Static charge

1. Mar 22, 2010

ctech4285

i been watching some lectures from OCW and the professor said something like a ball with a radius of 1m can not exceed 3MV because of the electric breakdown of air.

to me it sounds like you can not exceed a potential difference 3MV per meter. so it does not matter the physical size of an object just the voltage difference over a distance is less.

2. Mar 22, 2010

Andrew Mason

It is the strength of the electric field, not the potential of the ball that determines when the air breaks down.

You can use Gauss' law to determine the electric field:

$$\int E\cdot dA = \frac{q}{\epsilon_0}$$

$$E = \frac{q}{4\pi r^2\epsilon_0}$$

The electric potential outside the sphere is the same as if you were dealing with a point charge:

$$V = \frac{q}{4\pi r\epsilon_0}$$

which means that E = V/r

So you can see that for a given voltage of a conducting sphere, the field strength varies as 1/r.

AM

3. Mar 22, 2010

Naty1

http://en.wikipedia.org/wiki/Electrical_breakdown