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How to calculate electric charge with a given electric potential?

  1. Mar 31, 2013 #1

    An iron plate with a mass of 1 kg surrounded by vacuum has an electric potential of +20 kV. How do I calculate the charge, Q of the iron plate?

    I have all the information I need, right?
  2. jcsd
  3. Mar 31, 2013 #2
    You also need the shape of the plate (thickness, diameter). And I don't know if you are talking about self capacitance here or if the plate is near another plate.
    The charge of the plate is the electric potential times it's capacitance.
  4. Mar 31, 2013 #3
    Thanks for your reply!

    The iron plate is not near another plate, so I'm talking about self capacitance.

    But I don't have the capacitance.

    Is the capacitance a value based on the shape and the electric properties of the material? How do I calculate it?

    Say that the shape of the 1 kg iron plate is a square with an area of 0.225 m^2. That makes the thickness 0.0025 m since the density of iron is 7874 kg/m^3.
  5. Apr 1, 2013 #4


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    It is going to be pretty complicated expression because it is not a highly-symmetric object. (For example, a sphere would give a nice answer). But for your problem, there is still an exact answer. Assuming the charges in the object are fixed, we than have the electrostatic equation:
    [tex]\nabla^2 \phi = - \frac{\rho}{\epsilon_0} [/tex]
    This is the 'exact answer' I was talking about. From here, it gets a bit complicated, because the charge distribution rho does not follow a simple symmetry. Of course, you could take the limit of some point near the surface of the object, in which case, you can get a simple (approximate) answer.
  6. Apr 1, 2013 #5
    Thanks for your reply Bruce, and happy Easter!

    Is there a difference in charge distribution for a positively charged object compared to a negatively charged object? The iron plate would be positively charged if that makes any difference.
  7. Apr 1, 2013 #6
    Did you look at the wikipedia page for self capacitance? For a circular disc it's approximatly 8εr. r is the radius of the disc and ε = 8.85× 10−12 F/m. The thickness doesn't matter as long as the radius is much bigger than the thickness. So a thin big plate will have a larger capacitance than a small thick one.
  8. Apr 1, 2013 #7


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    the only difference in the potential for a positively or negatively charged plate is that the sign of the potential would be opposite. (using the convention that the potential at infinity is zero). Also, DrZoidberg brings up a good point. (p.s. I'm a futurama fan too). I guess that for a circular disk, you can use a cylindrical coordinate system, and take the approximation that the disk is very flat, then I guess it is not too difficult to do the calculation. I have not done this calculation myself, but I would guess that is the general idea.
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