# What is the magnitude of the electric field produced by the disk

1. Jan 27, 2005

### bbbbbev

A disk of radius 1.4 cm has a surface charge density of 4.9 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 12 cm from the disk?

I tried solving this problem in the same way you would solve a similar problem with a ring instead of a disk, using the equation E = ((k)(z)(surface charge density))/r, where r = sqrt(z^2 + R^2). z is the y-component of r, the distance between the charges, and R is the radius, which is the x-component of r. But I just realized that that equation solves for E_x, I think. I don't know how to solve for E because I don't know theta or E_y. Could someone please explain this problem!? I feel like I have no idea what is going on with the situation laid out in the problem. I'm lost. I would really appreciate any kind of help you could give!

Thanks, Bev

2. Jan 27, 2005

### dextercioby

$$d\vec{E}=\frac{dq}{4\pi\epsilon_{0}a^{2}}\frac{\vec{a}}{|\vec{a}|}$$ (*)

Where "dq" is the infinitesimal electric charge of a surface element of the disk "dS" which is found at the distance "a" from the point at which u wish to compute the electric field...

The ratio involving the vector in the RHS of (*) is the unit vector along the "a" direction.

Use the symmetry of the problem (axial symmetry) and then the necessary double integral to find your result.

Daniel.

3. Jan 27, 2005

### bbbbbev

Thanks, that helped a lot. I at least *think* I am setting the problem up correctly, but I only have one more chance to get the answer right, so I was wondering if you could just look at my work and see if it looks right. Here is my work (R = radius of disk):

dE_x = (k(dq))/a^2
dE_x = (k(surface charge density)(ds))/a^2
dE_x = (k(surface charge density)(R)(dtheta))/a^2
dE_x = dEsin(theta) = (k(surface charge density)(R)(dtheta)sin(theta))/a^2

Then I integrated both sides with the bounds 90 and -90, and I got:

dE_x = ((k(surface charge density)(R))/a^2)(cos(90) - cos(-90))
dE_x = (2k(surface charge density)(R))/a^2
dE_x = 8.45e4

But I'm not sure that that is the final answer (or that I am even doing the problem right). I think dE_y = 0, so the electric field is just pointing directly to the right? Thanks so much for taking the time to help me. I really do appreciate it.

Thanks, Bev

4. Jan 27, 2005

### dextercioby

Forget about the numbers,let's see what u did wrong with the letters...
Draw a picture...Chose that tiny domain "dS" and attach "dq" to it...
$$dq=\sigma dS$$ (**)=(2)

I decided to label numbers for relations...Else,i would have to "paint" too many stars... :tongue2:
Let's call the radius disk by "R".Let's call the distance from the origin to "dq" $\rho$.Let's call the angle made by the vector radius $\rho$ with the Ox axis (on which i chosed the point in which i compute the field at the distance "x",NOT "z") by [itex] \varphi [/tex]

Therefore:
$$dq=\sigma \rho \ d\rho \ d\varphi$$ (3)

From the first tringle,by applying the cosine law u get
$$a^{2}=\rho^{2}+x^{2}-2\rho \ x \cos\varphi$$ (4)

Now project the vector E on the Ox axis and make use of some geometry...
$$dE_{x}=(\frac{x-\rho \cos\varphi}{a})\cdot dE$$ (5)

Now assemble (5),(1/*) (3) & (4) into one expression...

Daniel.

P.S.Tell me what u get.

Last edited: Jan 27, 2005
5. Jan 27, 2005

### bbbbbev

I don't get it. I don't really know what all those expressions mean. I'm sorry. I tried to draw a picture, but I can't tell where I'm supposed to put everything, so the equations aren't making much sense to me. I'm sorry, I just don't know what I'm doing.

6. Jan 27, 2005

### dextercioby

I'm sorry,but i cannot help you mre than i already did...I need some COOPERATION,because i won't be solving the problem for you.

Daniel.

7. Jan 27, 2005

### bbbbbev

Ok, thanks for the helping as much as you did. I'm not trying to be uncooperative; I just don't understand what you are talking about with all those symbols. Like, I don't know which symbol stands for which thing. I don't know what you mean when you say "the first triangle," and i don't know what angle you are talking about using that symbol for, etc. I think I should probably forget about getting it right on this homework assignment and ask my teacher about it later in person rather than doing this over the internet. Thanks for helping me, though, I didn't mean to sound unappreciative.

8. Jan 27, 2005

### MathStudent

Consider another approach, which is a little less mathmatical (I think dextercioby is too smart sometimes for us mortals)

Choose your surface element ds to be a thin ring of radius r.
Then the area of ds is approximately its circumference multiplied by its width ( outer radius - inner radius / which well denote dr )
so you have $$dS=2\pi rdr$$
now how much charge is on ds?
The charge = area multiplied by the charge density (since sigma = Charge/Area )
$$dq=\sigma dS$$
$$dq=\sigma 2 \pi r dr$$

(now you just need to sum up the E from all the rings of radius 0 to R )
so you can plug in dq for the equation you already have for the E of a ring for points on the central axis, and integrate from 0 to R (or whatever the radius of the disk is ,, I forgot)

maybe this approach is easier,, maybe its not
hope it helps
-MS

Last edited: Jan 27, 2005
9. Jan 27, 2005

### bbbbbev

Thank you so much!! That sounds like it makes sense. I'll try it!

10. Jan 27, 2005

### dextercioby

It is easier,but unfortunately incorrect...You have an angle dependence...You cannot make the integral for "phi" and win up with 2"pi"...Take anothe look at it.Maybe make a drawing...

Daniel.

11. Jan 27, 2005

### MathStudent

I am assuming that her equation for the electrical field of the ring already takes care of the angle dependence and thus it represents the E component of the ring that is parallel to the axis through its center

12. Jan 27, 2005

### dextercioby

1.Sorry,i didn't take that into account.Then your method is valid... For this problem.

2.Is he a she???

Daniel.

13. Jan 27, 2005

### MathStudent

hmm... I figured Bev was short for Beverly (scratches head) dunno

If its actually Bevis ( or something of that nature ) than I apologize

Last edited: Jan 27, 2005
14. Jan 29, 2005

### bbbbbev

Yeah, you're right. Bev is short for Beverly. I'm a she.