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Electric field problem

  1. Feb 13, 2006 #1
    An equilateral triangle with sides 66 m has charges +5 [tex] \mu C [/tex], -5 [tex] \mu C [/tex], and -9.3 [tex] \mu C [/tex] at its vertices and charge + 18.6 [tex] \mu C [/tex] at the midpoint on its side between the vertices with charges [tex] +5 and -5 \mu C [/tex]. What is the magnitude of the x-component of the electric field at point P, the point forming a second equilateral triangle with +5 [tex] \mu C and -5 \mu C [/tex]? Answer in units of N/C.
    Sorry I don't have a picture but it's 2 equilateral triangles that look like a mirror image of each other.
    First I drew a picture of what the electric fields would look like. The -9.3 and 18.6 [tex]\mu C [/tex] would not have a y-component, so I ignored those for now.
    so I found that
    k * 5 x 10^-6/ 66^2 * 33/66 = 5.17
    and k * -5 x 10^-6 / 66^2 *57.2/66 = 8.95 (since it's absolute value of the charge)
    So E_x= 14.12... which isn't right.. help please?
     
  2. jcsd
  3. Feb 14, 2006 #2
    Ok I figured out what I was doing wrong... the second part of the problem asks for the magnitude of the y-component of the electric field at point P.
    E_1 = k * 5 x 10^-6 / 66^2 * 57.2/66 = 8.93
    E_2 = 8.93 (same set up as E_1)
    E_3 = k * 18.6 x 10^-6 / 57.2^2 = -51.2, since the field would be downward
    E_4 = k * 9.3 x 10^-6 / 114.4^2 = 6.40
    8.93+8.93-51.2+6.40= -26.94
    This isn't right.. can someone help me?
     
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