1. Limited time only! Sign up for a free 30min personal 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!

Change in electric potential

  1. May 20, 2009 #1
    http://users.on.net/~rohanlal/elec.jpg [Broken]

    for question b, wouldn't there be no change in electric potential since
    electric potential for a particular point is determined by the sum of the electric potentials for
    each particular charge about that point, and if a charge is moved to point p
    and the eq for electric potential is kq/r then that charge would be 0 since r would be 0.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 20, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    In b) they are asking for the change in potential energy. This is not the same as asking for the electric potential - i.e. the voltage.

    The change in potential energy is going to be expressed as the work required to move the charge.

    For b) then what you want to explore is work given by ...

    W = qΔV

    The voltage at ∞ is 0, so ... ΔV = simply the voltage at P.
     
  4. May 20, 2009 #3
    According to my lecture notes electric potential is the same as potential energy.

    the answer to this problem requires the initial electric potential of the system which is found by

    http://users.on.net/~rohanlal/pot.jpg [Broken]

    to be subtracted from 3.0x10^-6
    which gives a final answer of - 1.89 x 10-2

    but i don't understand why this is the case
     
    Last edited by a moderator: May 4, 2017
  5. May 20, 2009 #4

    diazona

    User Avatar
    Homework Helper

    Your lecture notes are wrong, then...

    Mathematically, electric potential is
    [tex]V = k\frac{q}{r}[/tex]
    whereas electrical potential energy is
    [tex]U_E = k\frac{q_1 q_2}{r}[/tex]
     
  6. May 20, 2009 #5

    LowlyPion

    User Avatar
    Homework Helper

    It's not subtracted.

    The Work is the charge times the change in voltage potential.

    What they want is Work.

    And if you multiply the charge by the Voltage, then you get the answer indicated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Change in electric potential
Loading...