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Homework Help: Potential Difference between Concentric Circles

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Three concentric circles of radii 1.5m, 2.5m, and 4m are filled with a gas that breaks down in electric fields greater than 1.6 x 10^7 volt/meters. What is the highest potential difference that can be maintained between the innermost circle and the outermost circle. (Hint: the middle circle must be maintained at a potential such that breakdown of the gas is about to occur on its outer surface as well as on the surface of the inner most circle).

    2. Relevant equations
    Change in potential from A to B = [itex]\int E \cdot dl [/itex]

    3. The attempt at a solution
    I calculated the net charge that the inner most circle needed to have to produce an electric field of 1.6 * 10^7 V/m at its surface. I also calculated net charge needed by the 2nd circle so that the sum of the two fields would be 1.6 * 10^7 V/m at the outer surface of the second circle (as per the hint given). However, when I then integrate the net electric field from 1.5 m to 4.0 m, I don't get the answer given in the book.

    The answer given in the book is 31.1 MV for reference. The book also claims that 2nd circle must be maintained at a potential of 18.8 MV (I think relative to the outermost circle).
  2. jcsd
  3. Jan 15, 2010 #2
    Problem solved. I realized that the "circles" in question were actually supposed to be cylinders (of unspecified length). In that case, the field generated by the cylinders drops off as 1/r, not 1/r^2 as I had assumed (see http://www.davidpace.com/physics/em-topics/capacitance-cylinders.htm#em6eq2 [Broken])

    Thus if E = Q / (2 Pi e0 L r), let s = Q / (2 Pi e0). For the first cylinder, s1 must be equal to 2.4 * 10^7 so that on its surface the field is 1.6*10^7. Then s2 for the second cylinder is 1.6*10^7, so that above its surface the field is 1.6*10^7. Then we integrate s1 from 1.5m to 2.5m and then integrate (s1+s2) from 2.5m to 4.0m. The sum of these two integrations is 3.106*10^7 volts ~ 3.11 * 10^7 volts = 31.1 MV
    Last edited by a moderator: May 4, 2017
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