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Calculating Electric field strength

  1. Oct 5, 2005 #1
    Alright everyone, normally im prett good with electrostatics, butlately...I've been in a serious rut. My physics professor has asked us to varify the electric field at a distance r from a cylindrical gaussian surface, under three conditions. 1) with a constant chrage density, 2) with a charge density of A/r and 3) with a charge density of Ar, where r is a variable distance and A is a constant.
    Now in aware that in order to get the charge within a surface, i need to integrate the density by the differential volume which is equal to r*dr*dphi*dz. This happens to be a triple integral. I even now the Gaussian law and such. The problem lies in the final algebraic remanipulation and substitution. Im getting either anomalies or nonsensical answers. Will someone show me the LIGHT!?
     
  2. jcsd
  3. Oct 5, 2005 #2
    why not put the tripple integral in maple or matlab? or do you have to evaluate by hand?
     
  4. Oct 5, 2005 #3

    robphy

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    Is this an infinitely long cylinder?
    If so, is there some symmetry you can exploit?
    You may be able to translate your problem into a simpler one.
     
  5. Oct 6, 2005 #4
    NO no. I'm afriad I have to do the math by hand and the electric field is evaluated only within a closed surface. This is a finite cylinder which apparently makes the mathe easier. I've been working on it still but the answers sometimes differ. I dunno whether it is more elegant to leave the desnity in the answer or to write the E-field as a function of Q,r,L and so on. I'm almost there, I just need another kick start. Help will be greatly appreciated.
     
  6. Oct 7, 2005 #5

    robphy

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    This is what I picture about your problem. Tell me if this is correct.

    Since the charge density is only a function of r (not z or theta), you have been given an infinitely long cylindrically symmetric charge distribution. [If the charge distribution is finite, you need to specify, say, a length L. Then, of course, it's not technically cylindrically symmetric since the charge density depends on z.]

    When using Gauss's law, you determine the electric field ON the [mathematical] gaussian surface... not some distance from it. You probably are determining the electric field from the center of some charge distribution.

    If it is an infinitely long cylindrically symmetric charge distribution, then the usual method is to choose a finite-length cylinder as your gaussian surface. Because of cylindrical symmetry and translational symmetry [my first comment was a hint], the electric field has a nice form. That is why a finite-length cylinder is chosen. Of course, the enclosed charge in this cylinder is finite.

    Is this your physical situation?
    If so, then your triple integrals are reducible to a single integral with respect to r.
     
  7. Oct 7, 2005 #6
    Righteous, I finally got it. Thanks a lot, the symmetrical analysis helped a lot. Just so you know, the cylinder was of a finite Length L and I was to find the field inside the cylinder. So the Whole thing was confined o concentric cylinders. I feel stupid. I realized thattwo of the three integrals reduced out to 2piL, which makes sense. Once again, thanks so much
     
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