How Does Charge Density Variation Affect Integration in a Non-Conducting Sphere?

[orthovector]

Homework Statement

A non conducting solid sphere with radius r_1 has charge density \rho_E = \rho_o \frac{r_1}^{r}
what is the charge enclosed for 0 < r < r_1 inside the non conducting sphere?

Homework Equations

\frac{q_{enc}}^{\frac{4}^{3}} \pi r^3}} = \rho_E = \frac{dq_{enc}}^{4 \pi r^2 dr}

(1) \frac{4}^{3} \pi r^3 \rho_E = q_{enc}
\frac{4}^{3} \pi r^3 \rho_o \frac{r_1}^{r} = \frac{4}^{3} \pi r^2 \rho_o r_1 = q_{enc}

\frac{8}^{3} \pi \rho_o r_1 r dr= dq_{enc}
WHY CAN'T I TAKE THIS INTEGRAL TO FIND ENCLOSED CHARGE?
\int_{0}^{r} \frac{8}^{3} \pi \rho_o r_1 r dr = \int_{0}^{r} dq_{enc} = Q_{enc}I KNOW I MUST put \rho_E = \rho_o \frac{r_1}^{r} with dq_{enc} = 4 \pi r^2 dr befofe i take the integral, but I'm not sure why (1) does not work.

 

[gabbagabbahey]

\frac{q_{enc}}^{\frac{4}^{3}} \pi r^3}} = \rho_E = \frac{dq_{enc}}^{4 \pi r^2 dr}

Why would this be true?

How can \rho_E= \frac{q_{enc}}{\frac{4}{3} \pi r^3} and \rho_E= \frac{dq_{enc}}{4 \pi r^2 dr} both be true?

That would imply dq_{enc}=\frac{3q_{enc}}{r} which would mean you have some with unit of charge on the LHS equal to something with units of charge over distance on the RHS...clearly one, or both of those equations is wrong!

(1) \frac{4}^{3} \pi r^3 \rho_E = q_{enc}

This would be true if \rho_E was a constant; but it clearly isn't since it varies like 1/r.

Instead of messing around with all these jumbled equations, go back to the definition of volume charge density...what is that?

You should see that the charge enclosed by a volume \mathcal{V} is always given by the equation q_{enc}=\int_{\mathcal{V}} \rho_E dV where dV is the differential volume element and \rho_E is the charge density in that region.

In this case, \rho_E is not constant throughout the volume and so you can't take it outside the integral. That means that q_{enc}\neq\rho_E *\text{Volume}.