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

• neroE
In summary: If you want to be really consistent, you write ##d\vec E = {1\over 4\pi\varepsilon_0} \; {1\over r^2}\;{\vec p - \vec r' \over |\vec r|^2 } \; dA## and then integrate over the area of the disk.In summary, the conversation discusses the approach to solving a question involving a charged disk of radius R and a uniform electric field at a specific point. The conversation delves into using surface integration to solve the problem, and the correct notation for the variables and equations is also discussed. It is concluded that the problem can be solved using double integration rather than vector surface integration
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.

#### Attachments

• 1.pdf
190.3 KB · Views: 84
Hello @neroE ,

neroE said:
which I found in various electricitiy and magnetism books
neroE said:
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##

##\ ##

nasu, MatinSAR and 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 occuring (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.

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##.

MatinSAR, vanhees71 and neroE
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##

neroE said:
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.

##\ ##

MatinSAR, neroE and vela
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.

neroE said:
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)##

neroE said:

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, vanhees71 and MatinSAR
Alright, thank you very much @BvU
I learnt a lot of things from this thread & some latex.

BvU

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