Induced charge in conductor

  • Thread starter mishima
  • Start date
  • #1
543
33
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.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
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.
 

Related Threads on Induced charge in conductor

  • Last Post
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
20
Views
6K
Replies
5
Views
401
  • Last Post
Replies
1
Views
1K
Replies
1
Views
618
  • Last Post
Replies
2
Views
1K
Replies
2
Views
526
Top