# Using Divergence Theorem

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.

• 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 $\ds{\color{#f00}{R > r}}$, $\ds{\vec{\mrm{E}}\pars{R} = -\nabla_{\vec{R}}\Phi\pars{\vec{R}} = 4\pi r^{2}\,{\vec{R} \over R^{3}} \not= \vec{0}}$.