Find the charge distribution from the given E-field (spherical)

[goohu]

Homework Statement

see picture

Relevant Equations

1) $\nabla \times E = 0$

2) $\rho = \epsilon_0 \nabla \cdot E$

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?

 

[BvU]

Why not proceed as if you got a yes and a yes as answers and see what comes out !

 

[goohu]

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) $\rho = \epsilon_0 \alpha e^{-\lambda R} (\lambda R + 1) \frac{1}{ R^2}$

 

[BvU]

Don't understand a) at all ...

 

[goohu]

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.

 

[Delta2]

From my textbook I've learned that a field that is not conservative is not an E-field

Slight (but important) correction: A field that is not conservative is not a static(time independent) E-field

 

[Delta2]

Slight correction (i think) for b) of post #3 i am getting the parenthesis as$(1-\lambda R)$ instead of $(1+\lambda R)$

 

[goohu]

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

 

[Delta2]

well I seem to calculate $$\epsilon_0\alpha\frac{1}{R^2}\frac{\partial (Re^{-\lambda R})}{\partial R}$$ where do i go wrong?

 

[goohu]

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

 

[Delta2]

I did it in wolfram, wolfram agrees with me. I did it in cartesian coordinates there though, here is the link, warning one might get headache in cartesian coordinates
https://www.wolframalpha.com/input/...{-\lambda\sqrt{x^2+y^2+z^2}}/({x^2+y^2+z^2})"
Question: The $\hat a_R$ in the problem statement is the radial unit vector in spherical coordinates right ?

 

[goohu]

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