1. Sep 10, 2006

benndamann33

Question: The electric field on the surface of an irregularly shaped conductor varies from 56.0 kN/c to 28.0kN/c. calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is (a) greatest and (b) smallest.

I am stuck as far as how to get an exact number. The only thing I concluded was that at the largest radius of curvature, charge density is smallest, and I called this lamda. Because electric field ranges to double the smallest value, I concluded at the smallest radius of curvature the charge density is 2 lamda. Is this totally incorrect?

2. Sep 10, 2006

Kurdt

Staff Emeritus
The charge density is proportional to 1/(radius of curvature at that particular point).

3. Sep 10, 2006

benndamann33

I understand that, but I'm just given the information above, which is simply an electric field, so I couldn't find the exact charge density any more than what I have above, could I?

4. Sep 10, 2006

Kurdt

Staff Emeritus
Ok correct my mistake I totally read that wrong. The equation relating the potential and the electric field just outside the surface is:

$$v=aE=\frac{a\sigma}{\epsilon_0}$$

where a is the radius of curvature at a particular point and sigma is the charge density.

5. Sep 10, 2006

benndamann33

thanks man, I was overlooking that for some reason, appreciate it.

6. Sep 10, 2006

Kurdt

Staff Emeritus
sorry about the original post just didn't have my head screwed on.