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Homework Help: Two Point Charges Net Electric Field

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Two point charges lie on the x axis. A charge of -2.5 µC is at the origin, and a charge of +8.5 µC is at x = 10.0 cm.

    What is the net electrical field at points x = +4 and at x = -4?

    2. Relevant equations

    E = kq/r^2

    3. The attempt at a solution

    For x = +4...

    E= 9x10^9(8.5x10^-6)/.06^2 = 21250000
    E= 9x10^9(-2.5x10^-6)/.04^2 = -14062500

    Net E = 7187500 (which is wrong)

    How do I solve this? Thanks.
     
  2. jcsd
  3. Feb 1, 2010 #2
    use absolute values for charges and then add/subtract them up qualitatively.

    the -2.5uC will make the E field LARGER at point x=+4 so you would add it instead of subtract.
     
  4. Feb 1, 2010 #3
    That didn't work.
     
  5. Feb 1, 2010 #4
    I'm pretty sure it should work. What's the correct answer for both positions?
     
  6. Feb 1, 2010 #5
    It doesn't give me the right answer until I submit the right answer. I entered 35312500 for x = +4 and it says that is incorrect. (I submit the answers online)
     
  7. Feb 1, 2010 #6
    the answer is 3.5*10^7 after sig figs.
     
  8. Feb 1, 2010 #7
    It says that is wrong too :yuck:
     
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