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Equipotential surface

  1. May 29, 2006 #1
    "For a point charge of +3.50 microC what is the radius of the equipotential surface that is at a potential of 2.40 kV?"

    For a point charge, the voltage = kq/r, where k =9*10^9 N*m^2/C^2, q is the charge, and r is the radius. Solving for r, I get r = 13.1 m. But the answer is 12.6m...
  2. jcsd
  3. May 29, 2006 #2
    Your method is correct.
    The difference isn't all that alarming and has probably risen in your approximate value of k .
  4. May 29, 2006 #3
    ok, thanks!
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