Recent content by madafo3435

ohh I get it, I need to think a lot more about it since it is not obvious to me. thanks for your help

Your reasoning confuses me. It is not clear to me which elements have the same potential, and I do not see why this leads to no current flowing through R. I also assume that p.d. refers to partial discharge, but I am not related to that concept.

My reasoning is as follows: with the short circuit present, a 0 resistance can be considered between the terminals. Then, since the only objective of the short circuit current ##I_{SC}## is to give information about the Thevenin equivalent resistance and the open circuit voltage (the existence...

I was wrong writing hahaha :-p Thank you very much for your help! I also thank Delta2 for your contributions.

I see it much clearer now! Then we agree that the potential difference will be the same in both hemispheres and, following my reasoning raised at the beginning, the capacitance would be: ##2\pi \epsilon _0(1+K)(\frac {r_ar_b}{r_b-r_a})## ??

interesting observation, his contribution enlightened me

yes, I'm sorry. I wanted to say: if we want to find the potential difference due to only one of the hemispheres, we see that the field is different in each hemisphere, because one hemisphere has dielectric and the other does not. Then, when calculating the potential difference due to different...

I'm fine with what you say, but I'm still confused about something. I understand that the entire conductor is equipotential, but this is due to all the field generated by the two spheres and the dielectric. But if we want to find the power due to only one of the hemispheres, we would be...

radially yes, but the field that the dielectric can generate confuses me :confused:

The capacitance must be found, that is: ##C=\frac {Q}{V_a-V_b}##. Let ##\vec{E_d}## and ##\vec{E_v}## be the electric fields due to the zone with dielectric and the zone without dielectric respectively. In the case of a spherical capacitor with a vacuum between its plates, it is easy to...

I am satisfied with your considerations . thanks for your help. PeroK PeterDonis @etotheipi

thanks @jbriggs444, @etotheipi and @PeterDonis. Your considerations have helped me to understand my problem much better. But I still have problems seeing because the field inside the solid is going to be 0. I understand that initially the charge is distributed in such a way that some time later...

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Thanks for your appreciation

Because the direction of ##\vec{E}## is given by ##cos(\theta )\imath + sen(\theta )\jmath ## and due to the fixedness and uniformity, the as a component in ##\imath ## total is canceled Do you agree with me?