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What is the net electric field at x = + 2.0 cm?

  1. Jan 30, 2005 #1
    Can someone please help me with this question?
    Two point charges lie on the x axis. A charge of + 3.5 µC is at the origin, and a charge of -3.5 µC is at x = 10.0 cm. What is the net electric field at x = -2.0 cm? What is the net electric field at x = + 2.0 cm?

    for the first one i did
    [tex]k q/r^2[/tex]
    [tex](9 x 10^9) (3.5 x 10^-6)/(0.02m)^2 = 78750000 N/C[/tex]
    [tex](9 x 10^9) (-3.5 x 10^-6)/(0.12m)^2 = -2187500[/tex]
    [tex]78750000 - 2187500 = 76562500N/C[/tex]
    can someone please tell me what i'm doing wrong?
     
  2. jcsd
  3. Jan 30, 2005 #2
    what seems to be the problem?
    do the solutions have a different answer?:S
     
  4. Jan 30, 2005 #3
    yeah when i entered this value in, i got it wrong
     
  5. Jan 30, 2005 #4
    I just tried the first question, and I got 7656250N/C (to the left). I'm sure that's the answer...
     
  6. Jan 30, 2005 #5
    how do you get that?
     
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