Electric field of a conductor

Discussion in 'Physics & Math' started by kingwinner, Feb 7, 2007.

  1. kingwinner Registered Senior Member

    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
    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!
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  3. James R Just this guy, you know? Staff Member

    Use the first one, but make sure you get the charge density right!

    Remember that the total charge is spread over TWO surfaces, not one.
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  5. kingwinner Registered Senior Member

    Why should I use the first one? Isn't E=eta/(2*epsilon_o) the electric field of a plane of charge??

    So 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??
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  7. James R Just this guy, you know? Staff Member

    It is the field of an infinitely-thin plane of charge, or the field where the plane is taken to be a non-conductor. Apply Gauss's law to a thin conducting plane with charge density eta on one side and you get E = eta/epsilon_0.

    This is quite a subtle point, which many people get confused about. The problem essentially lies in how eta is defined for a plane with two sides.

    Yes, half on the top, half on the bottom. There's no mention of finite thickness in the problem statement, so assume there are no sides (or that they are infinitessimally thin). If there are sides, then you have a much more complex problem, since you'll need to worry about fringing fields.
  8. kingwinner Registered Senior Member

    So E = eta/epsilon_0 is for conductors and E = eta/(2*epsilon_0) is for non-conductors with uniform charge density, am I right?

    I am one of those people who get confused by this...

    How actually is eta defined for a plane with 2 sides? I don't quite get it...

    And how come when I recall using the formula E = eta/epsilon_0 to find the electric field for an infinite plane of charge (for non-conductors, I guess), I NEVER divide the total charge by 2 to get eta? I am very confused on when to divide the total charge by 2 and when not to...

    What if the same question says "assume the plate has no thickness" and it mentions nothing about the "top surface" and "bottom surface" business? (so I guess the top surface is the same as the bottom surface in this case) Then can I assume the plate is 2-dimensional and use the total charge of 1.0x10^10 electrons (i.e. not dividing it by 2) to find eta?

    And in this case, which formula should I use?

    Thanks for your help!
  9. James R Just this guy, you know? Staff Member

    The important difference here is conductor vs. non-conductor.

    If you take a set amount of charge and place it on a conducting plate, what does it do? First, it all goes to the surfaces of the plate. Second, it spreads across both surfaces of the plate (we're ignoring sides by assuming the plate is very thin), so that half the charge is on one side and half on the other. If the total charge put on the plate is Q, and the plate area is A, then the charge density on each side of the plate is Q/2A.

    Because the plate is a conductor, the field inside is zero. So, consider a Gaussian cylinder with one end inside the conducting material and one end above the centre of the plate. Applying Gauss's law, we have:

    \(\epsilon_0 \oint E.dA = EA = \frac{Q}{2A}A\)

    which gives a field above each side of the plate of

    \(E = \frac{Q}{2A \epsilon_0} = \frac{\sigma}{\epsilon_0}\)

    where sigma is the charge density on one side of the plate:

    \(\sigma = Q / 2A\)

    Now consider a non-conducting plate. We place charge Q on it. The charge can't flow around, so it just stays where we put it. We choose to spread it "evenly" over the plate, which again we assume to be very thin. We can think of the charge as sitting in the "middle" of the plate, half way between its two surfaces. The field inside the non-conducting plate is NOT zero, unlike the conducting case.

    The charge density on the plate in this case is

    \(\sigma = Q/A\)

    (NOT Q/2A), because the charge is assumed to be spread evenly over the plate, but the plate does not have two distinct sides any more. The same charge produces electric field on both sides of the plate.

    To apply Gauss's law, take a cylinder that is centred on the plate. Field lines penetrate the ends of the cylinder on both sides. We get:

    \(2EA\epsilon_0 = \frac{Q}{A}A\)


    \(E = \frac{Q}{A \epsilon_0} = \frac{\sigma}{\epsilon_0}\)

    Notice that the expressions for the field are both given by

    \(E = \frac{\sigma}{\epsilon_0}\)

    Only the value of sigma is different in the case of a conducting plate with two surfaces, compared to a non-conducting plate with the charge in the centre.
  10. kingwinner Registered Senior Member

    Many thanks for your detailed, understandable explanation. I am feeling much better now after reading your post. However, I still have some questions:

    For a conducting plate, will the charges on the TOP surface create an electric field below the BOTTOM surface? Why or why not?

    The charge is distributed uniformly over the non-conducting plate. And also, a plate always have 2 sides: top and bottom, isn't it? Why can you assume that there is only 1 side?
  11. James R Just this guy, you know? Staff Member

    For a conducting plate, the charges on the top surface DO NOT create an electric field below the bottom surface. The conductor itself effectively "shields" the bottom surface from the effects of the charge on top. The field inside a conductor in electrostatic equilibrium is always zero.

    For a non-conductor, the field from charge on top penetrates the material and adds to the field from the charge on the bottom.

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