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!

Change in electric potential energy

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A tire of radius 32.5 cm has a surface that is conducting. A charge of 4.89×10-8 C is then placed on the surface. The potential on the tire's surface is 1.35E+03 V. Later, suppose that some of the air is let out of the tire, so that the radius shrinks to 10.5 cm, the new potential on the tire's surface is 4.19E+03 V.

    What is the change in electrical potential energy during the shrinking process?

    2. Relevant equations

    Change in electric potential energy = q (change in electric potential)

    3. The attempt at a solution

    Delta U = qDelta V) = (4.89e-8C)(4190J/C-1350J/C) = 1.4e-4 J

    But that doesnt work.
     
  2. jcsd
  3. Feb 14, 2010 #2
    There is an equation for V, just electric potential, that involves charge and distance (r).
     
  4. Feb 14, 2010 #3
    V = KQ/r and then I can solve for the potential for both cases again...

    Still doesnt help find the change in potential ENERGY
     
  5. Feb 14, 2010 #4
    Ok maybe I am reading this wrong but it says that the surface of the tire already has an electric potential. Then R is decreased, I guess the pressure released does work, and brings the surface to a new potential. So you need to find what the initial charge on the tire was and you can do this because you have V and R.

    You know what... I am now not sure.
    Its easy if all we are worried about is the point charge on the surface. But I am assuming, maybe incorrectly that the tire already has charge and we are also bringing all those charges in a circle closer together...

    Sorry, and I will await anothers response
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook