PX157 - B5b - electric field from a uniformly charged sphere

ρ = \frac{Q}{V} = \frac{Q}{\frac{4}{3} π a^{3}}

step 1: symmetry
- $\vec{E}$ points radially outwards if $Q > 0$ , ie: $\vec{E} = E \hat{r}$
- $| \vec{E} |$ is constant on a sphere centred on the charged sphere, ie: $E (r)$
step 2: gaussian surface
- sphere of radius, $r$ , centred on the charged sphere
$\begin{align*}$ \vec E \cdot d\vec S &= E , \hat r \cdot dS , \hat r = E , dS \
\therefore ;\oiint \vec E \cdot d\vec S &= \oiint E , dS = E , 4\pi r^{2}
\end{align*}$$
step 3:
- if $r > a : Q_{e n c l .} = Q$
- if $r \leq a : Q_{e n c l .} = ρ V_{G S} = ρ \frac{4 π}{3} r^{3} = Q (\frac{r}{a})^{3}$
step 4:$$\begin{align*}
E , 4\pi r^{2} &= \begin{cases}
\frac{Q}{\epsilon_{0}} & r>a \
\frac{Q}{\epsilon_{0}}(\frac{r}{a})^{3} & r \leq a
\end{cases}\
E &= \begin{cases}
\frac{Q}{4\pi r^{2}\epsilon_{0}} & r>a \
\frac{Q}{4\pi a^{2}\epsilon_{0}}(\frac{r}{a}) & r \leq a
\end{cases} \
\therefore \vec E &= \begin{cases}
\frac{Q}{4\pi r^{2}\epsilon_{0}} \hat r & r>a \
\frac{Q}{4\pi a^{2}\epsilon_{0}}\left(\frac{r}{a}\right)\hat r & r \leq a
\end{cases} \
\end{align*}$$