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

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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[itex]\frac{20ε}{r^3}[/itex]
for point (3,-4,1)(cm) r=[itex]\sqrt{26}[/itex]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
 
on Phys.org
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.