Recent content by Gustav

Okay I see, thank you for the explanation!

Okay, now I understand. As for the calculation of the volume density charge, is it correct?

But why did we change the ## \sigma_b ## from ## r ## to ##R##?

I have already calculated the polarisation that is $$ \mathbf{P} = \frac{\rho_f r}{2} \left( 1 - \frac{\epsilon_0}{\epsilon} \right) \hat{r} . $$ I tried to use the following formulas to calculate the density bound charges. For the surface bound charge I got: $$ \sigma_{b1} = \mathbf{P} \cdot...

Now that I calculated the polarisation $$ \mathbf{P} = \frac{\rho_f r}{2}\left( 1- \frac{\epsilon_0}{\epsilon} \right) \hat{r} $$ I was able to calculate the bound charges $$ \sigma_b = \mathbf{P} \cdot \hat{n} = \frac{\rho_f r}{2} \left( 1 - \frac{\epsilon_0}{\epsilon} \right) $$ $$ \rho_b = -...

Oh okay, I think I understand now.

Yes, that part I understand, but why can't A' be equal to A, even though the divergence in A is a non-zero value?

No I meant how can they be related? Since they have the same curl there must be a relation between them?

For our vector potential ##\vec A##, we got a non-zero value for the divergence and the curl. But now when calculating for the Coulomb gauge ##\vec A'## we need to have a non-zero value equal to the one in ##\vec A## for the curl, but this times the divergence needs to be equal to zero. How can...

I just want to make sure if I understood this correctly, so we converted from Cartesian coordinates to cylendrical coordinates because it will make the calculations easier? But how is it possible to determine whether one should convert to cylinderical or any other coordinates?

Because then I will have: 1) Curl: $$ \vec B = \nabla \times \vec A' = -\frac{2C}{s}\hat{\phi} $$ 2) Divergence: $$ \nabla \cdot \vec A' = 0 $$

Oh okay, so does it work if $$ \vec A' = C \ln \left( \frac{1}{s^2} \right)\hat{z} ? $$

Okay, I am with you so far. But does that mean I need to think of an expression for A' or is there a way to calculate it?

Does it mean that it is the same as the vector potential A?

Yes, but how can I solve for it?