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-   -   Charging by conduction with and odd net charge? (http://www.physicsforums.com/showthread.php?t=656873)

Legaldose Dec4-12 09:35 PM

Charging by conduction with and odd net charge?
 
Hello, I've just got a quick question about charging an object by conduction. I've been studying for my final and I came across a problem that I can't really see how to solve.

Say you have two conducting spheres, A and B. They are both exactly the same in every way except for their electric charge. Say A is electrically neutral and B has a charge of -3e. Now what happens when they touch? I was under the impression that when they come into contact they would share the charge equally, but you can't have a charge of -(3/2)e. So what happens?

Thanks, and happy physics!!

berkeman Dec4-12 10:09 PM

Re: Charging by conduction with and odd net charge?
 
Quote:

Quote by Legaldose (Post 4184740)
Hello, I've just got a quick question about charging an object by conduction. I've been studying for my final and I came across a problem that I can't really see how to solve.

Say you have two conducting spheres, A and B. They are both exactly the same in every way except for their electric charge. Say A is electrically neutral and B has a charge of -3e. Now what happens when they touch? I was under the impression that when they come into contact they would share the charge equally, but you can't have a charge of -(3/2)e. So what happens?

Thanks, and happy physics!!

Small numbers like that aren't realistic, but I'm pretty sure it would be random which sphere the one extra electron would end up on. Think thermal randomness...

Naty1 Dec5-12 12:44 PM

Re: Charging by conduction with and odd net charge?
 
one would think the three electrons were initially equidistant from each other on the one sphere. Touching the spheres together would enable at least one of the electrons to move to the second sphere.....so I like the 'random' explanation already provided. But an argument might be made that in most cases, two would remain on the initial sphere.


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