# Non uniformly charged circular plate

1. Jun 15, 2012

### klabautermann

1. The problem statement, all variables and given/known data
Find the electric field above the center of a flat circular plate of radius a when the charge distribution is br$^{2}$δ(z) for r≤a

2. Relevant equations

3. The attempt at a solution
i didn't get the intgeral resulting form coulomb's law working, so i tried with the potential, the result i got is:

θ(z)=$\frac{b}{4ε}_{0}$[$\frac{2}{3}$(z$^{2}$+a$^{2}$)$^{\frac{3}{2}}$-2z$^{2}$(z$^{2}$+a$^{2}$)$^{\frac{1}{2}}$+$\frac{1}{2}$z$^{3}$]

i have my doubts...

Last edited: Jun 15, 2012
2. Jun 15, 2012

### SammyS

Staff Emeritus
Show how you got that, so we can help. (It's also a requirement for this Forum, that you show some work.)

3. Jun 15, 2012

### klabautermann

calculating the potential in polar coordinates leads to:

θ(z)=$\frac{1}{4πε}$∫$\frac{r'^{2}δ(z')}{(z^{2}+r'^{2})^{\frac{1}{2}}}$r'dr'd$\varphi'$dz'

bei integrating over z' i get rid of the delta function, integrating over phi yields a factor of 2π. now substituting r'$^{2}$ by x gives a differential dx=2r'dr' wich kills the last r'. now we are left with:

$\frac{b}{4ε}$∫$\frac{x}{(z^{2}+x)^{\frac{1}{2}}}$dx

substituting again with y=z$^{2}$+x. now the integral can be easily evaluated and after substituting back i got the result from the original post.
by the way the primed variables are the source variables.

4. Jun 15, 2012

### rude man

This is just an infinitely thin disc centered on the x-y plane at z = 0, since
σ(r) = ∫σ(r,z)δ(z)dz.

So I would not use potential but just integrate an annulus of charge at r = 2πrσ(r)dr from r=0 to r=a. You're only required to find the field along the z axis, thank goodness.

5. Jun 15, 2012

### klabautermann

hi!
thanks for your reply! it is quiet clear to me that it is just an infinitly thin charged disk. i tried what you are suggesting, but i wasn't able to integrate that thing. so i treid to avoid the integral by taking a detour over the potential, also my calculations are restricted of course to the symmetry axis as you can see. but thanks anyway!

6. Jun 15, 2012

### vela

Staff Emeritus
I think you messed up the algebra near the end. I got
$$V(z) = \frac{b}{6\epsilon_0}\big[\sqrt{a^2+z^2}(a^2-2z^2)+2|z|^3\big]$$

7. Jun 15, 2012

### klabautermann

that is very possible! thanks for checking, i will go over it again!

8. Jun 15, 2012

### klabautermann

excuse me, would you mind telling me how you got this result, i cant find the error i made.

9. Jun 15, 2012

### rude man

If you give up with potential, let me know! I maintain the integration is easy.

10. Jun 15, 2012

### gabbagabbahey

You need to be a little careful when determining the field from the potential. Is knowing the potential on the $z$-axis really sufficient for finding the field on the $z$-axis? How do you know how the potential varies with respect to $x$ and $y$? If you don't know that, then how can you properly calculate its gradient, and hence the electric field?

Unless you want to calculate the potential everywhere, I'd suggest following rudeman's advice and calculating the electric field directly. If you get stuck, post your work and explain where you are stuck.

11. Jun 16, 2012

### vela

Staff Emeritus
I evaluated the integral by hand (and Mathematica gives me the same answer). Post your work if you can't find your mistake.

12. Jun 16, 2012

### vela

Staff Emeritus
In this problem, you really do only need to know how the potential varies as z changes, but either method is pretty straightforward.

13. Jun 16, 2012

### klabautermann

good morning everybody,

again, it is quiet clear to me that i only need to determine the field along the z-axis. it is also clear to me that it doesn't matter if i calculate the potential or the e-field.

vela: i assumed you did the integral by hand, but what i wanted to know what did you do differently (the third entry states clearly what i did).

gabbagabbahey: why would i be interesseted in how the field varies with x and y. i am only interessted in the field along z, so yes the potential along the z axis gives me all the information i need. by symmetry, the field can only have a z-component on the z axis.

14. Jun 16, 2012

### vela

Staff Emeritus
I used the same method you did. If you need help spotting your error, you need to post the details of your work.