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

AI Thread Summary
The discussion centers on the legitimacy of a static electric field derived from a charge distribution, emphasizing that a static electric field should have a curl of zero. Initial calculations suggested a non-zero curl, leading to confusion about the conservative nature of the field. Corrections were made regarding the charge density formula, indicating a misunderstanding in the sign of a term. The importance of calculating the divergence of the electric field rather than its gradient was highlighted, with references to using computational tools for verification. The radial unit vector in spherical coordinates was confirmed as part of the problem's context.
goohu
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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?
 

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Why not proceed as if you got a yes and a yes as answers and see what comes out !
 
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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} ##
 
Don't understand a) at all ... 🤔
 
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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.
 
goohu said:
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
 
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Slight correction (i think) for b) of post #3 i am getting the parenthesis as## (1-\lambda R)## instead of ##(1+\lambda R)##
 
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
 
well I seem to calculate $$\epsilon_0\alpha\frac{1}{R^2}\frac{\partial (Re^{-\lambda R})}{\partial R}$$ where do i go wrong?
 
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Seems I am wrong , I calculated ##\nabla E ## instead of ## \nabla \cdot E##
 
  • #12
Yeah I think you got it right. Its in spherical coordinates, I attached the picture of the problem in the opening post.
 
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