(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider charge distribution [itex]\rho = \frac{A}{r}[/itex] with spherical symmetry, for [itex] 0 \leq r \leq R[/itex], and [itex]\rho = 0[/itex] for [itex] r > R[/itex], and A is a constant. Find the Electric Field in all of space. Check your answer obtaining [itex]\rho[/itex] from your answer.

2. Relevant equations

Gauss's law:

[tex]\oint \mathbf{E} \cdot \mathbf{dS} = \frac{q_{enc}}{\epsilon_{0}} [/tex]

or

[tex]\mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho}{\epsilon_{0}}[/tex]

3. The attempt at a solution

So, this should be pretty simple.

I should get the field using Gauss's law in it's integral form, since there is a spherical symmetry.

[tex]\mathbf{E} = \frac{q_{enc}}{4 \pi \epsilon_{0} R^{2}}[/tex]

And then I find the [itex]q[/itex] by integrating the charge density:

[tex]q_{enc} = \iiint\limits_V \rho r^{2} sin(\theta) dr d\theta d\phi[/tex]

[tex]V = \left\{ (r, \theta, \phi) | 0 \leq r \leq R, 0 \leq \theta \leq \pi, 0 \leq \phi \leq 2\pi\right\}[/tex]

[tex]q_{enc} = \iiint\limits_V \frac{A}{r} r^{2} sin(\theta) dr d\theta d\phi[/tex]

[tex]q_{enc} = 4 \pi A \int r dr[/tex]

[tex]q_{enc} = 2 \pi A R^{2}[/tex]

Substituting in the Electric field:

[tex]E = \frac{A}{2\epsilon_{0}}[/tex]

But this has to be wrong.

By doing Gauss's law in the differential form, I got [itex]\mathbf{\nabla} \cdot \mathbf{E} = 0[/itex], instead of finding the charge density [itex]\rho[/itex].

Where am I going wrong here?

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# Homework Help: Electric field due to a charge density.

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