Electric Field due to a disk of radius R in the xy-plane

[neroE]

Homework Statement

N/A

Relevant Equations

Coulomb's law mainly.

Hello,
This question, which I found in various electricitiy and magnetism books (e.g. Introduction to electrodynamics grif.).

There are many variations of this question, I am mainly interested in the following setup of it:
-Suppose there is a charged disk of radius R lying in the xy-plane, and the electric field is uniform. What is the electric field at the point P(0,0,z)?

I know this can be solved by first considering a ring, deriving a formula for it, and then summing the infinite number of rings that make up a disk (via integration), and thus, the desired result is yielded. However, I am more interested in how to approach this using surface integration?I started with the parametrization G(u,v) = <ucos(v),u*sin(v),0> and found the normal unit vector to be <0,0,1>
where: 0<u<R and 0<v<2*pi

But, I am not sure how to continue from there, I attached a pdf showing my subsequent steps, which I am not sure if it is valid or not.
Please help, and thank you in advance.

 

[BvU]

Hello @neroE ,

$\qquad$ !​

which I found in various electricitiy and magnetism books

Probably not : instead of

and the electric field is uniform

do you mean 'and the charge is uniformly distributed' $\qquad$ ?You also want to

• explain the symbols you are using ($G$ ?) in words or in a picture
• write legibly -- or, better: use $\LaTeX$
• read the PF guidelines

$\$

 

[neroE]

Hi @BvU ,
Yes, sorry, the charge is uniformly distributed.
I did not use any symbols except the common three symbols:
r: radius that varies in the disk (I.e.starts from r=0 till r=R)
theta: which is the angle that kind of describes the ring/disk (I.e. starts from 0 to 2pi).
G: This is just to indicate a mapping is occurring (I.e. G(r,theta) means parametrizing the surface using r and theta, and then: G(r,theta) = <rcos(theta),rsin(theta),0> means x=rcos(theta), y=rsin(theta), z=0.
I apologise if this notation is not that common, I read it in a calculus textbook before.
Lastly, I know some latex; however, I may not be able to write the pdf I attached completely in latex as I did not learn that much latex. So, please excuse my first post not being in latex, and hopefully my next posts will be in Latex as I will be learning it.
Thank you in advance.

 

[vela]

Your expression $dr\,d\theta$ for $dA$ is wrong. It needs to have units of area. You forgot to multiply by the Jacobian, which will give you a factor of $r$.

 

[BvU]

Ok, and I suppose

is the Coulomb constant, a.k.a. $\displaystyle {1\over 4\pi\varepsilon_0}$ .

( re notation: I use separate symbols for the vector in the disk and the vector from $dA$ to point P )

You have $\vec r = \vec p -\vec r'$ with $\vec p =<0,0,z>$ and $\vec r'= <r'\cos\theta,r'\sin\theta,0>$ and $\hat r = {\vec r \over |\vec r|}$
And then I can follow $$d\vec E = {1\over 4\pi\varepsilon_0} \; {1\over r^2}\;{\vec p - \vec r' \over |\vec r| } \; dQ$$
where $dQ = \sigma dA = \sigma r' dr'\, d\theta$ (*)

Note that $dQ$ is a scalar !

( (*) I have the impression you overlooked the factor $r'$ ? Nice to see vela agrees...)​

In sort, I agree with your

(quoting $\LaTeX$ really is a lot more comfortable ).

Except that it looks as if you think $dA$ is a vector and then in the subsequent

you come up with dot product $\vec r \cdot \vec n$

which I am not sure if it is valid or not

I don't think it's valid, but here it works because you pick out the $z$ component, which is the only one that doesn't cancel from symmetry.

$\$

 

[neroE]

Hi,
I learnt some latex, not the best but at least better .

Thanks @vela and @BvU.
I think my main initial problem was treating the problem as a vector surface integration question, and in this, we always convert F*dA(vector) to scalar surface integral by parametrization and finding the normal vector (which is kinda the "jacobian"). But, as both of you have stated, I just realized that it is just a double integration problem and no need for vector fields I guess.
Thanks in advance!

 

[BvU]

Hi,
I learnt some latex, not the best but at least better .

Great ! Would be even better to embed it in the post: in-line math enclosed in double $and displayed math enclosed in double More tips: use \cos and \sin and$\theta$instead of$(\theta)##

I think my main initial problem was treating the problem as a vector surface integration question, and in this, we always convert F*dA(vector) to scalar surface integral by parametrization and finding the normal vector (which is kinda the "jacobian"). But, as both of you have stated, I just realized that it is just a double integration problem and no need for vector fields I guess.

Yep. As stated, $dQ = \sigma \, dA = \sigma \,r'\,dr'\, d\theta$ , ALL scalars. So you can't replace $r'\ (=|r'|)$ by $<r'\cos\theta,r'\sin\theta,0>$

(1) not correct to replace $dA$ by a vector
(2) you take a dot product where you should multiply with $|r'|$ (and in the denominator you write $r^2$ instead of $r'^{\, 2} \$)

But I think we start repeating things .

$\$

 

[neroE]

Alright, thank you very much @BvU
I learnt a lot of things from this thread & some latex.