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Electric potential difference

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data
    three point charges are located on a circular arc as shown:
    a) find the electric potential at the centre of the arc (point P)
    b) find the electric potential energy of a 25 pC point charge placed at P


    Diagram: [ r= 4 cm = 0.04 m]

    http://i1097.photobucket.com/albums/g356/wildpisces/Physics/23-p-023-alt.gif

    3. The attempt at a solution

    a) so i know the verticle components of the tension of Q1 and Q3 will cancel out therefore.

    Q1 = (3 x 10^-9)(cos30)
    Q1 = 2.598 x 10^-9 C

    Q3 = (3 x 10^-9)(cos30)
    Q3 = 2.598 x 10^-9 C

    Q2 = -2 x 10^-9 C

    Qtot= -2 x 10^-9 C + 2.598 x 10^-9 C + 2.598 x 10^-9 C
    Qtot= 3.18 x 10^-9 C

    Vp = KQtot/d
    Vp = (9x10^-9 NM^2/c^2)(3.18 x 10^-9 C)/(0.04m)
    Vp = 715.5 v

    b) Eel = + K Qtot q / d
    Eel = (9x10^-9 NM^2/c^2)(3.18 x 10^-9 C)(25x 10^12 C)/(0.04m)
    Eel = + 17 nJ

    so it will take 17 x 10^-9 to pull the P point?
     
  2. jcsd
  3. Dec 5, 2012 #2

    haruspex

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    The question is about potential (a scalar), not field. There's no cancelling out of components.
     
  4. Dec 5, 2012 #3
    ok i see... because potential is scaler
    but is my process. correct..
    i took this up with my tutor...
    he gave me this process..
    but i just want to make sure. im doing it okay..
     
  5. Dec 5, 2012 #4

    haruspex

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    No, it's much easier. Just calculate the potential due to each charge separately (q/4πε0r) and add them up. Since the r's are all the same it's particularly easy.
    (I make the answer about 900V.)
    Some of the signs look wrong on your powers of 10, but they don't seem to have worked through into your answers, so I'm guessing that's an error in writing the post.
     
  6. Dec 5, 2012 #5
    so, my process is not correct?
    and u need to use the equation
    What is that equation .. i have never seen that before...
     
  7. Dec 5, 2012 #6

    haruspex

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  8. Dec 5, 2012 #7

    SammyS

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    [itex]\displaystyle\frac{q}{4\pi\varepsilon_0 r}\ \ [/itex] is the same as [itex]\displaystyle \ \ k\frac{q}{r}\ \ [/itex] and it's not an equation (there's no equal sign), it's a mathematical expression or formula.
     
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