Recent content by srnixo

But each time the constant of integration must be calculated, even if above or below the plane layer, the constant of integration is different and therefore the statement is different. the first part of the potential expression for both ( above or below the plane layer is the same) which is...

For the inside, when choosing V=0 : the constant of integration C1 will indeed be 0 so the expression of V(M) inside will be : V(M)= -ρ. z²/2ε0 For the outside: we have two cases: above the plane layer z>e and below the plane layer z<-e : → for the first case which is above the plane layer...

When V=0 which implies z=0, the term z² in the expression for the potential inside the layer also becomes zero. So, the potential V(M) inside the layer (-e < z < e) at z = 0, will be equal to zero? right?

Here is the exercise: And these are my attempts: This is for the first question about the electric field. (I know I'm missing the drawing, which is a drawing of the plane layer of thickness 2e with a cylinder on it as a GAUSS SURFACE ). As for the second question, I'm not sure about it, so I...

Alright, Thank you so much for your efforts. This is so helpful.

Could you Mr @Delta2 clarify for me something! Is this formula (equation1) which you provided only used when the object is at rest in the origin or moving along a curved path? In other words, It should not be used when the object is in motion?

I've done some more research and found out that the tangential and normal coordinate systems do not have a fixed origin (the origin moves with the object). Therefore, the position vector of the object will constantly change to reflect its changing position and will not be equal to ZERO. It has...

I found it on a website in Arabic, as far as I remember. they said that: R(t) represents the radius of curvature in tangent direction and P(T) is the radius of curvature in the normal direction at that point. so that the equation (1) provides a vector that accurately describes the position and...

Thank you so much Mr @Delta2 . Your explanation is excellent and precise. Now I understand well about this matter. Thank you again.

Mr @haruspex : You said it's useful to use curvilinear coordinates but, As I mentioned previously, this lesson was presented by the professor, So I can't change coordinates like i want. He chose ( n-t coordinates and Cartesian coordinates). In addition, for the first question about position...

First question: Does the position vector exist in the tangential and normal coordinate system in curvilinear motion? ↧↧↧ Regarding my teacher's response, he confirmed the existence of the position vector formula but was unable to provide it. Upon returning home, I attempted to find its formula...

I got it now, @nasu , @haruspex , @berkeman Thank you so much for your consideration, and thanks for all these explanations to help me I appreciate it ❤️

I repeated the experiment five times, and obtained the first table. Although I know that the speed of sound is not directly related to distances, but in the exercise, to calculate the speed of sound you need to consider distances because it is Vs=2.D/∆T . I got confused because the second table...

That link doesn't provide any benefit because it only explains how to conduct the experiment, and that is already mentioned at the beginning of the exercise. Additionally, believe me, I don't know what the conclusion is because, before conducting the experiment, I ensured all necessary...

Alright. Mr @nasu , not just for the homework , but for my general knowledge as a student of physics, is it true that the speed of sound in enclosed spaces, like houses or adjacent rooms, remains the same [ as the average speed of sound in dry air at 20 degrees Celsius which is approximately 343...