# Charge from field, cylindrical symmetry

1. Sep 10, 2012

### ur5pointos2sl

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: Sep 10, 2012
2. Sep 10, 2012

### Simon Bridge

In cylindrical coordinates $\vec{p}=(r,\theta,z)$, the div vector is
$$\vec{\nabla}=\left ( \frac{1}{r}\frac{\partial}{\partial r}r, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z} \right )$$

Since $E(r,\theta,z)=E(r)$, you needed: $$\frac{1}{r}\frac{\partial}{\partial r}\big ( rE(r) \big ) = \frac{1}{\epsilon_0}\rho(r)$$

Last edited: Sep 10, 2012
3. Sep 10, 2012