How to calculate electric charge with a given electric potential?

[Bobcent]

Hi!

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?

 

[DrZoidberg]

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.
http://en.wikipedia.org/wiki/Capacitance#Self-capacitance
The charge of the plate is the electric potential times it's capacitance.

 

[Bobcent]

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.

 

[BruceW]

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:
\nabla^2 \phi = - \frac{\rho}{\epsilon_0}
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.

 

[Bobcent]

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.

 

[DrZoidberg]

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.

 

[BruceW]

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.