Uniformly charged disk and the E field some distance Z from the center

In summary, the conversation discusses finding the electric field of a uniformly charged disk at a distance z from its center. The equation for the E field is given as E = 2\pikσz(1-\frac{z}{\sqrt{z^{2}+R^{2}}}), but a typo is identified and the correct expression is E = 2\pikσ(1-\frac{z}{\sqrt{z^{2}+R^{2}}}). The binomial expansion of (R/z) is then discussed, with the final conclusion that the E field decreases as 1/z for large distances.
  • #1
anban
20
0

Homework Statement



Hi,

I have a problem that describes a uniformly charged disk and the electric field a distance z from the center.

I have found an equation that describes the E field at any point z already. Now I have to find out how the E field decreases as z increases-- as 1/r^2, I assume, but I am not sure.

Homework Equations



E = 2[itex]\pi[/itex]kσz(1-[itex]\frac{z}{\sqrt{z^{2}+R^{2}}}[/itex]) where R is the radius of the disk and z is the distance away from the disk.

The Attempt at a Solution



I know I have to do a binomial expansion of (R/z). I think I am having mathematical issues rather than physical issues.

Before the expansion, I need to get some term (R/z). Does taking out a z^{2) from the denominator leave a z^{4} out front? Or a z^{3} out front? If I can figure this out then I know the rest.
 
Physics news on Phys.org
  • #2
##\sqrt{z^2+R^2} = \sqrt{z^2(1+(R/z)^2)}## and ##\sqrt{(a b)} = \sqrt{a}\sqrt{b}##
 
  • #3
anban said:
E = 2[itex]\pi[/itex]kσz(1-[itex]\frac{z}{\sqrt{z^{2}+R^{2}}}[/itex])

I believe you must have a typo in this expression. The dimensions on the right do not match the dimensions of electric field.
 
  • #4
Good catch: The expression should be E = 2[itex]\pi[/itex]kσ(1-[itex]\frac{z}{\sqrt{z^{2}+R^{2}}}[/itex]).

So, the denominator of the last term can be rewritten as z[itex]\sqrt{1-(R^{2}/z^{2})}[/itex]. After a binomial expansion, I got that the E field decreases as 1/z.
 
  • #5
anban said:
I got that the E field decreases as 1/z.

I don't think that's correct. Check your binomial expansion simplification. Your original idea that the disk should behave as a point charge for large distances is right.
 

Related to Uniformly charged disk and the E field some distance Z from the center

1. What is a uniformly charged disk?

A uniformly charged disk is a circular disk with a constant charge density, meaning that the charge is evenly distributed across the entire surface of the disk.

2. How is the electric field (E field) of a uniformly charged disk calculated?

The electric field of a uniformly charged disk can be calculated using the formula E = (σ/2ε0)(1- cosθ), where σ is the charge density, ε0 is the permittivity of free space, and θ is the angle between the point of interest and the center of the disk.

3. What is the direction of the E field at a point some distance Z from the center of a uniformly charged disk?

The E field at a point some distance Z from the center of a uniformly charged disk is perpendicular to the surface of the disk and points away from the disk.

4. How does the E field vary with distance from the center of a uniformly charged disk?

The magnitude of the E field decreases as the distance from the center of the disk increases. This relationship follows an inverse square law, meaning that the E field decreases by a factor of 1/r2 as the distance from the center increases.

5. How is the E field affected if the uniformly charged disk has a larger radius?

If the radius of the uniformly charged disk increases, the magnitude of the E field at a given distance from the center will decrease. This is because the charge is spread out over a larger surface area, resulting in a lower charge density and therefore a weaker E field.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
551
  • Introductory Physics Homework Help
2
Replies
36
Views
583
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
344
  • Introductory Physics Homework Help
Replies
6
Views
316
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
275
  • Introductory Physics Homework Help
Replies
5
Views
7K
Back
Top