Calculate the field of a sphere with a homogenous distributed charge using Gaussian theorem and the 1st Maxwell Equation, flux integral calculation

Question

Calculate the field of a sphere with a homogenous distributed charge $\rho_0$ using Gaussian theorem and the 1st Maxwell Equation:

$$ \oint \limits_{S} \vec{E} \cdot d \vec{S} \underset{\substack{\uparrow \\ \text { Gauss }}}{=} \int \limits_{V} (\vec{\nabla} \cdot \vec{E}) dV \underset{\substack{\uparrow \\ \text { 1. Maxwell }}}{=} \frac{1}{\epsilon_{0}} \int \limits_{V} \rho dV=\frac{Q(V)}{\varepsilon_{0}} $$

I don't have problems with the volume integral:

Let $R$ be the radius of the sphere:

$$Q(V)=\int \limits_{V} \rho dV=\rho_o \int \limits_{0}^{R} r^2 d r\int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{\pi} d \theta \sin \theta= \rho_0 \frac{4\pi }{3}R^3 $$

In spherical coordinates the field is given by $\vec{E}(\vec{r}) =E_{r} \vec{e}_{r}+E_{\theta} \vec{e}_{\theta}+E_{\phi} \vec{e}_{\phi}$

I thing because of the radial symmetry, I can assume: $$\vec{E}(\vec{r}) =E_{r} \vec{e}_{r}$$

with $\vec{e}_{r}=[(\sin \theta) (\cos \phi),(\sin \theta) (\sin \phi),\cos \theta]^t $.

Our convention is that the normal surface vector points outward i.e. $d \vec{S}\| \vec{e}_{r}$.

How do I calculate: $\oint \limits_{S} \vec{E} \cdot d \vec{S}$?

Answer 1

Let's assume the sphere is centered at $0$ and has radius $R$. Then the outward pointing normal vector at $\vec{x}\in S$ is $\vec{x}/R$ (since $\vec{x}$ has lenght $R$). You want to find $\vec{E}$, which, as you argued by spherical symmetry, has constant magnitude on the entire sphere. Thus $$\int_S \vec{E}\cdot d\vec{S} = \int_S \vec{E}\cdot\frac{\vec{x}}{R} dS = \int_S E dS =E\int_SdS=E\cdot 4πR^2,$$ where, in your notation, $E = E_r$.

Answer 2

To calculate $\iint{\vec{E}.d\vec{S}}$, we use the area element of the spehrical surface $S$ in the spherical coordinates, which is: $$d\vec{S} = \vec{e}_r\,r^2 \sin(\theta)d\theta d\phi$$ Therefore,

$$ \begin{aligned} \iint_S{\vec{E}.d\vec{S}} &= \iint_S{ E \,\vec{e}_r . \vec{e}_r \, r^2 \sin(\theta)d\theta d\phi}\\ &= \iint_S{ E\,r^2\sin(\theta)d\theta d\phi}\\ &= E\,r^2\iint_S{ \sin(\theta)d\theta d\phi}\\ \end{aligned} $$ where $r$ is the radius of the sphere. Both the radius and the field magnitude are constant. The last integral is simple, and done on the whole surface: $$ \iint_S{\sin(\theta)d\theta d\phi} = \int_0^{2\pi}d\phi\int_0^{\pi}{ \sin(\theta)d\theta}=4\pi $$