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Electric Field on the Axis at Certain Point

  1. Sep 27, 2009 #1
    Two point-charges of +8 nC and -17 nC lie on the x-axis at points x = 0 and x = +10 m, respectively. Find the electric field on the axis at point x = +5.0 m. Assume vacuum. Answer in N/C.

    E= k0(q./r^2)

    Ok, this should be pretty simple, but for some reason it's not. First, I converted the nC to C: +8E-9 and -1.7E-8. I assume the radius is +5 which would then be +25 because it's squared. K0 is 8.99*10^9, right? Then I just plugged in my values:

    E= -1.7E-8/25 + 8E-9/25= -3.6E-10
    -3.6E-10 * 8.99E9= 3.2364

    This answer is wrong though. I've tried entering it in as both a positive and negative number, but that doesn't make a difference. Where am I going wrong? This is the only way I can see to solve the problem.
     
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 27, 2009 #2

    Delphi51

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    Homework Helper

    Note that the E field due to the positive left charge is to the right.
    The E field due to the negative charge on the right is to the right as well.
    So you add the two fields.

    Leaving out the k and putting the brackets in the wrong place confused me, maybe you as well. Look at it this way:
    E = kQ/R^2 + kQ/R^2
    = k*8E-9/5^2 + k*17E-9/5^2
    = k/25*25E-9
    = 8.99 N/C
     
  4. Sep 27, 2009 #3
    Thanks! I would've never guessed to add them. I'll take out those brackets too. Now that I look at it, they are kind of confusing.
     
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