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Gauss law differential form.

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Space distribution of electric charge is limited by to planes. Charge has the same intensity in planes parallel to these planes, but in dependence of the x coordinate, charge density is being distributed like:

    [itex] \rho (x)=A\cdot x\cdot (d-x)[/itex] d is distance between those to planes, where A is a constant.

    Find the electric field strength in all points in space.

    2. Relevant equations

    [itex]div \vec{E}=\frac{\rho }{\epsilon } [/itex]

    And since we have only x coordinate, partial derivatives of y and z will be 0.

    3. The attempt at a solution
    [PLAIN]http://pokit.org/get/3da968a3cd6dab87fbe51fcfdcd6b116.jpg [Broken]

    Here are the solutions.

    I get here:

    [itex] |\vec{E}|= (2d-3x)\cdot \frac{A\cdot x^{2}}{6\cdot \epsilon}+C_{1} [/itex]

    Then I used my intuition, which I think is WRONG. I found the stationary points of

    [itex] \rho (x)=A\cdot x\cdot (d-x)[/itex] which resulted in d/2.

    Then i figured if density is most dense or most thin there Electric field would be 0.

    I got that constant you see in the picture [itex] C_{1}=- \frac{A\cdot d^{3}}{12\cdot \epsilon} [/itex]

    What I don't get is how do we calculate the field outside the planes. Those solutions, are constants. But I can't figure out how did he get to that solution.

    Any help?

    P.S. Why are my LaTex fractions so small? How do I make them bigger?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 8, 2011 #2
    I think you are on the right track here. However to get C, you do not need stationary points, just ask what x has to be for rho to be zero (hence E = 0).

    Now let us assume you have calculated the field inside the charged volume.
    Outside the planes I would then use a gaussian cylinder with one end in the space between the planes and use the integral form, appealing to cylindrical symmetry.

    By the way there seem to be a number of typos in the question so I am not sure I fully grasp the geometry so am reading between the lines.
  4. Oct 8, 2011 #3
    These are infinitely long planes, they don't have charge on them, they are limiting charge between them. Having said that, charge density is a function of x, between those two planes.

    In that sense, how do I calculate field outside those planes. As you see field outside is that constant, with different signs. I don't know how to get that result.
  5. Oct 8, 2011 #4
    Hi Bassalisk,

    Here is how I would solve the problem.

    Firstly, I would change the coordinate system in order to exploit the symmetry of the problem. Defining the variable [itex]z=x-d/2[/itex], the charge density becomes:
    [tex]\rho(z)=A\left[\left(\frac{d}{2}\right)^2-z^2\right].[/tex]To solve the electrostatics problem, I use Gauss' law in integral form (cf. D.J. Griffiths, Introduction to Electrodynamics):
    [tex]\oint{\bf{E}\cdot d\bf{a}}=\frac{Q_{enc}}{\epsilon_0},[/tex]where the left hand side represents a surface integral over a closed surface, and [itex]Q_{enc}[/itex] is the charge enclosed in within that surface.

    Now, consider a "Gaussian pillbox" with area [itex]a[/itex] which straddles the [itex]z'=0[/itex] plane (I'm appending a prime to the [itex]z[/itex] coordinate here since it will be a dummy variable in the integration later), and extends from [itex]-z[/itex] to [itex]z[/itex], where for the moment we assume that [itex]|z|<d/2[/itex]. By symmetry, [itex]\bf{E}[/itex] points away from the plane [itex]z'=0[/itex] (upward for points above, downward for points below). Thus, the top and bottom surfaces yield [tex]\int{\bf{E}\cdot d\bf{a}}=2a|\bf{E}|,[/tex] whereas the sides contribute nothing.

    The enclosed charge [itex]Q_{enc}[/itex] is [tex]Q_{enc}=a\int_{-z}^{z}{\rho(z')dz'}.[/tex]Combining the results above, we find that for [itex]|z|<d/2[/itex], the electric field is given by [tex]\mathbf{E} =\frac{A}{\epsilon_0}\left[\left(\frac{d}{2}\right)^2z-\frac{z^3}{3}\right] \mathbf{a_z},[/tex]where [itex]\mathbf{a_z}[/itex] is a unit vector pointing in the [itex]z[/itex]-direction. Since our "Gaussian pillbox" does not enclose any extra charge beyond [itex]|z|=d/2[/itex], we have that for [itex]|z|>d/2[/itex],
    [tex]\mathbf{E}=\mbox{sign}(z)\frac{2}{3} \frac{A}{\epsilon_0} \left(\frac{d}{2}\right)^3 \mathbf{a_z}.[/tex]These equations specify the field both between and outside the planes. You can convert it to your 'asymmetric' coordinate system by substituting [itex]x=z+d/2[/itex].

    Hope this helps.

    Last edited: Oct 8, 2011
  6. Oct 8, 2011 #5
    Thank you for your reply. But honestly I don't understand what you did. Is your result right? Have you checked it with my solution provided?

    And I think, you must solve this problem by divergence theorem of electric fields. By "must" I mean that any other method is not allowed.

    Can you help me where I'm currently stuck?
  7. Oct 8, 2011 #6
    Hi Bassalisk,

    Thanks for reminding me to check the solution. My solution is indeed equivalent with the one you give, which can be verified by substituting [itex]z=d/2[/itex] into my equation (or [itex]x=z+d/2[/itex] into your equation).

    Although my method using the integral form is the standard procedure taught in the classic reference by Griffiths, I guess you could also reason in a different way. Since the field points in the [itex]x[/itex]-direction, the differential form of Gauss' law reduces to [tex]\frac{\partial E_x}{\partial x}=\frac{\rho(x)}{\epsilon_0},[/tex] where [itex]E_x[/itex] is the component of the electric field in the [itex]x[/itex]-direction.

    The field [itex]E_x[/itex] can be found by integrating the above equation, where the constant of integration is determined by the boundary condition [itex]E_x=0[/itex] at [itex]x=d/2[/itex]. (This follows from the observation that the charge distribution is symmetric about this plane).

    The fact that [itex]E_x[/itex] is constant outside the two planes simply reflects the fact that [itex]\rho(x)[/itex] is zero in this region.

  8. Oct 8, 2011 #7
    Yes I get that. But how do I get that constant, including the sign of that electric field. Check the solution. How do I get that? How do I get that result? I see that outside the planes, the field is that constant I got from Ex=0 etc...

    Its all fine if you look at the picture, but how do you do that mathematically?
  9. Oct 8, 2011 #8
    Hi Bassalisk,

    Mathematically, [itex]\rho(x)[/itex] is a piecewise defined function which is equal to [itex]Ax(d-x)[/itex] for [itex]0<x<d[/itex] and zero otherwise.

  10. Oct 8, 2011 #9
    How do you get this (minus)sign then?
    [PLAIN]http://pokit.org/get/ee177c4e4a2b16477fa07098b4d0d124.jpg [Broken]

    And I still don't understand how do you this get formula for field outside the planes, even with that, what you said in this post. How do I derive it?
    Last edited by a moderator: May 5, 2017
  11. Oct 8, 2011 #10
    Finally I got it. Thank you, you have been very helpful.
  12. Oct 8, 2011 #11

    Specifically, your field component in the [itex]x[/itex]-direction, [itex]E_x[/itex], is given by the formula [tex]E_x(x)=\frac{1}{\epsilon_0}\int_{d/2}^x{\rho(x')dx'}[/tex] where [itex]\rho[/itex] is the piecewise defined function given in the previous post. Choosing the lower limit at [itex]d/2[/itex] ensures the boundary condition mentioned earlier is satisfied.

    If [itex]x>d/2[/itex], then you are integrating "from left to right" and get a positive result (assuming the integrand is positive). However, if [itex]x<d/2[/itex] you are integrating "from right to left" and get a negative result.

    Last edited: Oct 8, 2011
  13. Oct 8, 2011 #12
    Looks like you have been guided correctly by Kurt so I will not add anything here but just correct something in an earlier message that was an incorrect statement on my part. I said or rather implied that when rho was zero the E field was zero which is of course wrong, it is the derivative of E that is zero. Your intuition was correct and the electric field is zero in the plane that lies midway between the two outer planes. This is because of the symmetry of rho. The stationary point was where the quadratic charge d/n peaked. Sorry if my comment confused you.
  14. Oct 8, 2011 #13
    No problem, I saw what you meant and I figured the typo. Thank you.
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