Electric field due to a charge density.

[FLms]

Homework Statement

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

Homework Equations

Gauss's law:

\oint \mathbf{E} \cdot \mathbf{dS} = \frac{q_{enc}}{\epsilon_{0}}
or
\mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho}{\epsilon_{0}}

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.

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

And then I find the q by integrating the charge density:

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

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

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

q_{enc} = 4 \pi A \int r dr

q_{enc} = 2 \pi A R^{2}

Substituting in the Electric field:

E = \frac{A}{2\epsilon_{0}}

But this has to be wrong.
By doing Gauss's law in the differential form, I got \mathbf{\nabla} \cdot \mathbf{E} = 0, instead of finding the charge density \rho.

Where am I going wrong here?

 

[Vipho]

\mathbf{E} = \frac{q_{enc}}{4 \pi \epsilon_{0} R^{2}}Is that really Electric Field in all of space? What're electric field inside and outside sphere?

 

[FLms]

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


Is that really Electric Field in all of space? What're electric field inside and outside sphere?

Taking r = a ; a < R, I get:

q_{enc} = \iiint\limits_V \rho dV
q_{enc} = 4 \pi A \int r dr
q_{enc} = 2 \pi A a^{2}

More generally: q_{enc} = 2 \pi A r^{2}

And the total charge Q_{tot} is 2 \pi A R^{2}

So:

q_{enc} = \frac{Q_{tot} r^{2}}{R^{2}}

Now:

\oint \mathbf{E} \cdot \mathbf{dS} = \frac{q_{enc}}{\epsilon_{0}}
\mathbf{E} = \frac{q_{enc}}{4 \pi \epsilon_{0} r^{2}}
\mathbf{E} = \frac{Q_{tot} r^{2}}{4 \pi \epsilon_{0} r^{2} R^{2}}
\mathbf{E} = \frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}}

Still getting the same result here, and \mathbf{\nabla} \cdot \mathbf{E} here still is zero, since R is a constant.

For r > R:

q_{enc} = Q_{tot} = 2 \pi A R^{2}

\oint \mathbf{E} \cdot \mathbf{dS} = \frac{Q_{tot}}{\epsilon_{0}}
E 4 \pi r^{2} = \frac{2 \pi A R^{2}}{\epsilon_{0}}
E = \frac{A R^{2}}{2 \epsilon_{0} r^{2}}

I believe here I do have \rho = 0, because \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^{2}}) = 0 everywhere, except the origin.

Shouldn't the EF outside the sphere behave like the EF of a point charge?

 

[FLms]

Anyone?

 

[Vipho]

I think A must be zero, because at r\rightarrow 0, \rho \rightarrow \infty

 

[Mindscrape]

A isn't zero, A=Q/(4pi(R^2/2)

The outside Efield does look like a point charge, check yourself. Also, div(1/r^2)=4pi delta(r), not 0.

The field inside is constant, but that doesn't mean it's divergenceless. Are you sure you're using spherical divergence formula?

 

[FLms]

How about this:

\rho (\mathbf{r}) isn't exactly defined at the origin; it explodes at the point r = 0.

So the integral would have to be from a small radius a to a general radius r \leq R

Using this, I got:

q_{enc} = 2 \pi A (r^{2} - a^{2})

The total charge, however, still is Q_{tot} = 2 \pi A R^{2}

q_{enc} = \frac{Q_{tot}}{R^{2}} (r^{2} - a^{2})

\oint \mathbf{E} \cdot \mathbf{dS} = \frac{q_{enc}}{\epsilon_{0}}

E = \frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}} (1 - \frac{a^{2}}{r^{2}})

Calculating the divergence os \mathbf{E}, I got:

\mathbf{\nabla} \cdot \mathbf{E} = \frac{1}{r^{2}}\frac{\partial}{\partial r} (r^{2} \frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}} - \frac{Q_{tot} a^{2}}{4 \pi \epsilon_{0} R^{2}})

\mathbf{\nabla} \cdot \mathbf{E} = \frac{Q_{tot}}{2 \pi \epsilon_{0} R^{2} r}

Substituting Q_{tot} = 2 \pi A R^{2}

\mathbf{\nabla} \cdot \mathbf{E} = \frac{A}{\epsilon_{0} r} = \frac{\rho (\mathbf{r})}{\epsilon_{0}}

I got one last question: Is it possible to write \rho (0) = 4 \pi A \delta (\mathbf{r})?

 

[FLms]

A isn't zero, A=Q/(4pi(R^2/2)

The outside Efield does look like a point charge, check yourself. Also, div(1/r^2)=4pi delta(r), not 0.

The field inside is constant, but that doesn't mean it's divergenceless. Are you sure you're using spherical divergence formula?

Yes, I found out that I was miss calculating the Divergence of the Electric Field.
Too bad I was writing my other post while you answered it.
Thanks anyway.

By the way, about the Divergence of \frac{\hat{r}}{r^2}, it is equal to zero everywhere, except the origin. And in general form, it is written as 4 \pi \delta(\mathbf{r}).
(Griffiths' "Itr. to Electrodynamics" section 1.5.3 and problem 1.16)

 

[Vipho]

Oh, I think electric field inside sphere
\mathbf{E}= E\hat{\mathbf{r}}=\frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}}\frac{\mathbf{r}}{r}<br />, (\hat{\mathbf{r}}=\frac{\mathbf{r}}{r})
Hence,
\mathbf{\nabla}\cdot \mathbf{E}= \frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}}\mathbf{\nabla}( \frac{\mathbf{r}}{r})<br />
So
<br /> \mathbf{\nabla}\cdot (\frac{\mathbf{r}}{r})=\frac{2}{r}<br /> and Q_{tot} = 2 \pi A R^{2}<br />
So, we have
<br /> \mathbf{\nabla}\cdot \mathbf{E}= \frac{Q_{tot}}{4 \pi \epsilon_{0} R^{2}} \frac{2}{r}=\frac{A}{\epsilon_{0}r}=\frac{\rho}{ \epsilon_{0}}<br />

 

[Chestermiller]

The electric field for r < R is E =(A/2ε₀₎ i_r

The electric field for r > R isE =(A/2ε₀₎ (R²/r²)i_r

In spherical coordinates, the divergence of E is A/(rε₀) for r < R and 0 for r > R.