# Surface Charge Density (Electric Fields)

1. Jul 19, 2015

### Jimbob999

A disk with a uniform positive surface charge density lies in the x-y plane, centered on the origin. The disk contains 2.5 x 10-6 C/m2 of charge, and is 7.5 cm in radius. What is the electric field at z = 15 cm?

I have used the formula:

http://edugen.wileyplus.com/edugen/courses/crs7165/halliday9781118230725/c22/math/math121.gif
(if this picture doesn't appear it is E=o/2e0 x (1-(z/sqrt(z^2+r^2)))

Using 2.5x10-6C/M2 as my surface charge density, I get a final result of 1.5x10^4 N/C

However it isn't one of the answer options, although it is exactly half of one of the options, so I'd lean towards that if I was guessing.

Here are the possible options:

30 N/C

300 N/C

3000 N/C

3.0 x 10^4 N/C

3.0 x 10^7 N/C

2. Jul 19, 2015

### DeldotB

Your equation is good. (assuming o/(2e0) is σ/(2ε0)
Is this supposed to say 7.5 cm in diameter? Or is the radius acually 7.5 cm? did you convert cm to m?

3. Jul 19, 2015

### Jimbob999

I copied the question exactly as it is written. I did convert to m.

4. Jul 19, 2015

### Staff: Mentor

The question does not state a thickness for the disk, and it's unlikely to be comprised of conductive material if its uniformly charged in that shape. I guess you have to assume that it's much thinner than it is wide. How many (flat) sides does a disk have? Does that give you any ideas?

5. Jul 19, 2015

### Jimbob999

Oh, so given a disk has two sides, then I would just x2 my answer. What would I do if I was given a thickness?
Does the equation only tell you the electric field of one particular side of an object normally?

6. Jul 19, 2015

### Staff: Mentor

You would have to take into account the actual distance of each surface from the location of interest.
You should check the derivation of the formula. Applying a formula without knowing what it actually represents can be risky.

I can tell you that the formula $E = \frac{δ}{2 ε_o}$ yields the field strength on either side of a uniform sheet of charge.

7. Jul 19, 2015

### Jimbob999

Thanks

8. Jul 19, 2015

### DeldotB

The disk is of finite radius. It is not an infinite sheet.

I have done the derivation of that formula many times. It's for one side of a uniform conducting disk of surface charge density σ.

Edit: I get the same answer. Notice if you doubled your answer you would have 3.0x10^4. I don't see why you would double it though unless there is other given information.

Last edited: Jul 19, 2015
9. Jul 19, 2015

### DeldotB

Your equation is correct in the limit that r→∞

10. Jul 19, 2015

### Staff: Mentor

Yes, hence the (1 - ...) term that accompanies the $\frac{δ}{2 ε_o}$.

The point was to recognize the portion of the formula that corresponds to an infinite sheet has the "2" in the denominator for a reason.

11. Jul 19, 2015

### DeldotB

Ok, misunderstanding. I thought you were mentioning a formula to be used in this instance.
P.s nice Tardis, I have one on my desk in front of me.

12. Jul 19, 2015

### Staff: Mentor

Okay, so assuming that the disk is actually infinitely thin (like a disk punched out of a sheet of charge), then Jimbob999's answer would be correct, and the supplied list of possible answers does not include it.

This can happen sometimes due to a typo in the problem parameters, or when an author tweaks a problem to make it "new" and fails to update the answer key. Sometimes the problem can be a problem

13. Jul 19, 2015

### Jimbob999

So it wouldn't be twice my answer as there are two sides to the disk?

14. Jul 19, 2015

### Staff: Mentor

Not if the disk is only a single layer of charge. Come to think of it, I suppose they would have called it a cylinder if it had vertical thickness. That's me not thinking