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!

Induced charge in conductor

  1. Nov 28, 2012 #1
    Say you had a rectangular piece of conductor perfectly balanced on a central pivot point. Around it, you put something that causes an electric field throughout the region. On one end of the balanced piece, atomic cores get exposed as the electron sea rushes to the opposite side via induction. So would the piece now turn on the pivot, since one side was heavier than the other (one side has more electrons than the other)? Or does the force on the electrons somehow keep it up and oriented in the same direction despite weight differences?

    I know this would probably never be able to be done in real life because of how small the mass is, I'm just wondering.
  2. jcsd
  3. Nov 28, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Lets tidy up to make this easier to think about:

    You have a conducting beam which is initially balanced and you apply an external uniform electric field horizontally along the length of the beam. This causes a shift in the distribution of electrons to one side of the beam - which would shift the center of mass of the beam away from the pivot point and so it should tip up.

    That seems reasonable.
    You could imagine a negatively charged pivot and a non-conducting beam with a positive charged slug constrained to move along it's length. In this case you can construct a free-body diagram for the situation ... there would be a force ##qE## from the field, another ##kq^2/x^2## back along the beam, then there is gravity ##mg## acting down.

    Have fun.

    A version of this experiment has been done in real life - famously, by Millikan - using oil droplets in free-fall instead of a charged slug on a beam.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook