Recent content by curious_mind

No no, I understand there can be lot of things if we consider microscopic view, there will be Quantum Electrodynamics and so on, I understand that. I just want to stick around macroscopic picture for the time being. But do you mean Graphene is a 2D conductor which can have surface charge...

Yes I am already familiar with it, but frankly I do not see any applicabilty or even connection of image method with this problem. Is there any ?

Yes, but my confusion is not Gauss' law. The thread is helpful to me because it depicts capacitor metallic conducting plate having non-zero thickness. I am still amazed, that there cannot be 2D metallic conductor or conducting shell with zero thickness having non-zero surface charge density !?

Oh my bad, really sorry.😂 https://www.physicsforums.com/threads/finding-electric-field-between-two-conducting-plates-using-gauss-law.1053894/

Ok I think I do understand your point. I will get back again after some pondering. I just looked at diagrams in this link, which are having finite thickness. EDIT : There was a wrong link I cited previously, here is the link...

Then what about conducting plates of capacitor ? We take them infinitesimally thin long metallic plates, right ?

You mean that it is not ideal to consider thickness of the shell zero ? Even if it would, it would no longer be a conductor ?

Well, If we take hollow sphere having two surfaces, then I understand that the situation will be similar to charge put inside a cavity of a conductor. And the answer will be followed. But that precisely is my dissatisfaction. What if we take thickness of the shell to be zero. An ideal situaiton...

I have read Griffiths' Chapter 2 sections on Conductors. According to it, (if I understood it correctly) if the charge is put inside the cavity of a conductor, then the equal and opposite total charge will be induced surrounding the cavity. This charge and the total charge induced surrounding...

I see, I think I understood now your argument. Thank you.

Well, I have to clarify I am talking with respect to first type written diagram. In the typewritten solution, oscillation happens in the vertical direction. In hand written illustration, oscillation happens in horizontal diagram. It is just that I found caveats in vertical typewritten...

Diagram is correct. Yes. I think the problem is symmetrical in this regard. Not really, I think. Springs deflect in both horizontal and vertical components. Only Vertical spring oscillates in vertical component. I assume you are considering x-axis to be horizontal. I am not sure how you think...

So, the ratio of these two would be Vertical displacement is to length of spring which is equal to ##\cos \theta ## is that you want to say ? Which would mean if elongation is ##\Delta l## and vertical displacement is ##x##, then for very small ##x##, it would be ##cos \theta = \dfrac{x}{\Delta...

Thanks for the diagram but Can you explain what you want to say ?

Yes, I have tried the same way. But in case of elongation of center spring C, do not we require distance ##H+ \Delta H## ? And so, length of elongated spring A or B will be ##( H+\Delta H)\csc(\theta + \Delta \theta)##?