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I really have trouble with solving this, please help me

  1. Oct 20, 2009 #1
    http://img194.imageshack.us/img194/1559/45954311.png [Broken]

    Positive electric charge Q, total, are distributed uniformly on a semicircle
    with radius a. What is the electric field (in moderation and direction) in the center
    curvature (ie, point P)?

    please help me, its the last i have to solve in order to finish my assignment, im really desperate

    Ps: im from egypt, sorry for my english :(

    ps1: this is from the book "university physics by D. Young"

    thanks everyone in advance
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi desparate! Welcome to PF! :smile:

    (btw, it's modulus! :wink:)

    Find the field at P due to the charge between θ and θ + dθ, and integrate over -π/2 ≤ θ ≤ π/2. :smile:
     
    Last edited: Oct 20, 2009
  4. Oct 21, 2009 #3
    Re: Welcome to PF!

    can you tell me what the direction of the electric field is in P?
     
  5. Oct 21, 2009 #4

    tiny-tim

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    Hi desparate! :smile:

    I've just noticed that the diagram doesn't match your question …

    you asked for the field at the centre of curvature, which is O, not P. :confused:

    The field at O seems a far more likely question (and symmetry will give you the direction there). :wink:

    Before we go any further, can you please check which point is intended? :smile:
     
  6. Oct 21, 2009 #5
    doesnt it say P? (ie, P)?

    so i think it has to be P
     
  7. Oct 21, 2009 #6

    tiny-tim

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    Well, I'm highly doubtful.

    (btw, is the book by Hugh D. Young?)

    ok … then the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P.

    You'll need to split it into x and y components before integrating. :smile:
     
  8. Oct 21, 2009 #7
    yes its from this book, i really dont know why we are doing this kind of physics im not studying physics or something like that

    can you please draw something? i m not good at english, and i cant really understand what you mean by saying "the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P."

    thanks for your help
     
  9. Oct 21, 2009 #8

    tiny-tim

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    Yes, choose a point A at an angle θ from the middle, and a point B very close to A, at an angle θ + dθ, where dθ is very small.

    Then the charge of the section AB will be (dθ/π)Q,

    and since dθ is so small, we can assume that it is a point charge … that is, that the whole of AB is at the same distance (AP) from P.

    So the field will be of strength (dθ/π)Q/(AP)2, along the direction of AP.

    Carry on from there. :smile:
     
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