1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yep, looks good. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?