itex]q/3\epsilon_0[/itex].
     
  16. Nov 14, 2005 #15
    Galileo, I am NOT considering the charge which was at the corner of your cube to be inside. Please read my post carefully. I have said that the flux contribution is one third of what it would be out of each face when the charge would be at the center. So I think you misunderstood what I said. So I'll try and explain by another method:

    Suppose we draw a bigger cube with a length equal to twice the length of the cube we've been given so that the charge now lies at the center of the bigger cube while still at a corner of the smaller one. Now, for the bigger cube, the flux out of each face is [itex]\frac{q}{6\epsilon_{0}}[/itex] and any face of this bigger cube can be thought of as containing four sub-faces each of which is identical to a face of our original small cube. The flux out of any such subface is simply one fourth of this value, i.e.

    [tex]\frac{q}{24\epsilon_{0}}[/tex]

    Now this is a fairly standard problem and it doesn't have a nonunique answer. In fact, many people seem to have discussed it on PF.

    Anyway, we're drifting from the main topic. I tried to come back to it in my last post on this thread.
     
  17. Nov 14, 2005 #16

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I agree with that treatement of the problem, however, I am puzzled by its conclusion: by Gauss' law there is a charge of magnitude q/24 inside the cube. How peculiar.
     
  18. Nov 14, 2005 #17

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I was about to say the same thing, only... the opposite! Consider a Gaussian surface passing through a point charge. Near the charge, the flux is zero, because the field is parallel to the surface (hence perpendicular to the normal vector). Directly at the charge, the field is undefined (division by 0), so there is a discontinuity in the scalar function [itex]\vec{E}\cdot \hat{n}[/itex].
    But the integral of a function which is continuous except at a finite number of points (continuous by parts) is convergent. As a result, the integral converges to a finite value. :smile:
     
  19. Nov 16, 2005 #18
    OK, first of all, I'm sorry that I haven't posted much recently, I feel very ill at the moment. So I came back to this thread today and caught up (sorry to quote an entry from ages ago).

    By looking at the equation I submitted, you can see that a value of r = 0, cannot be put into the equation. And because electric field MUST have a direction (oweing to the fact that it is a vector quantity), there MUST be a magnitude for displacement.

    The electric field at the location of a point charge is undefined. period.
     
  20. Nov 16, 2005 #19
    It is undefined and so you can freely say that it has an undefined direction as well (one of the comforts that mathematics provides..just call it arbitrary or undefined!). So far as its magnitude is concerned, it is infinite. Sticking to the "undefined" word for its magnitude doesn't sound a good idea to me.

    And by the way, the unit vector [itex]\hat r[/itex] does not go to zero as r tends to zero. Remember, a zero vector (thats what [itex]\vec{r}[/itex] is at the location of the point charge) has arbitrary direction...no unique direction. So the unit vector of a zero vector is not zero...its just not defined. But that doesn't make it zero.

    The original issue that we were discussing has not been addressed in recent posts. For those of you who have just checked in, the original query was whether one can chose a Gaussian surface with a finite number of point charges ON it (lying on it...is to be distinguished from being enclosed within it). If not, then is it because the flux is undefined for such a situation or is it because the field is undefined at the location of the point charge--which also happens to be a point on our surface. For simplicity, lets just consider one point charge lying ON a Gaussian surface and that no charges are inside our outside this surface. So the field is solely due to this point charge.

    Cheers
    Vivek
     
  21. Nov 16, 2005 #20
    By definition, a unit vector cannot be zero. It must also have a direction, if no direction is defined, it isn't a unit vector.

    I don't understand why you think there is a field at the location of a point charge. Say there was, then there would be a field (magnitude infinite) pointing in an arbitrary direction, but there would also be a field of infinite value in the opposite direction:

    [​IMG]
    I've tried to draw a diagram of this concept. For every field in one direction there would be an equal and opposite field. So the overall would be 0. An electric field can't exist at a location of a point charge.
     
    Last edited: Nov 17, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?