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Homework Help: Electric field as a function, find potential diff

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    the electric field between two conducting surfaces is given by an expression:

    E(x) = 9x^2 <i hat> where x is any location between the conductors. if one conductor is at x = 1 meter and the other is at x = 3 meter, what is the potential difference between them


    2. Relevant equations

    electric potential difference deltaV = V_B - V_A = - [<integral> E*dx] where x is distance

    3. The attempt at a solution

    i'm guessing i am supposed to do do the integral with the limits being from x = 1 to x = 3 so:
    deltaV = [9x^3/3]x from x = 1 to x = 3, so ((9*27)/3)*3 - (9/3)*1 = 243 - 3 = +240 volts

    i am not too sure about the correctness of my integration, but even still, was my approach correct?

    thanks
     
  2. jcsd
  3. Jun 9, 2008 #2

    rl.bhat

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    deltaV = [9x^3/3]x
    this is wrong. deltaV = [9x^3/3]
     
  4. Jun 9, 2008 #3
    i thought my integration was wrong, i had it the way you have before, but doubted myself, well i redid the intrgral and did the calculation:

    deltaV = [9x^3/3] from x = 1 to x = 3, so ((9*27)/3) - (9/3) = 81 - 3 = 78 volts

    correct now?
    thanks
     
  5. Jun 9, 2008 #4

    Redbelly98

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    Yep, looks good. :smile:
     
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