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!

Net Charge within a Gaussian Cube

  1. Feb 8, 2016 #1

    CARNiVORE

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    The figure shows a closed Gaussian surface in the shape of a cube of edge length 1.80 m. It lies in a region where the electric field is given by E= (2.86x + 3.82)i + 7.18j + 8.36k N/C, with x in meters. What is the net charge contained by the cube?

    b1d1950e8edc5fb67434db18ccf2f352.png

    2. Relevant equations
    ε0 × ΣΦ = Σqenc
    Φ = ∫E⋅dA

    3. The attempt at a solution
    I'm not even sure if the above equations are the correct ones to use, actually. I've never seen a problem like this and I'm not sure how to even begin. I know, however, that the y- and z-components of the electric field on this cube are constant. That is as far as I am, unfortunately.

    Thanks so much for your help.
     
  2. jcsd
  3. Feb 8, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    You're on the right track. Take a look at your first Relevant equation. How might you go about finding the net flux exiting the cube?
     
  4. Feb 8, 2016 #3

    CARNiVORE

    User Avatar
    Gold Member

    That's good to know!

    So, I made my first attempt. The net flux is only relevant in the x-component (again, since Ey and Ez are constant), meaning that ΣΦ = Ex * ∫dA

    If that is true, then ε0 * Ex * ∫dA = Σq. I substituted in the value of ε0; the value of Ex at x=1.80 m, which is approximately 8.9; and, I substituted s^2 (where s is the length of a side of the cube) for ∫dA. Multiplying these three numbers together left me with 2.5725E-10 Coulombs, which was incorrect.

    If I had to guess, I'd say that there is an error in the way that I am expressing the implication of the E-field on this problem. Actually, I'm a bit confused on the area, as well. Am I solving for the net flux correctly?
     
  5. Feb 8, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    You need to be careful about the signs (directions of the vectors). Note that there are two faces of interest, one at x = 0 and one at x = -1.80 m. You also need to take into account that the area normal vectors face outwards from the face. That's why there's a dot product in the integral in Gauss' law...check that the field vectors and area normal vectors in the same directions in both cases.
     
  6. Feb 9, 2016 #5

    CARNiVORE

    User Avatar
    Gold Member

    Hey, I'm sorry for the late response - I want you to know I haven't quit this problem or this thread (the homework isn't due until Sunday, I just like to get a headstart). For my own well-being, I've decided to take a break from physics today, lol. Thank you for your help - and again, I'll get back to this tomorrow, so stay tuned! :P
     
  7. Feb 10, 2016 #6

    CARNiVORE

    User Avatar
    Gold Member

    What I got from that, is that:

    1. I should account for twice the area I've been looking at so far, and
    2. The overall charge of the system should be negative.

    You also need to take into account that the area normal vectors face outwards from the face.

    I have to admit, I'm not sure what this means. Could you elaborate?
     
  8. Feb 10, 2016 #7

    gneill

    User Avatar

    Staff: Mentor

    When applying Gauss' law with an actual Gaussian surface it is convenient to have a mathematical way to distinguish the direction that the flux is passing though the surface. When using vector calculus to do your sums a standard method is to use a vector normal to the surface to represent the "outward" facing direction, and and to assign an area element to it. That way you can integrate the flux passing though the surface as the integral of the field vector dotted with the area element vector.

    ##\Phi = \int \vec{E} ⋅ d\vec{A} ##

    The dot product takes care of selecting the components of the vector field that are perpendicular to the surface in question and handles sorting out the sign of the contribution.

    Suppose you had a field that entered the cube in one face (perpendicular) and left the cube on the opposite face (also perpendicular). On entry the field vector would be oriented in the opposite direction of the surface area normal vector (which always points outwards) so the dot product would yield a negative value. On exit the field is oriented in the same direction as the surface normal so it would have a positive value. The net contribution would be zero if the field had the same magnitude entering and leaving.

    In your case the situation is simple enough to take advantage of the geometry: a cube with easy to calculate surface areas and oriented conveniently perpendicular to coordinate axes. That and the uniformity of the field over the "important" faces. But your electric field vectors are not so nicely oriented. They have multiple components. You need to select the component of the field that is perpendicular the a given face. That's where you can manually do the work of a dot product: figure out which vector component of E is aligned with the given face and whether it is entering or leaving the cube through that surface, and assign the appropriate sign to the product of area by that component.

    Regarding the sign of any net charge, that will be decided by the net flux. Be sure to pay attention to the coordinates of the cube faces. Is the cube sitting on the positive x axis or the negative x axis? What value of x will you plug in for back face?
     
  9. Feb 11, 2016 #8

    CARNiVORE

    User Avatar
    Gold Member

    Actually, I think I understand at least most of what you're saying . Again, Ex is the only field that has any affect on the net flux, and the cube is sitting on the negative x-axis.

    That means the two squares that exist in the yz-field (that are affected by the net flux) are at x = 0 and x = -1.80 meters. The flux that must be considered in this problem is that which affects these two squares. Is that right, or am I off on my components?
     
  10. Feb 11, 2016 #9

    gneill

    User Avatar

    Staff: Mentor

    It is right. You should be able to justify being able to ignore the other faces (in case you ever get a similar question where it asks for it).
     
  11. Feb 11, 2016 #10

    CARNiVORE

    User Avatar
    Gold Member

    I know you can disregard the other faces because their net flux is constant - I know this because the E-field is constant for all components except x.

    I am really having trouble with this, though. I substituted -1.80 m for both values and got -7.6E-11, and also tried substituting 0 for one and -1.80 m for the other, getting -3.8E-11. These were both wrong.

    I know it was pretty stupid to try these, but I got impatient. I've only got one attempt left, so I'll be more careful at this point - I just really don't get what I'm doing wrong at all. It seems simple, but I keep getting the wrong answer.
     
  12. Feb 11, 2016 #11

    gneill

    User Avatar

    Staff: Mentor

    Show your calculations in detail. You've not supplied units with the numbers you've quoted. What are the fluxes you've found for each face (with appropriate signs)?
     
  13. Feb 12, 2016 #12

    CARNiVORE

    User Avatar
    Gold Member

    Will do:

    ε0 * ∫E*dA = Σq

    ε0 * (2.86x + 3.82) * ∫dA = Σq

    ε0 * (2.86x + 3.82) * x^2 = Σq

    At this point, I tried substituting my values in for x. I feel I'm missing something simple.
     
  14. Feb 12, 2016 #13

    gneill

    User Avatar

    Staff: Mentor

    Okay, first note that the face areas do not depend upon x. It is a coincidence due to the fact that the gaussian surface is cubical that the side length happens to be the same as one of the coordinates of interest. So the area of each face of interest is the same, namely (1.80 m)2. Thus the normal vectors representing those areas will have that magnitude, but opposite signs (one points along the +x direction, the other along the -x direction).

    The x-components of the E-field vector at those faces, on the other hand, do depend on the value of the x coordinate according to the given information. So compute the contributions to the sum separately for each face, taking into account the signs of the area vectors and field components. If the area vector and field component vector are in the same direction then the contribution is positive: flux is crossing from the inside to the outside of the gaussian surface. If the two point in opposite directions the contribution is negative: flux is crossing from outside to inside through the surface. Once you sort out the signs of the contributions, sum them for a net flux leaving (or entering) the gaussian surface. If the sum is zero there is no net charge contained within the surface. If it is positive there is a net positive charge, negative: negative charge. Multiply the net flux by ε0 to obtain the value of the charge.
     
  15. Feb 13, 2016 #14

    CARNiVORE

    User Avatar
    Gold Member

    All right, I see what you mean, and I think I've finally got it:

    Σq = ε0 * E∫dA
    Σq = ε0 * ((2.86x + 3.82)*(1.8)^2 + (3.82)*(1.8)^2)
    Σq = ε0 * (-4.30272 + 12.3768)
    Σq = ε0 * (8.07408)
    Σq = 7.1488E-11

    How's that look? I'm actually confident with this answer.
     
  16. Feb 13, 2016 #15

    gneill

    User Avatar

    Staff: Mentor

    Close. You need to set the sign of the area element so that the vector representing it is leaving the cube. You've attributed a positive sign to both area terms. How can that be? One face faces the +x direction, the other the -x direction.
     
  17. Feb 13, 2016 #16

    CARNiVORE

    User Avatar
    Gold Member

    I think I see what you mean - since the field leaves the cube from the yz square at x=0:

    Σq = ε0 * ((2.86x + 3.82)*(1.8)^2 - (3.82)*(1.8)^2)
    Σq = ε0 * (-4.30272 - 12.3768)
    Σq = ε0 * (-16.67952)
    Σq = -1.4768E-10

    This makes more sense, actually - what do you think?
     
  18. Feb 13, 2016 #17

    gneill

    User Avatar

    Staff: Mentor

    Yes, the field leaves the cube at the x = 0 face. So both the area vector and the x-component of the field there should have positive values, as the area faces outward along the +x direction.

    At the other face, where x = -1.80, the field also is leaving the cube. The area vector there should point along the negative x-axis, and so should the field component (which it does as the x-component is negative there). That is, both will be negative, so that when they're multiplied a positive value results.

    Both faces should yield positive values for the flux, since flux is leaving both faces. Looking at the cube from the point of view of the negative y-axis:
    upload_2016-2-13_17-17-21.png

    ##E_x## is positive when x = 0, and negative when x = -1.80.
     
  19. Feb 13, 2016 #18

    CARNiVORE

    User Avatar
    Gold Member

    So I had it backwards, then? The net charge should be positive 1.4768E-10 Coulombs rather than negative. Instead of negating the value at x=0, I should've negated the value at x=-1.8, which makes both values positive.

    That actually makes sense, since the flux comes from within the cube. I'm not sure if that's what you were getting at, though.
     
  20. Feb 13, 2016 #19

    gneill

    User Avatar

    Staff: Mentor

    You need to compare the directions of the area and field vectors to see if the contribution is positive (leaving) or negative (entering). Always make a sketch of the vectors at the surfaces. So your result now is correct.
    Yes, that's what I was pointing out. Note that in some other situation one of the fluxes might have been entering while the other was leaving. You need to always compare those vectors!
     
  21. Feb 13, 2016 #20

    CARNiVORE

    User Avatar
    Gold Member

    Thanks so much for your help again! This was really difficult for me - I'll be sure to study up on questions similar to this in my textbook.

    I just have one question left, though - you say that, in some other situations, one of the fluxes could be entering while the other is leaving. If the cube is uniformly charged, how could this be possible?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Net Charge within a Gaussian Cube
  1. Net charge of cube (Replies: 1)

Loading...