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Electromagnetics charge distribution

  1. Mar 18, 2014 #1
    [itex][/itex]1. The problem statement, all variables and given/known data

    In fact, there are two problems, and I want to know whether my solutions are right or not.

    1- Two charged line with density of 15 n c/m along the x and y axes ([itex]x\pm\infty[/itex], [itex]y\pm\infty[/itex]), Find the Electric field at:
    (a) (0,0,4)
    (b) (0,5,4).

    2- A cylinder with radius [itex]\rho=8cm[/itex] and surface charge distribution of [itex]\rho_{s}=5e^{-20|z|} nc/m^{2}[/itex], determine the electrical flux which crosses a part of this cylinder limited by [itex]1<z<5[/itex], [itex]\frac{pi}{6}<\phi<\frac{pi}{2}[/itex]

    2. Relevant equations

    [itex]E=\frac{\rho_{l}}{2\pi\epsilon b}[/itex] (Electric field of a uniformly charged line)
    [itex]\rho_{s}=\frac{Q}{A}[/itex]
    [itex]\psi (Flux)=Q (Charge)[/itex]

    3. The attempt at a solution

    1- (a) For the uniformly charged line, I am only interested in the distance between the line and the point, which is 4.
    And because the two line are in the same distance from the point, and they have same density:
    [itex]E=2\frac{150×10^{-9}}{2\pi\epsilon 4}=134.8(N/m)[/itex]

    (b) The distance is [itex]\sqrt{(4)^{2}+5^{2}}[/itex], and then apply the previous formula?

    2- [itex]\int^\frac{\pi}{2}_\frac{\pi}{6} \int^{0.05}_{0.01}=5e^{-20|z|} z dz d\phi[/itex] ?
     
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 18, 2014 #2

    rude man

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    problem 2:

    first, there is an epsilon missing somewhere.

    second, your left integral makes no sense.

    Still, you seem to be on the right track. Equate the total free charge in the volume described to the total flux crossing ALL the surfaces defining the cylinder section. There are 5 separate areas to consider.

    Special note: fortunately, the problem asks for the flux thru ALL the surfaces defining the cylinder section. At least, that's my interpretation. Lucky for you because otherwise the problem would be a lot more difficult.
     
  4. Mar 18, 2014 #3
    I don't see where is that missing epsilon, I built my solution on that [itex]\rho_{s}=\frac{Q}{S}=\frac{dQ}{dS}[/itex], [itex]dQ=\rho_{s}dS[/itex], [itex]Q=\int \rho dS[/itex] , where ds in this case [itex]\rho d\phi dz[/itex]

    I made a mistake for the integral, I think it should be [itex]\int^\frac{\pi}{2}_\frac{\pi}{6} \int^{0.05}_{0.01}=5e^{-20|z|} \rho dz d\phi[/itex], where [itex]\rho=8cm=0.08m[/itex], isn't?
     
  5. Mar 18, 2014 #4

    rude man

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    Your integral still does not make sense. Every integral needs a differential element in it.

    The expression you need is ε∫∫E*dS = Qfree.
    Vectors are in bold. The asterisk signifies the vector dot product.

    The surface integral is over the closed surface defining the volume; the free charge is that inside that volume.

    I think you're mixing D and E up. For D, ∫∫D*dS = Qfree is correct.
     
  6. Mar 18, 2014 #5
    OK, but any luck in the first problem?

    I've found this problem (in the attachment) with its solution:
    RP = (1, 2, 3) − (1,−2, 5) = (0, 4,−2)
    Why did it assume that line defined by y=-2 and z=5 have the point (1,-2,5)? why not (0,-2,5)?
     

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  7. Mar 19, 2014 #6

    BvU

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    Re your problem 1b: the distances to the two lines aren't the same any more. You have to add two E vectors with different magnitudes and different direcetions.

    RE problem 2.17: You want the distance to the line, not just to any point on the line.
     
  8. Mar 19, 2014 #7
    Can you confirm the following for problem 1b?

    According to the following problem:
    http://forum.allaboutcircuits.com/cache.php?url=http://i.imgur.com/VBlEM4Z.jpg?1 [Broken]

    Rzp=(1,2,3)-(0,0,3)=(1,2,0)

    I assumed that the vector (0,0,3) was chosen with respect to the point of the interest, since the z component is 3.

    So, depending on that, and returning to our problem:

    Ryp=(0,5,4)-(0,5,0)=(0,0,4)

    We can see that only z-component is exist, so:
    [tex]E=\frac{\ 5 \times 10^{-9}}{2 \pi \epsilon (4)}=22.47_{az} (V/m)[/tex]

    Rxp=(0,5,4)-(0,0,0)=(0,5,4)
    The distance is [tex]\sqrt{41}[/tex] and unit vector is [itex] \frac{5_{ay}+4_{az}}{\sqrt{41}} [/itex], so:
    [tex]E=\frac{\ 5 \times 10^{-9}}{2 \pi \epsilon \sqrt{41}} \times \frac{5_{ay}+4_{az}}{\sqrt{41}} =10.95_{ay}+4.76_{az}(V/m)[/tex]

    _______________

    And for problem 2.17:

    How did you know that the x component is 1? I'm told in the problem the line is defined by y=-2 and z=5, why didn't defined it by (0,-2,5) or (x,-2,5)?
     
    Last edited by a moderator: May 6, 2017
  9. Mar 19, 2014 #8

    BvU

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    1b: looks good. Final answer is adding the two.

    The new problem you brought in: any point on the line has (z, 0, 0). Distance from 1,2,3 to such a point is √ (1-0)2+(2-0)2 +(3-z)2. The distance to the line is the minimum of that, which is reached for z =3.

    2.17: on the line, x can be anything. Distance from 1,2,3 to a point (x,-2,5) on the line is √ (1-x)2+(2-(-2))2 +(3-5)2. Distance to the line is minimum of that, which is for x = 1.
     
  10. Mar 19, 2014 #9
    Problem 1b: That's nice. :)

    For the new problem and problem 2.17: What I've understood that I can get my desired point on a line depending on the point of interest, which was (1,2,3) in the new problem and hence I've ended up with 3 in the z coordinate of the line.

    Thank you very much, I really appreciate your help.
     
  11. Mar 19, 2014 #10

    rude man

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    You can also look at it this way: your y=-2, z=5 line runs parallel to the x axis so a line connecting a point (a,b,c) to the y=-2, z=5 line must have the same x=a value on the y=-2, z=5 line in order to be perpendicular to the y=-2, z=5 line.

    Think 2-dimensions: let the line be y=3 and the point be (4,5). The if you draw a line from this point to the line y=3 it's obvious that the perpendicular line must intersect the y=3 line at x=4, right?

    So then d^2 = (4 - 4)^2 + (5 - 3)^2 etc. See the same 4's? That's why you have the same 1's in your 3-dimensional problem.
     
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