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} ##...
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...