Recent content by goohu

for a) integration limits were a to 2a.

Yeah I think you got it right. Its in spherical coordinates, I attached the picture of the problem in the opening post.

Seems I am wrong , I calculated ##\nabla E ## instead of ## \nabla \cdot E##

a) we are integrating from a to 2a. ## V = -\int E dl = -\int \frac{R^3 \rho}{3\epsilon_0 R^2} + \int \frac{a^3 \rho}{3\epsilon_0 R^2} = - \frac {\rho}{3 \epsilon_0} \int R dR + \frac{a^3 \rho}{3 \epsilon_0} \int \frac{1}{R^2} dR = -\frac{\rho a^2}{2\epsilon_0} + \frac{a^2 \rho}{6\epsilon_o} ##...

I missed a minus sign. I got it to ## \rho = -\epsilon_0 \alpha e^{-\lambda R} (\lambda R + 1) \frac{1}{ R^2} ## using quotient rule

Actually I recalculated the curl of E and I got it to 0. This is actually the condition for the E-field to be legitimate (since it is conservative). From my textbook I've learned that a field that is not conservative is not an E-field. Sorry about the confusion.

I already did , I just wanted to confirm. For: a) ## \nabla \times E = (0, 0, -5)##. So it is legitimate since it is not 0. b) ## \rho = \epsilon_0 \alpha e^{-\lambda R} (\lambda R + 1) \frac{1}{ R^2} ##

a) Static charge distribution should result in a static electric field? Legitimacy should be checked with curl of E = 0? b) Using the second equation should give is the answer?

There is the equation for V. I assume both shells has the same charge density (a bit unclear to me). ## V = \int \frac{\rho_v}{4\pi\epsilon_0R} dV = \int \frac{\rho_v}{4\pi\epsilon_0R} R^2 sin\theta dRd\phi d\theta = \frac{\rho_v}{\epsilon_0} \int R dR = \frac{\rho_v a^2}{2\epsilon_0} ##...

Sorry for the late reply. b) The electric field inside a conductor (in this case 0 to a) is 0. So V should be a constant. Not sure if we need to go further than this? I am unsure because of the wording in the task : "using the same assumptions as those of part (a)." The assumption being "the...

R is a constant? Can you explain why? E(R) seems to be a function depending on R, for me it seems like its a variable? I don't understand. What do you mean with wrt? For b) if its a conductor you could use Gauss law I guess. C = Q/V I'm mostly familiar with using it to determine the...

a) I think you find V by just integrating E in regards to R. Then we integrate from the point of interest, which is a, to the 0 potential which is (R = 2a)? b) If the same logic as a) applies here as well then we should integrate from the point of interest to the 0 potential. This should be...

a) Just using the equations gives us: surface charge density = ## \rho_{\rho s} = kR^2 ## volume charge density = ## \rho_\rho = -4kR ## b) I am not sure here but the Q on the shell is the same as within. If that's the case we can use gauss law to find Q which I guess is the total charge. ##...

Attempt at solution: a) Since I need help with b) this section can be skipped. Results : ##ρ_{psa} = -Pa ## ##ρ_{psb} = Pb ## ##ρ_{p} = \frac {-1}{R^2} \frac {∂(R^2PR)}{∂R} = -3P ## b) This is where I am unsure (first time using gauss law for P) so I need some confirmation here: ## \int...

My bad, for the circle I think it should be : ## V = \frac{1}{4\pi \epsilon_0} * \rho_s * \int_0^a \frac{2\pi R}{R} \ dR ##