Homework Help: Zero Potential In Uniform Electric Field

1. Apr 4, 2006

daletaylor

Hi,

I actually have two questions...

When dealing with a uniform electric field, where do we normally chose to 0V to be?

Also, if we have a charged plate, we can treat it as a charged plane to use Gauss' Law right? Now, if we figure out the magnitude of the field created by that plane, we end up with:

(SCD)/ 2(Permittivity constant)

where (SCD) is the surface charge density of the plane.

Now because the magnitude of this field does not depend on the distance from the plane, the field has the same magnitude at every point in front of it, right?

Do does this mean that if I hung a charged sphere from the ceiling and took a metal plate and put sufficient charge on it, and pointed it at the sphere it would move to a point where is was hanging with a certain angle to the vertical? And furthermore it would stay at that same angle no matter how far I went with the plate?

Thanks,
Dale

2. Apr 4, 2006

Andrew Mason

'Volts' are units of measure of the potential difference between two points. There is no such thing as a point of 0V except in relation to another point.

Not unless it is an infinitely large plane.

If the sphere is much smaller than the plane, it will see a uniform electric field. Assuming that if the field is turned off the sphere hangs vertically, when the field is turned on it would experience a horizontal force equal to qE. This would cause it to swing horizontally and vertically until the forces of gravity, tension and electricity balanced (or until the string breaks). To then move the plate relative to the charge, you would have to do work against the electrical force between the plate and charge, so the electrical energy of the sphere relative to the plate would change. But the force on the charge would not change.

AM

3. Apr 4, 2006

daletaylor

Thank you very much!

Just out of curiosity, how would one calculate the electrical energy between the plane and the sphere?

There is a formula for the potential energy,

U = [(Ke) * q1 * q2] / r

but this seems to be only for point charges. How would you do this for a plane of charges?

Thanks,
Dale

4. Apr 4, 2006

Hootenanny

Staff Emeritus
The potential of a single plane is given by basically repeating the point charge equation and leads to the following intergral;

$$\int^{a}_{-b} \frac{kdq}{r}$$

The potential on the surface of a sphere is identical to that of a point charge.

-Hoot

 latex still isn't working so i've included an image from hyperphysics;