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Guassian Surface Decision?

  1. Jul 2, 2007 #1
    Guassian Surface Decision????

    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    My thoughts/attack for the question is first to find the charge
    density throughout the insulator

    which i caculate to be : 4*10^-9 C/m^3

    Now, use a quassian surface, such that using the volume and thus
    charge density above, of this guassian shape, I can calculate the
    charge enclosed by it.

    Then, at all points on the guassian surface, the Electric field will
    be equivalent and therefore using the surface area, it should collapse
    down to something nice like

    E= q/[(epsilon 0) * (SA of Shape)]

    BUT....the problem is I cant get the correct answer, I GET
    ~0.9v/M....whereas the answer is 0.68V/m.

    Im findind it difficult to select a correct guassian surface for the calculation, has anybody got any thoughts/tips/recommendations about how I shuld alternative attack or a guassian surface I shhould choose???

    Thanks for your time!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 2, 2007 #2

    Doc Al

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    Staff: Mentor

    It's not clear to me what Gaussian surface you are using, nor by what reasoning you think the field would be the same everywhere on its surface.

    Instead, think of this as composed of planes of charge--use superposition.
     
    Last edited: Jul 2, 2007
  4. Jul 3, 2007 #3
    im sorry , i dont quite understand how you consider it is composed of planes of charge....My understanding (and im in no way saying im correct,just trying to show you my thinking) is that the electric field radiates outwards in every direction from a charge.

    HENCE...what I was meaning, is on the surface of a symmetrical shape, ie circle, squre etc, at each point on the surface of this guassian "symmetrical" shape, the Electric field should be constant (this is why I thought you could take it out the front of the intergal sign
    in the equation I provided in my last post)

    Thats why I was thinking the field would be the same everywhere on its surface. Hence, by finding the volume of this symmetrical shape, and thus the charge it encloses, I would consider this overall charge as a point charge, and then calculated my answer....unfortunatly wrong


    Would you be able to possibly show me the dimensions/calculations I could use, because I still cant get the answer.

    Thanks again.
     
  5. Jul 3, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    It's certainly true that the electric field radiates outward uniformly from a point charge, but this is not just a point charge. The reason you can view this configuration of charge as planes of charge is that the width is much smaller than the area, thus it's a good approximation to the field from an infinite plane of charge. Hint: Consider the planes above and below the point P.

    But you certainly can use Gauss's law as well--if you choose the correct Gaussian surface. So let's stick with that approach.

    Choose a surface that has the point P on it and yet has symmetry with respect to the charge distribution. How would you draw such a surface? How much charge will it contain? Where will it be centered? What shape would it have? How would the field point on the surface?

    So far, so good.

    What Gaussian surface did you choose? (The field will not be that of a point charge.)
     
  6. Jul 3, 2007 #5
    i think I understand why my approach wont work, I dont think Im able to choose a "symmetical" guassian surface which encloses all the charge of the slab, and which also has "point P" on its surface.

    What I previously selected was a smaller cube within the slab itself, but in doin so, I have neglected the vector components of the elctric field, which would also contribute to the electric field at point P, from those other charges located just outside the surface.

    For instance, I used a small cube within the slab, with P on its surface,

    hence

    Volume of Cube = (0.001)^3 = 1*10^-9 m^3

    SA of Cube = 6*10^-6 m^2

    Q(enclosed) = (4*10^-9)*(1*10^-9) = 4*10^-18 C

    E = Q(enc)/(epsilon naught * SA)
    = 0.075 V/m

    WHICH IS INCORRECT AND I UNDERSTAND WHY lol, as I said previously, I cant juts neglect the rest of the charge outside this cube, as it most definitely will contribute to the eletric field at point P.

    So, do you know I a guassian surface which makes this mthod possible, or is it indeed invalid for his Q?

    But now, how do you do this plane of charges calcultion??

    Cheers for your help Doc AL
     
  7. Jul 3, 2007 #6

    Doc Al

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    Staff: Mentor

    Since "point P" must be on the surface of the Gaussian surface, that surface cannot contain all of the charge.

    The only thing wrong with your choice of a Gaussian surface is that you did not chose one with the needed symmetry. If your cube is 1 mm on a side, then it will not be evenly placed in the charge distribution and thus opposite sides will have different electric field values. Your cube is 1 mm from one side of the charge distribution, but 3 mm from the other.

    Pick a Gaussian surface with is evenly placed. Try it!

    Don't give up on using Gauss's law--once you "see the light" the solution will be easy.

    But you can also view that charge distribution as being a 4mm thick slab of charge below P plus a 1mm thick slab above P. These "slabs" are thin and flat--easily approximated by an "infinite" sheet of charge.
     
  8. Jul 4, 2007 #7
    Okay...im getting closer with your help...which is great!!:)

    This time i followed your suggestion, and cough cough attempted again.


    I understand the equivalent 1mm from both the top and bottom of the slab now. Heres the question, do also make it 1mm from the other sides?

    Or better still, for my guassian surface, ie a cube, wht will the 3rd dimension be of length?

    Since I get A cube (and just quickly, the shape must be a cube or circle, but not rectangle right??, since on the surface of the rectangle they will be different legnths from the centre?- im abit onfused by this bit thats all)

    Vol = 2.7*10^-8
    SA=5.4*10^-5
    Q(enclosed)=1.08*10^-16

    Therfore E = 0.226V/m (which i substantially closer!!!yay...so just to clear up the last bit and I should be there!
     
  9. Jul 4, 2007 #8
    um my cube was (3mm)^3 sorry
     
  10. Jul 4, 2007 #9

    Doc Al

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    Staff: Mentor

    Good! So your Gaussian volume has a height of 3mm.
    Definitely not! Keep your volume close to the center, where you know the field will be uniform across the top and bottom of your Gaussian surface. Near the edges of the slab, the field will start deviating from straight lines.

    It doesn't have to be a cube, just some shape that has a uniform field on the key surfaces. The top/bottom of that Gaussian surface can have any shape you like, but the sides better be vertical.
    OK. You chose a cube--no problem.
    Problem here! You need to calculate the net flux through your Gaussian volume. The direction of the field matters in calculating the flux through each of the six surfaces. Hint: Only two sides of your cube will have non-zero flux. The electric field is not perpendicular to all six sides!
    Good.

    Correct your mistake above and you'll be fine.

    Got it. :wink:
     
  11. Jul 5, 2007 #10
    Doc Al your a champ!...i really appreciate all your help!

    You highlighted the reason why i have gone wrong from the start, i will show you my thinking:

    I imagined the set of charges within the slab as a point charge at the centre of the Guassian Surface in this manner:

    [​IMG]

    which then for a cube I imagined it would look like this

    [​IMG]

    , and thus assummed, with this method of thinking that therfore each face(in being a square) will have the same amount of flux passing through it.

    You have told me that only 2 faces do, this must mean the field actually looks like this(sine that way the flux passing is perpendicular to these 4 "white faces" and thus =0)!!!!

    [​IMG]

    ....arggg...brain explosion lol...why is this so???...is it because i simplify the problem too much...ie making it a point charge??
     
  12. Jul 5, 2007 #11
    oh and just quickly.....of course your correct... when using only the 2 faces, of which, both have surcae area of 3*10^-6 m^2, then the calculation gives exactly the answer...


    ie

    (1.08*10^-16) / ((8.85*10^-12)*(2(3*10^-6)^2)) = 0.68V/m!!!!!!!!!!!
     
  13. Jul 5, 2007 #12

    Doc Al

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    Staff: Mentor

    Your fundamental error is thinking that the uniform distribution of charge--which extends beyond the chosen Gaussian surface--can be replaced by the field of a point charge within the Gaussian surface. No way!

    It's true that the net flux only depends on the total charge contained in the Gaussian surface--Gauss's law tells us that. But the magnitude and direction of the electric field depends on the distribution of all the charges. Of course, by taking advantage of symmetry we can use Gauss's law to figure out the field for a few simple charge distributions.

    One such charge distribution is an infinite sheet of charge. The field is everywhere perpendicular to the plane (how can it point in any other direction, since the plane looks the same in all directions?). What makes a charge distribution a "sheet of charge", versus a blob of charge, is that it's width is much smaller than its area. That's why in this problem the thickness is only 5mm, while the length is 2m. (The diagram is not drawn to scale!)

    So, as long as we stick close to the middle of this charge distribution, and stay away from the ends, the field near point P must look like the field from a stack of charged sheets. (That's why point P is chosen as smack in the middle of the large area. If point P were near the edge, you could not assume symmetry--a uniform field--and apply Gauss's law.) In the middle of the charge distribution, the field must be zero--since there's equal charges above and below. As you move from the middle, up towards point P, the field must point upwards (and on the opposite face of your Gaussian cube the field must point downwards. At all points near the center of the large area, the field is vertical (per your original diagram), so the four sides of your cube that are aligned vertically get zero flux.

    I hope that makes a little sense.

    You might want to read up on how Gauss's law can be applied to various charge distributions. Try this: Applications of Gauss' Law
     
  14. Jul 5, 2007 #13
    thanks for all your help!!! greatly appreciated...deserve a medal for the lenght of this convo lol
     
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