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

In summary, the conversation discusses finding the electric charge density per unit volume at a given point using the electric field intensity and the divergence operator in spherical coordinates. The correct approach is to use aR as the unit vector in spherical coordinates, rather than ar as mistaken in the initial attempt. This mistake could have led to an incorrect solution, as there are multiple possible solutions for the electric charge density at the given point.
  • #1
Nero26
21
1

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
 
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  • #3
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=-[itex]\frac{40ε}{r^3}[/itex],so where am I doing wrong?
Thank you for your response.
 
  • #4
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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  • #5
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.
 
  • #6
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.
 

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

1. What is charge density?

Charge density is a measure of the amount of electric charge per unit volume of a material. It is typically denoted by the symbol ρ (rho) and has the unit of coulombs per cubic meter (C/m³).

2. How is charge density related to electric field intensity?

Charge density and electric field intensity are directly proportional. This means that as the charge density increases, so does the electric field intensity, and vice versa.

3. Can charge density be negative?

Yes, charge density can be negative. This typically occurs when there is an excess of electrons in a given volume, resulting in a negative charge density.

4. How can I find the charge density if the electric field intensity is given?

To find the charge density when the electric field intensity is given, you can use the equation ρ = ε0E, where ε0 is the permittivity of free space (8.85x10⁻¹² C²/Nm²) and E is the electric field intensity in N/C.

5. What are the practical applications of finding charge density?

Finding charge density is important in many areas of science and technology, including electronics, electromagnetism, and electrochemistry. It helps in understanding the behavior of electric fields and in designing various devices such as capacitors and batteries.

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