Answer to Question #211525 in Electricity and Magnetism for Rocky Valmores

Question

Use gauss’s law to find the electric field inside and outside a spherical shell of radius R, which carries uniform surface charge density σ. Compare your answer to problem using Coulomb’s law.

Answer 1

As $\sigma$ is uniform, we remark that in spherical coordinates (with the origin placed at the center of the shell) the electric field $E(r,\theta,\phi)$ will only depend on $r$ and will be directed towards $\vec{e}_r$ (unit vector in the direction of a radius-vector of a point). Therefore, for a point at the distance $r$ from the center of the shell let us apply the Gauss law to a spherical surface of radius $r$ centered at the origin. In this case we have :

$\int_{sphere}\vec{E}(r)\cdot d\vec{S} = \frac{Q_{inside}}{\varepsilon_0}$

As $d\vec{S}=dS \cdot \vec{n}$ (where $\vec{n}$ is an orthonotmal vector to the surface of the sphere) and $\vec{E}(r)=E(r)\cdot \vec{n}$, we conclude that the integral on the left is equal to $E(r)\cdot 4\pi r^2$.

Now let us distinguish two cases : $r>R$ and $r<R$ (electric field is not well defined in the case $r=R$). In the first case ($r>R$) the charge inside the region bounded by the spherical surface cointains all the charge : $Q_{inside}=\sigma\cdot 4\pi R^2$. Therefore, we have

$E(r)\cdot 4\pi r^2 = \frac{\sigma \cdot 4\pi R^2}{\varepsilon_0}$

$E(r)=\frac{\sigma R^2}{\varepsilon_0 r^2}$

In the second case ($r<R$) we find that the charge in the region bounded by the spherical surface is zero (as the shell carrying the charge is outside this surface) and thus we have

$E(r)=0$

Combining two results we conclude that

$\vec{E}(r)=\begin{cases} 0 ,& r<R \\ \frac{\sigma R^2}{\varepsilon_0 r^2} ,& r>R \end{cases}$

Now to use the Coulomb's law, we need to simply calculate the integral (up to rotating the coordinate system, we can suppose that the point where we calculate the field has $\theta=\phi=0$) :

$\vec{E}(r) = \int_{shell} \frac{\sigma (\vec{r}-\vec{r'})}{4\pi\varepsilon_0|\vec{r}-\vec{r'}|^3} dS$

Now using the fact that we know that $\vec{E}$ is directed towards $\vec{r}$ we can simply calculate the integral of the projections on $\vec{r}$ :

$E(r) = \int_{shell} \frac{\sigma \frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|}\cdot \frac{\vec{r}}{r}}{4\pi\varepsilon_0 |\vec{r}-\vec{r'}|^2} dS$

Now passing to the spherical coordinates (where $dS=R^2 \sin{\theta}d\theta d\phi$) we have

$E(r)= \int_{shell} \frac{\sigma}{4\pi\varepsilon_0}\cdot \frac{r-R\cos \theta}{((r-R\cos \theta)^2+R^2\sin^2\theta)^{3/2}} \cdot R^2 \sin\theta d\theta d\phi$

$E(r)=\int_{-1}^1 \frac{\sigma R^2}{2\varepsilon_0} \frac{r-Ru}{((r-Ru)^2+R^2-R^2u^2)^{3/2}}du$, where $u=\cos\theta$

$E(r)=\frac{\sigma R^2}{2\varepsilon_0} \int_{-1}^1 \frac{r-Ru}{(r^2+R^2-2Rru)^{3/2}} du$

$E(r)=\frac{\sigma R^2}{4\varepsilon_0 r} \int_{-1}^1 \frac{R^2+r^2-2Rru-R^2+r^2}{(r^2+R^2-2Rru)^{3/2}}du$

$E(r)=\frac{\sigma R^2}{4\varepsilon_0 r}[\frac{-\sqrt{r^2+R^2-2Rru}}{Rr}+\frac{(r^2-R^2)}{Rr \sqrt{r^2+R^2-2Rru}}]_{-1}^1$

Now let us first study the case $r<R$ :

$\sqrt{r^2+R^2-2Rr}=R-r$,

$\sqrt{r^2+R^2+2Rr}=R+r$,

$E(r)=\frac{\sigma R^2}{4\varepsilon_0 r}(\frac{2}{R}-\frac{2}{R})=0$

And finally if $r>R$ :

$\sqrt{r^2+R^2-2Rr}=r-R$,

$\sqrt{r^2+R^2+2Rr}=r+R$,

$E(r)=\frac{\sigma R^2}{4\varepsilon_0 r}(\frac{2}{r}+\frac{2}{r})=\frac{\sigma R^2}{\varepsilon_0 r^2}$

Combining the two :

$\vec{E}(r)=\begin{cases} 0 ,& r<R \\ \frac{\sigma R^2}{\varepsilon_0 r^2} ,& r>R \end{cases}$

Therefore, we obtain the same results using the Coulomb's law and Gauss' law.