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E-Field and Voltage"

  1. Sep 24, 2014 #1
    1. Looking to find E-field and Voltage (in pic)



    2. Coulomb's Law, E=q0V, E=integralde



    3. Basically i tried to split the quarter ring into halves and then integrate. It worked i got the right answer. But my professor did it in 1 step? Something about just integrating over DV? I dont get how or why since when you do E-field integration you have to account for the splitting up of dq into lamdadl, etc like i did (also in a pic).

    Also how did she get the same result in Number 2? I got 2 times the result because there are twice as many rings?
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2014 #2

    gneill

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    Staff: Mentor

    Hello AGGENGR, Welcome to Physics Forums.

    In future, please be sure to use the posting template provided when you begin a new thread in the homework section of Physics Forums.

    Usually one splits up a charge distribution for integration in order to take advantage of some underlying symmetry that simplifies the calculation. This is often the case when you are summing vector quantities (like E fields) and a clever choice allows you to ensure that matched pairs of dq's from each portion cancel out one or more of the resultant's components.

    This is not the case in scalar situations where its just scalar values adding. Electric potential is an example of a scalar sum, and there's no real advantage to splitting the problem into pieces.

    Regarding your "2x" result, note that the same total charge is spread over the two quarter rings. So each quarter ring gets half the amount of charge that the original quart ring had.
     
  4. Sep 24, 2014 #3
    See thats what my prof said and it sort of made sense then so i went with it (Note to self, if physics makes sense right away, ya done screwed up!:(). When i go onto calculate it seems weird.? Ok let me try to explain what i seem to know.

    "Because E-Field is a vector, you must account for 'dq-splitting' when calculating it in these situations, but since Voltage is a scalar you can just integrate over dq? This should only be possible if they give you the charge q, right? Say the dont give you the charge but give you charge density. Then it gets real, and you have to take the long route? Still seems fuzzy"


    I figured the second one later in the evening. The 2 goes away bc of the symmetry.

    Thanks!!
     
  5. Sep 24, 2014 #4

    gneill

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    Staff: Mentor

    You *can* just integrate over the entire span of the charge. But then you need to break the problem into separate x and y components to account for the fact that you are summing vectors. By taking advantage of symmetry to show that one of those sums (vector component of the resultant) will be zero if you divide the problem up cleverly, then you save yourself a lot of pointless work.

    Charge density works the same as total charge. In fact, when you're given the total charge the first thing you do is determine the charge density over the geometry in order to form your dq for the integration.
    :D
     
  6. Sep 25, 2014 #5
    Ok i can see how you can possibly just go through with the integration and skip "those" steps, but is there a rule to when you can and cant? I mean the other way works fine for me, cut i really do wanna know when that can be done? I can see sort of why.
     
  7. Sep 25, 2014 #6

    gneill

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    Staff: Mentor

    There's no hard and fast rule. You have to appraise the problem as given and determine if you can see any shortcuts that symmetry might provide. It is most prevalent when you are dealing with summing vector quantities.

    My suggestion is to always be on the lookout for symmetry in a given problem. Quite often recognizing an underlying symmetry will lead to a big savings in effort, so it's worthwhile putting it on the list of things to check for each problem. Eventually it will become intuitive and something you do almost without thinking.
     
  8. Sep 27, 2014 #7
    Seems to be clearer now. Just gotta practice!
    Thanks!
     
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