Parallel Component for E over charged disk

[Living_Dog]

(NOTE: I am doing this to possibly re-enter a graduate program and earn a PhD. This is not being done for class.)

DJ Griffiths, Prob. 2.6, solved.

My question is about the parallel component of the $$\vec{E}$$-field. If you integrate over cylindrical coordinates, then by symmetry this component cancels as you integrate around $$\phi$$. So my question is about that integral - will it integrate to zero, or are we to ad hoc cancel the $$\phi = 0$$ term for $$dq$$ with the $$\phi = 180$$ term for $$dq$$?

Here is the integral as I set it up:

$$2\pik\sigma\int{\frac{s^2}{(z^2 + s^2)^{\frac{3}{2}}}}ds$$,

where $$s, \phi, z$$ are the cylindrical coordinates.

Oh, and yes, I did try $$s = z \cdot tan(\theta)$$

and get $$\int{\frac{1}{cos(\theta)}}d\theta$$ plus another term which is handleable.

Now that I think of it, the integral should integrate to zero since it is for the full disk, namely: $$0 < \phi < 2 \pi$$. So it should naturally integrate to zero. ...no?

Any help would be appreciated.
-LD

 

[mighty2000]

Dear LD,

Thank you for your question. It is always great to see someone interested in pursuing a PhD in the field of physics. To answer your question, the integral for the parallel component of the \vec{E}-field will indeed integrate to zero. This is because of the symmetry of the problem, as you correctly pointed out. When integrating over cylindrical coordinates, the integral will naturally cancel out as you integrate around \phi. Therefore, there is no need to ad hoc cancel the \phi = 0 term for dq with the \phi = 180 term for dq.

As for your attempt at using s = z \cdot tan(\theta), this is a valid substitution and it should lead to the correct answer. However, it is important to make sure that you are using the correct limits of integration for \theta, which should be from 0 to \pi/2. This will result in the integral \int{\frac{1}{cos(\theta)}}d\theta, as you have correctly stated.

I hope this helps to clarify your doubts. Good luck with your studies and pursuit of a PhD.