State Gauss' law of electrostatistics. A spherical charge distribution has a volume charge density given by $$\rho= \left\{ \begin{array}{lr} \rho_0\left(1-\frac{r^2}{a^2}\right) & 0 \leq r \leq a\\ 0 & r>a \end{array} \right. $$ where $\rho_0$ is a constant. Calculate the charge, $Q$, enclosed. Am I to apply the law to the next question? Please give me the necessary steps. Thanks.
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$$\oint \vec{E} \cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$$
This is Gauss' Law. And you are given the charge density as:
$$\rho = \rho_0 \frac{1-r^2}{a^2}$$
Remember that charge density is defined as:
$$\rho = \frac{Q}{V}=\frac{dq}{dV}$$
Hence, we have: $$dq=\rho \; dV$$
Well, we can find $dV$. We are talking about a sphere here, so it will be:
$$dV=4\pi r^2 dr$$
That will be the micro volume. You can draw a diagram to get this, but this is very simply the derivative of the volume of a sphere, which is $\frac{4}{3}\pi r^3$.
Plug everything back in to $dq$, including the density: $$dq = \rho_0 \frac{1-r^2}{a^2} 4\pi r^2 dr=\frac{\rho_0 4\pi}{a^2}(r^2-r^4) dr$$
From this, you can get $q_{enc}$. $q_{enc}$ is just the integral. Let's start with $0 \le r < a$. The charge will be:
$$q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^r (r^2-r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}-\frac{r^5}{5}\right)$$
The thing about Gauss's law is that, in most cases, the electric field won't be dependent on the area, hence:
$$\oint \vec{E} \cdot d\vec{A}=\vec{E} \cdot \oint d\vec{A}=\vec{E}\cdot\vec{A}$$
The area is very simple. It's just $4\pi r^2$ because it's a sphere. Therefore:
$$E(4 \pi r^2)=\frac{\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}-\frac{r^5}{5}\right)}{\epsilon_0}$$
Simplifying gives you:
$$\therefore E=\frac{\rho_0}{a^2\epsilon_0} \left(\frac{r}{3}-\frac{r^3}{5}\right)$$
This is for $0 \le r < a$. Similar thing for when it's greater than $a$, except you will need the sphere's full charge:
$$q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^a (r^2-r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{a^3}{3}-\frac{a^5}{5}\right)=\rho_0 4\pi \left(\frac{a}{3}-\frac{a^3}{5}\right)$$
We only integrate to $a$ because afterwards there's no charge. That's the sphere's total enclosed charge. However, the Gaussian surface's area is still $4 \pi r^2$, of course.
Hence we have:
$$E(4 \pi r^2)=\frac{\rho_0 4\pi \left(\frac{a}{3}-\frac{a^3}{5}\right)}{\epsilon_0}$$
Simplified to:
$$\therefore E=\frac{\rho_0}{\epsilon_0 r^2} \left(\frac{a}{3}-\frac{a^3}{5}\right)$$
Shahar
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