Finding the charge density,when Electric field intensity is given.

AI Thread Summary
The discussion revolves around calculating the electric charge density (ρv) given the electric field intensity (E) in free space. The user initially calculated ρv as -2.67 nC/m³ but found the provided answer to be -1.42 nC/m³, leading to confusion about their approach. It was clarified that the divergence operator was incorrectly interpreted, specifically regarding the use of cylindrical versus spherical coordinates. After reviewing the divergence operator, the user recognized the mistake in interpreting the unit vector, which was crucial for the correct calculation. The conversation highlights the importance of correctly applying mathematical concepts in electromagnetics to arrive at accurate solutions.
Nero26
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Homework Statement



Given, E=ar20/r2 (mV/m) in free space,find ρv at the point (3,-4,1)(cm).
where, ρv=Electric charge density per unit volume
E=Electric field intensity

Homework Equations



∇.(εE)=ρv

The Attempt at a Solution



ε∇.E= -2\frac{20ε}{r^3}
for point (3,-4,1)(cm) r=\sqrt{26}x10-2 m
∴ρv=-2.67nC/m3

But the answer given was -1.42nC/m3

I'm trying to know whether the approach was correct or not. I couldn't figure out any reason for the problem.

Thanks for your help.

PS:The problem is adapted from "Fundamentals of Engineering Electromagnetics" by David K Cheng, pp90,Ex-3.4,2nd Edition
 
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TSny said:

I still can't get it.Here, ar is the unit vector along radius of sphere. E is independent of θ and ø so their partial derivatives of E will be zero.
∴ ε∇.E=-\frac{40ε}{r^3},so where am I doing wrong?
Thank you for your response.
 
The expression you are using for the divergence in spherical coordinates is not correct. It is not just the radial derivative of the radial component. Did you check the links provided?
 
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Orodruin said:
The expression you are using for the divergence in spherical coordinates is not correct. It is not just the radial derivative of the radial component. Did you check the links provided?

TSny said:

Thanks all, Now I've got it.I was wrong in interpreting the question, ar is meant for cylindrical coordinates and aR for spherical coordinates in my book.
 
Nero26 said:
Thanks all, Now I've got it.I was wrong in interpreting the question, ar is meant for cylindrical coordinates and aR for spherical coordinates in my book.

Good thing, because I was about to point out that there is no unique solution if ar
had been the unit vector in spherical.

For example,
ρv = qδ(x)δ(y)δ(z), with q = 20/k, k = 9e9 SI, then ρv(3,-4,1 cm.) = 0.

An infinite number of alternative ρv would also exist, all satisfying ρv(3,-4,1) = 0.
 
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