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Electric Potential and Second Charge

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data
    The electric potential at a position located a distance of 18.6 mm from a positive point charge of 7.50×10-9C and 10.9 mm from a second point charge is 1.02 kV. Calculate the value of the second charge.


    2. Relevant equations
    q2 = r2 * (v - (k * q1 / r1)) / k


    3. The attempt at a solution
    v = 1.02 kV = 1.02 x 10^3 V = 1020 V
    q1 = 7.50 x 10^-9 C
    r1 = 18.6 mm = 0.0186 m
    r2 = 10.9 mm = 0.0109 m
    k = 9 x 10^9 Nm^2/C^2

    q2 = (.0109) * (1020 - (9E9*7.5E-9/.0186)) / 9E9
    q2 = -3.15E-9 nC

    I keep getting the same answer but the homework system says it's wrong!
     
  2. jcsd
  3. Sep 10, 2014 #2

    BvU

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    I get -3.16 10-9 C (not nC). Rounding off error ? Or is it the nC instead of the C ?
     
  4. Sep 10, 2014 #3
    Wow. I feel like a total idiot. IT IS C instead of nC! I figured it couldn't be a unit error because the homework system usually notifies us of that.

    Thanks so much!
     
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