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Magnitude of electric field

  1. Mar 6, 2017 #1
    I think I have this correct, could someone please verify? I compared answers with a friend who got 2.0 x 106 N/C to the left, and I can't figure out how they got that answer, so I'm hoping mine is right...

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

    A negative charge of 3.5 × 10–8 C experiences a force of 0.070 N to the right in an electric field. What is the field magnitude and direction?

    2. Relevant equations
    Electric field strength = f / q

    3. The attempt at a solution

    Electric field strength = f / q

    Electric field strength = 0.070 N / 3.5x10-8

    Electric field strength = 2x106 N/C To the left
     
  2. jcsd
  3. Mar 6, 2017 #2

    BvU

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    Gold Member

    Funny, ##2\times 10^6## N/C to the left looks to me to be the same as ##2\times 10^6## N/C to the left . Or do you mean you found ##2\times 106 = 212 ## N/C to the left ?
     
  4. Mar 7, 2017 #3
    Oh goodness, I got myself all mixed up. I should have written "I compared answers with a friend who got 0.02 N/C to the left, and I can't figure out how they got that answer, so I'm hoping mine is right..."

    Thank you for asking for clarification. I'm sorry that I didn't proof read my question closely enough!
     
  5. Mar 7, 2017 #4

    berkeman

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    Staff: Mentor

    Once you fix your exponent, this looks correct to me.
    Not sure how your friend could have gotten this. Maybe ask them to show your their work?
     
  6. Mar 9, 2017 #5
    Thank you!
     
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