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 seperately. 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. Since the thickness of the sphere walls remains unchanged, the force tearing the sphere per unit length must remain unchange 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...