Recent content by ermia

Sorry I meant charge density.

This is the picture of the problem. I attach my solution. I first used a trick with gauss's law to calculate the radial electric field at first order of r. ( where r is small ) ( we can assume ##small r=\delta r##) I used a cylinder at the center of the ring then i calculated the ##\hat{z}##...

That equation for radius has a "type" error. $$r =\sqrt \frac{S}{\pi} $$

I wrote some potentials but they were wrong. I used the cm of all fluid parts and I used the radius which is $$ \sqrt S/ \pi $$ .

Thanks. Is my answer right? $$ \sigma_{left }= 5 \rho a $$ $$ \sigma_{right} = 6 \rho a $$

Once I wrote gausses law for the left and write part separately. I found two different feilds in x=a. One is ##\frac{ \rho a} {\epsilon }## the other is ## \frac{2 \rho a}{ \epsilon }## thus we can conclude that we have a surface charge in x=a. Where am I wrong?

I have wrote all feilds and potentials and I want to find the constants. My first question is " when we say in the a<x<2a the potential is V(x)" then the potential in the a is V(a) or V(0) ( cause it is 0 in our new area) ? Second one is " when I want to write the gausses law for the point x=a I...

krotovs solution is based on energy conservation. My question is that why the solution didn't consider inductors energy? the question: The solution:

I uploaded it here. It has nothing special for me except the part I asked. [FONT=times new roman]Since the thickness of the sphere walls remains unchanged, the force tearing the sphere per unit length must remain unchange [FONT=arial]but why?

Why the pressures for two different spheres at the area middle of themselves should be equal to each other??

Why? And why this happens to electric field Why? t is not a differential to say $$A=2\pi r t$$ b.t.w: I don't have doubt about whole question. Just the part I asked. I thought the answer of my question was sth obvious cause all solutions in the internet or the book refused explaining it but I...

..

I tried gauss law. And the fact that if alpha is less than pi/2 we can say that we have two parts with angle alpha and one other part which has a normal field at the center. But non of them helped me answer. The problem's solution says that we can use the fact that our section has longitudinal...