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heeyyy, kolay gelsinn!! 🫠

sanada kolay gelsin!!

@@selin4393 bundan bi sonraki olan example 7.9’a bakabildin mi, line integral’in icinden cikamadim

@@ErenCengiz-xu9wx sinava mayista girdim cok guzel gecti:) sorulari sadece cozmeyi ogren eger zamanda kisitliysan

@@selin4393 hayir daha 3 ayim var, rahatım; çok sevindim senin adına, uk'de mi okuyorsun?

Yes that is a mistake

I guess because of symmetry, you only look at the electric field lines from the smaller to the bigger cylinder in terms of radius s and you get Qenc = λL and ∫E*da = E2πsL = λL/ε0 -> E = λ/(2πsε0) in ŝ-direction with ε0 i mean epsilon with subscript 0 ofcourse

I'm afraid you are correct. Sorry for the mistake

Although there is same example solved in book but still I don’t understand this question

Fixed, thank you

What do you mean by (s)?

Hello! I'm not sure I understand the question correctly, but I'll try to explain anyway. We are trying to find the total number of microstates for a system of two solids. One way would be to find all the microstate in each macrostate and add them together. This method is correct, but long. We can do the same by simply imaginingin that instead of having two solids we have one (with the number of oscillators equal to the total number of oscillators in the system) . By doing this we can find all the microstate at once. To re-cap, instead of finding multiplicity (microstate) of every single macrostate and add them together, we can find the total number by assuming that the system is one single solid. Hope this helps, let me know if you have any more questions.

I can help you

What step are you referring to?

When we are inside the wire where s<R, Why did we get: I free enc = I * ( pi*s^2 ) divided by (pi*R^2) Thank you.

I could be wrong but: I is the total current in the wire so if you want a tiny, infinitesimal current you divide I by pi*R^2 (the total cross section area). Then you need to multiply that infinitesimal current by the total cross sectional area you're looking at, a.k.a. (I/(pi*R^2))*pi*s^2. Where s=R outside the cylinder and s is just s inside the cylinder. Notice that the whole thing reduces to just the total current, I, outside the cylinder since we care about the entire current when our boundary is greater than R.

Hope that helps even though it's 8 months later.

@@115xXzombieXx115 i got it in time for my eletromag mid terms, in wich i was quite successful! thanks for your help

That’s given by the book. I believe I have a video about it too

You’re right, I made an algebra mistake. I’ll make sure to add it to the description

The plate with V0 goes from z= negative infinity to z= positive infinity. So it’s independent of z. Condition IV says that it’s dependent on x

How can I help you?

I can try, but that will have to be done during winter, is that okay with you?

By the way, if you have any exercises that you can’t solve I’ll be happy to make a video about them

@@ProjectPatimo I'm taking this course now... By winter I'll be done with the semester. And okay thats nice of you!! Is there a way to contact you better than youtube in case i have a question? Thank you in advance 🙏❤️

I’ll try my best, but I’m taking five courses this semester so my schedule is pretty packed with work as well. I feel like explaining how to do the problem is more important as this stuff is going to be used for other courses and you don’t need to understand the concepts but how to setup the problem. You can comment as much as you want here, if you can private message me I’ll share a contact information