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I was reading this article here and near the half of it, the author discusses a way to prove that potential is a body is constant is the state which minimizes potential of a body. I was able to follow the proof well enough till the last step:

$$\bbox[6px,border:2px solid red]{\frac{1}{4\pi\epsilon_o} \int_V \frac{\rho(r')\,dV' }{|r-r'|} + \phi_{\operatorname{ext}} (r) =\lambda}$$

The author arrives at this equation and states that from the above equation we can say that $\rho(r)$ is the distribution which makes the potential a constant and it is said that it completes the proof of original statement.. but I don't quite understand how it does.

I am looking for how we can understand that the above equation implies that the potential inside conductor is constant. Also how it doesn't imply that the potential outside the conductor is also constant.

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  • $\begingroup$ He/she is saying that the first term on the LHS is $\phi_{\text{inner}}$ so, the LHS is $\phi_{\text{inside}}$. That whole is equal to the constant $\lambda$ for every $\mathbf{r}$ inside the conductor. Or, is your question why the integral is equal to $\phi_{\text{inner}}$? $\endgroup$
    – Gilgamesh
    Commented Jan 31, 2021 at 7:44
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    $\begingroup$ Ohh @MarcoCiafa I got it now, the I was thinking by mistake was the total potential was given as the first term. Now I understand that it is just the potential caused by charges inside, thank you. $\endgroup$
    – Brian
    Commented Jan 31, 2021 at 7:46
  • $\begingroup$ How does this arguement change when we talk of points outside the volume of container @MarcoCiafa (as in how do we say that potential may or may not be non zero for outside conductor?) $\endgroup$
    – Brian
    Commented Jan 31, 2021 at 7:50
  • $\begingroup$ I think it's becuase $\phi_{\text{ext}}$ is only defined as the potential of the outer carges $\textit{inside}$ the conductor. The function that you are extremising is the energy stored, so the potential he/she is defining is only defined a priori inside becuase it's where it matters. In the expression for $U$, where first appears $\phi_{\text{ext}}$, he/she integrates over $V$, the region inside. You can say nothing about the outside, at least with that function. $\endgroup$
    – Gilgamesh
    Commented Jan 31, 2021 at 8:20
  • $\begingroup$ Please let me know if you understood what I said and if I should write and answer. $\endgroup$
    – Gilgamesh
    Commented Jan 31, 2021 at 15:43

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