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Electric field of a conductor

  • Thread starter kingwinner
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1) A thin, horizontal 0.10m x 0.10m copper plate is charged with 1.0x10^10 electrons. If the electrons are uniformly distributed on the surface, what are the strength and direction of the electric field 0.1mm below the centre of the bottom surface of the plate? (Note: copper is a conductor, so all excess charges will be on the surface)

Should I use the formula E=(eta)/(epsilon_o), or should I use E=(eta)/(2*epsilon_o) ? Why?
[E=electric field, eta=surface charge density]

Second question: to find the surface charge density
eta=Q/Area
Should I substitute 1.0x10^10 for the charge Q or should I substitute 5.0x10^9 (half of the electrons) as the charge Q?

I am really having some problem understanding how to calculate this...can someone help me? Thank you!
 

Answers and Replies

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well even if it's thin, it still has some thickness. I would consider it to be three dimensional. Does that help?
 
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So should I use the formula for the electric field of a plane of charge: E=eta/(2*epsilon_o) ??

Would there be half (1.0x10^10 / 2) of the electrons on the top surface and half electrons on the bottom surface? How about the sides??
 
1,270
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Can someone help me please?
 
1) is it a plane?
- plane is supposed to be infinite, right?
try to find expression for electric field on the surface of conductor using Gauss's law
 
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Should I use the formula E=(eta)/(epsilon_o), or should I use E=(eta)/(2*epsilon_o) ? Why?
[E=electric field, eta=surface charge density]
Your choice. Either approach will work if done correctly.

Since it's a conductor, the charge will distribute itself across all surfaces. So half the charge is on the top surface; half is on the bottom. (Ignore the thin sides!)

You can treat it as a conductor, using the appropriate formula for the field at the surface of a conductor, but be sure to use just the surface charge, which is only half the total charge.

Or you can just treat it as a plane of charge by lumping both surfaces together to get the total charge. (Or you can treat it as two planes of charge and add them up.)
 

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