# Showing that $\Phi=\frac1{4\pi\epsilon_0}\int_{r>a}\frac{\rho_1(\mathbf{x})}r\operatorname{dV}$ for potential $\Phi$ and density $\rho_1$

Let $\mathbf{E_1}$ and $\mathbf{E_2}$, with potentials $\phi_1$ and $\phi_2$ respectively, satisfy $\mathbf{E}=-\nabla\phi,\space \mathbf{\nabla}\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$ with two different charge distributions with densities $\rho_1$ and $\rho_2$.

Suppose that $\rho_1(\mathbf{x})=0$ for $|\mathbf{x}|\le a$ and that $\phi_1(\mathbf{x})=\Phi$, a constant, on $|\mathbf{x}|=a$.

Show that $$\Phi=\frac1{4\pi\epsilon_0}\int_{r>a}\frac{\rho_1(\mathbf{x})}r\operatorname{dV}$$

We have shown, and are to use the following:

For a point charge $\phi=\frac Q{4\pi\epsilon_0 r}$ (Assuming I have that right) and

$$\frac1{\epsilon_0}\int_V\phi_1\rho_2\operatorname{dV}+\int_{\partial V}\phi_1\nabla\phi_2\cdot\operatorname{d\mathbf{S}}=\frac1{\epsilon_0}\int_V\phi_2\rho_1\operatorname{dV}+\int_{\partial V}\phi_2\nabla\phi_1\cdot\operatorname{d\mathbf{S}}\space\space\space\space(\dagger)$$

I am having trouble with determining what it is that I have to get to be $\frac1{4\pi\epsilon_0}\int_{r>a}\frac{\rho_1(\mathbf{x})}r\operatorname{dV}$. I suspect all I need is that first couple of lines and I can probably go from there but I don't know what to set $\Phi$ equal to to begin with to work with.

We're also told that we may assume that the integrals over the sphere at infinity in $(\dagger)$ are $0$ but I don't see why we would be doing that.

Any help is appreciated, thanks

We begin with the expression

$$\frac1{\epsilon_0}\int_V\phi_1\rho_2\,dV+\oint_{\partial V}\phi_1\nabla\phi_2\cdot \hat n \,dS=\frac1{\epsilon_0}\int_V\phi_2\rho_1\,dV+\oint_{\partial V}\phi_2\nabla\phi_1 \cdot\hat n\, dS \tag 1$$

GIVENS:

We are given that $\phi_1=\Phi$ on the surface $|\vec x|=a$ and $\rho_1=0$ for $|\vec x|<a$.

We take $\phi_2=\frac{1}{4\pi \epsilon_0 r}$ and $V$ as the region $|\vec x|> a$.

EVALUATION OF VOLUME INTEGRALS:

Note that since $\rho_2$ is a point charge at $0$, its charge density, $\rho_2=0$ for $V$ (i.e.$|\vec x|>a$). Then, we assert that

$$\int_V\phi_1\rho_2\,dV=\int_{|\vec x|>a}\phi_1 \underbrace{\rho_2}_{=0}\,dV=0 \tag 2$$

$$\int_V\phi_2\rho_1\,dV=\frac{1}{4\pi \epsilon_0}\int_{|\vec x|>a}\frac{\rho_1}{r}\,dV \tag3$$

EVALUATION OF SURFACE INTEGRALS:

We are given that $\phi_1(|\vec x|=a)=\Phi$. Hence, we can write

\begin{align} \oint_{\partial V}\phi_1\nabla\phi_2\cdot \hat n \,dS&= \oint_{|\vec x|=a}\color{red}{\phi_1}\color{blue}{\nabla\phi_2}\cdot\color{purple}{\hat n}\,\color{green}{dS}\\\\ &=\color{red}{\Phi} \oint_{|\vec x|=a}\color{blue}{\nabla\phi_2}\cdot\color{purple}{\hat n}\,\color{green}{dS}\\\\ &=\color{red}{\Phi} \int_0^{2\pi}\int_0^\pi \color{blue}{\frac{-\hat r}{4\pi\epsilon_0a^2}}\cdot \color{purple}{(-\hat r)}\,\color{green}{a^2\sin(\theta)\,d\theta\,d\phi}\\\\ &=\frac{1}{\epsilon_0}\color{red}{\Phi}\tag 4 \end{align}

Finally, since $\rho_1=0$ inside $V$, we have

\begin{align} \oint_{\partial V}\phi_2\nabla\phi_1 \cdot\hat n\, dS&=\oint_{|\vec x|=a}\color{red}{\phi_2}\nabla\phi_1 \cdot\hat n\, dS\\\\ &=\color{red}{\frac{1}{4\pi \epsilon_0 a}}\oint_{|\vec x|=a}\nabla \phi_1 \cdot\hat n\,dS\\\\ &=\color{red}{\frac{1}{4\pi \epsilon_0 a}}\oint_{|\vec x|=a} \vec E\cdot \hat n\,dS\\\\ &=0 \,\,\text{there is no charge enclosed in}\,\,|\vec x|<a\tag 5 \end{align}

Using $(2)-(4)$ in $(1)$, we arrive at the coveted relationship

$$\bbox[5px,border:2px solid #C0A000]{\Phi=\frac{1}{4\pi\epsilon_0}\int_{|\vec x|>a}\frac{\rho_1}{r}\,dV}$$

• Thank you very much, I seriously appreciate the effort made to make your answer very clear Commented Apr 23, 2017 at 16:18
• @Aka_aka_aka_ak Well, you're welcome very much. I'm pleases that I could help! -Mark Commented Apr 23, 2017 at 16:22
• +1. Basically, the OP is asking for the $Image\ Method$. Commented Apr 25, 2017 at 23:32