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E field and charge density

  • #1
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Homework Statement


The electric field produced by a static electric charge Pe(R) at a point R in free space is give by,

E(R) = (R3+AR2) R[tex]\hat{}[/tex] , R<a

E(R) = B(a5+Aa4)R-2 R[tex]\hat{}[/tex], R>a

Where a, A and B are arbitrary constants and R spherical coordinate system radial vector (R locates the observation point at its tip).
Determine the associated charge distribution Pe(R)

Homework Equations





The Attempt at a Solution



By Helmholtz theorem,

E(R) is given by -[tex]\Delta[/tex][tex]\Phi[/tex]

Where,

[tex]\Phi[/tex] = (1/4piε)[tex]\int[/tex](Pe/|R-R'|)dv'

I am not sure how to make use of the R<a and R>a conditions. I am thinking, when its R>a, the volume integral is just r2sin θ dθ dφ dr
When R<a, the only the radius changes to (a-R)2?
 

Answers and Replies

  • #2
kuruman
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How is finding the potential going to help you? How about using Gauss's Law in differential form?

Also, the first equation for R<a is a bit confusing because it implies that E has units of R3. Is there a constant up front? It is important to know in order to determine whether the field is continuous at R = a. If it is not, you will have to worry about surface charge density at R = a.
 
  • #3
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Kuruman, I solved the problem by poisson's equation :
[tex]\Delta[/tex]2[tex]\Phi[/tex] = P(R)

Helmholtz theorem for static(DC) case, E(R) = -[tex]\Delta[/tex][tex]\Phi[/tex], so del.E = P(R)
I got zero for the R>A case.

Can you explain the a=R case?
 
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  • #4
kuruman
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Is the electric field continuous at R = a? In other words, does E(a+δ) - E(a-δ) equal zero in the limit δ goes to zero? If not, there is charge density σ on the surface of a sphere of radius a given by

[tex]\frac{\sigma}{\epsilon_0}=lim_{\delta \rightarrow 0}(E(a+\delta)-E(a-\delta))[/tex]

The expression that you posted for the electric field is ambiguous about the continuity of the electric field because it appears to be dimensionally incorrect. There must be some other constant multiplying R3 in the expression for R<a.
 
  • #5
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I checked the expression. It is correct. I seem to be missing something here.
How do you evaluate E(a+δ) and E(a-δ)
I took divergence of E for both cases (R<a and R>a). Got zero for R>a and got an expression for R<a.
I am lost as to how to evaluate it for R=a?
If I put R=a in the first expression, I get
E(a) = (a3+Aa2) [tex]\hat{R}[/tex]
 
  • #6
kuruman
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I checked the expression. It is correct. I seem to be missing something here.
The expression may be what you were given, but it does not make sense. We will accept that it is what it is and proceed from this point.
How do you evaluate E(a+δ) and E(a-δ)
For E(a+δ) plug in R = a+δ in the second expression
For E(a-δ) plug in R = a-δ in the first expression
I took divergence of E for both cases (R<a and R>a). Got zero for R>a and got an expression for R<a.
What is the divergence of E equal to according to Gauss's law? What does the expression that you got for R < a mean?
I am lost as to how to evaluate it for R=a?
Evaluate E(a+δ) and E(a-δ) as indicated above, then let δ = 0. You get two expressions. If they are the same then there is no surface charge density, otherwise there is.
 

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