What Is Wrong with My Approach to Solving This Electrostatics Problem?

[quasar_4]

Homework Statement

This should be simple: We have a charge density in some region of space, r< R (R is some known constant), that goes like 1/r. Everywhere else, the charge density is zero. I need to find the electric field and potential everywhere. The total charge is Q, assumed to be known.

Homework Equations

The Attempt at a Solution

I don't see why I can't just use Gauss's law to find E and integrate to find Phi.

For r>R, the problem is really simple: the total enclosed charge is Q, so
$$\vec{E(r)} = \frac{Q}{4 \pi \epsilon0 r^2} \hat{r}$$

potential is just $$\Phi = \frac{-Q}{4 \pi \epsilon0 r}$$

For r <R, we draw a Gaussian surface of radius r. Then Gauss's law should give

$$E (4 \pi r^2) = Q_{enc}/\epsilon0 = \frac{4 \pi}{\epsilon0} \int_0^r r^2 \frac{1}{r} dr = 2 \pi r^2 \rightarrow \vec{E(r)} = \frac{1}{2 \epsilon0} \hat{r}$$

Potential would be:

$$\Phi = - \int_R^r E \cdot dl = - \int_R^r \frac{1}{2 \epsilon0} dr = \frac{-(r-R)}{2 \epsilon0}$$

So, these don't seem correct to me at all. E has no r-dependence, and worse, the wrong units! What am I doing wrong here? Please help!

 

[phyzguy]

The charge density inside R is proportional to 1/r, but you need to calculate the proportionality constant, which will have units of charge/area (since this constant divide by r has units of charge/volume). This will fix the units. In fact the E-field is constant in magnitude (but not in direction) inside R.

 

[quasar_4]

Thank you!