Surface Charge Density (Electric Fields)

[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

 

[DeldotB]

Your equation is good. (assuming o/(2e0) is σ/(2ε0)

and is 7.5 cm in radius

Is this supposed to say 7.5 cm in diameter? Or is the radius acually 7.5 cm? did you convert cm to m?

 

[Jimbob999]

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

 

[gneill]

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?

 

[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?

 

[gneill]

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?

You would have to take into account the actual distance of each surface from the location of interest.

Does the equation only tell you the electric field of one particular side of an object normally?

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.

 

[Jimbob999]

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.

Thanks

 

[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.

 

[DeldotB]

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.

Your equation is correct in the limit that r→∞

 

[gneill]

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

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.

 

[DeldotB]

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.

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.

 

[gneill]

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

 

[Jimbob999]

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

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

 

[gneill]

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

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