Surface Area and electric field strength

1. Sep 10, 2014

Samson4

This is my first post but I have frequented the forum for a little while now. I tried to figure things out myself and often times I am lead here by google. So my question is this:

How is the electric field of an object altered when the surface area of the object is altered?
Example: Take a perfectly smooth sphere and charge it to 100 volts. Then take the same sphere and etch it to increase it's surface area 100 times. By that I mean similar to etching aluminium with hcl in capacitor production. When comparing the electric fields, are they different in anyway?

My first thought was that by increasing the surface area; therefore capacitance, you would have more electric field lines on the altered sphere. Since those must terminate on a conductor or at infinity, it would have a stronger electric field. I was thinking with the idea that every charge carrier gets an electric field line. I don't think that is correct.

Now I think the answer is that the electric fields are the same. However, the altered sphere would feel a stronger force if placed in an external electric field. Instead of individual charges getting field lines, I think it's more accurate to say that the electric field is a measure of total surface charge pressure.

Can you guys shed some light on this or point me to something that will help me understand?

Last edited: Sep 10, 2014
2. Sep 10, 2014

Staff: Mentor

Yes, at any distance from the object. As long at the object is spherically symmetrical, there is some distance away at which its electrical field will be indistinguishable from that of a point charge. (If this were not true, we would find $F=Cq_1q_2/r^2$ to be much less useful).

No. Although you increase the surface area by making the object rough or spiky, you also decrease the average charge density per unit area. Grovel through the surface integral and you'll end up with the same net force, as long as the applied electrical field is constant in the neighborhood of the object.

If you haven't already done so, check out Gauss's Theorem (google will find it). It's not quite directly applicable, but it is a very powerful tool for making the shape of the distribution of charge within a volume irrelevant outside that volume.

3. Sep 10, 2014

Samson4

Sorry, I tried to be very specific with my question. I forgot to mention that the new sphere would also be charged to 100volts. So it would have 100 times the charges on it's surface. You still answer that question even though I didn't explain it enough.

To really clarify, I'd like to give another example. We have two spheres, a and b. A is smaller than b; but a has higher surface area than b. A is placed inside b and charged to 100volts. They are insulated from each other. Gaussian theorem states that the outer sphere would create an equal charge to the inner sphere. How does sphere b's electric field compare to a's. Are they still the same and if so, what happens to the charge from all the extra carriers needed to bring sphere A to 100v because of it's higher capacitance.