Using Divergence Theorem

Question

Let $B_r$ denote the ball $|x|\le r$ in $\mathbb{R}^3$, and write $dS_r$ for the area element on its boundary $\partial B_r$. The electric field associated with a uniform charge distribution on $\partial B_R$ may be expressed as $$E(x)=C\int\limits_{\partial B_R}\nabla_x|x-y|^{-1}dS_y,$$ a) Show that for any $r<R$, the electric flux $\int\limits_{\partial B_r}E(x).\nu dS_x$ through $\partial B_r$ equals zero.

**b)**Show that $E(x)\equiv 0$ for $|x|<R$("a conducting spherical shell shields its interior from outside electrical effects").

For part a, I tried to to using divergence theorem as follows: $$\int\limits_{\partial B_r}E(x).\nu dS_x=\int\limits_{B_r}divE(x)dx= C\int\limits_{B_r}\int\limits_{B_R}div\nabla_x|x-y|^{-1}dS_ydx,$$ However, I don't know how to proceed next steps to prove the integrand is zero. By the way, could you please suggest me a method for part b)?

Thank you very much for your help.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \Phi\pars{\vec{R}} & \equiv \int_{0}^{2\pi}\int_{0}^{\pi} {1 \over \root{R^{2} + r^{2} - 2rR\cos\pars{\theta}}}\,\ \overbrace{r^{2}\sin\pars{\theta}\,\dd\theta\,\dd\phi}^{\ds{\dd S}} \\[5mm] & = \left.\vphantom{\huge A}-\,{2\pi r \over R} \root{R^{2} + r^{2} - 2rR\cos\pars{\theta}} \right\vert_{\ \theta\ = 0}^{\ \theta\ =\ \pi} = -\,{2\pi r \over R}\,\pars{R + r - \verts{R - r}} \\[5mm] & = \bbox[#ffe,10px,border:1px dashed navy]{\ds{\left\{\begin{array}{rcl} \ds{-4\pi r} & \mbox{if} & \ds{R} & \ds{<} & \ds{r} \\[2mm] \ds{-\,{4\pi r^{2} \over R}} & \mbox{if} & \ds{R} & \ds{>} & \ds{r} \end{array}\right.}} \end{align}

• When $\ds{\color{#f00}{R < r}}$, the potential $\ds{\Phi\pars{\vec{R}}}$ is $\ds{\vec{R}}$-independent such that the Electric Field $\ds{\vec{\mrm{E}}\pars{R} = -\nabla_{\vec{R}}\Phi\pars{\vec{R}} = \vec{0}}$.
• When $\ds{\color{#f00}{R > r}}$, $\ds{\vec{\mrm{E}}\pars{R} = -\nabla_{\vec{R}}\Phi\pars{\vec{R}} = 4\pi r^{2}\,{\vec{R} \over R^{3}} \not= \vec{0}}$.