Charge from field, cylindrical symmetry

  • #1
The problem states:

From the field with a radial cylindrical component only given by the following equations:

E(r)= (ρ0*r3)/(4 * ε0*a2) for r<=a

E(r)= (ρ0*a2)/(4*ε0*r2) for r > a

obtain the corresponding charge distribution in free space in which the equation is:

ρ(r) = ρ0*(r2/a2) (0<=r<=a)

So I know that dE(r)/dr = p(r)/ε0

After differentiating the first E(r) equation I come to (3/4)*ρ0*r2/a2.

It would be correct if the 3/4 weren't there but I'm not sure where I'm going wrong. Any help appreciated.
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
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In cylindrical coordinates [itex]\vec{p}=(r,\theta,z)[/itex], the div vector is
[tex]\vec{\nabla}=\left ( \frac{1}{r}\frac{\partial}{\partial r}r, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z} \right )[/tex]

Since [itex]E(r,\theta,z)=E(r)[/itex], you needed: [tex]\frac{1}{r}\frac{\partial}{\partial r}\big ( rE(r) \big ) = \frac{1}{\epsilon_0}\rho(r)[/tex]
 
Last edited:
  • #3
Very helpful! Thanks!
 

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