# Proper method of integration

1. Feb 22, 2009

### orthovector

1. The problem statement, all variables and given/known data
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?

2. Relevant 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.

Last edited: Feb 23, 2009
2. Feb 23, 2009

### gabbagabbahey

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!

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 $q_{enc}\neq\rho_E *\text{Volume}$.