My question is about the math involved in the concept of electric potential. Though this is a physics concept, it is basically a lot of vector calculus. The derivations below are based on the content of Ch. 2 in Griffiths' Electrodynamics.
I'm going to write quite a few steps in reasoning and the concept of electric field and electric potential before getting to the actual question. My question actually starts at the bold text below.
Electric potential is defined as
$$V(\vec{r})=-\int_{\vec{r}_0}^{\vec{r}} \vec{E}\cdot d\vec{r}\tag{1}$$
where $\vec{E}$ is an (electric) vector field. We are interested in finding this field.
If we knew $V$ then we could find $E$ by taking the gradient of $V$.
Typically, we know a charge density $\rho(\vec{r})$ (and in this example I am assuming the density is per volume) for a continuous charge.
Now, if our charge were a point charge at the origin of our coordinate system, the electric field at some point $\vec{r}$ would be
$$\vec{E}=k_e\frac{q}{r^2}\hat{r}\tag{2}$$
and the electric potential at that point (using $vec{r}_0=\vec{r}_\infty$; that is, some point with infinite radius; not sure exactly how to write this vector correctly) is
$$V(\vec{r})=-\int_{\vec{r}_\infty}^{\vec{r}} \vec{E}\cdot d\vec{r}\tag{3}$$
$$=-\int_{\vec{r}_\infty}^{\vec{r}} k_e\frac{q}{r^2}\hat{r}\cdot (dr\hat{r}+rd\theta\hat{\theta}+r\sin{\theta}d\phi d\theta)\tag{4}$$
$$=-\int_{\infty}^r k_e\frac{q}{r^2}dr\tag{5}$$
$$=k_e\frac{q}{r}\tag{6}$$
Now, in geneal the point charge could be located at a position $\vec{r}_s$ instead of the origin, and so we would have
$$\vec{E}(\vec{r})=k_e\frac{q}{\lVert \vec{r}-\vec{r}_s\rVert^2}\frac{\vec{r}-\vec{r}_s}{\lVert \vec{r}-\vec{r}_s\rVert}\tag{7}$$
$$=k_e\frac{q}{\eta^2}\hat{\eta}\tag{8}$$
Note that $\vec{\eta}$ is the separation vector between the positions of field point and source charge.
Now, the book I am reading proceeds to say that the potential in this case is
$$V(\vec{r})=k_e\frac{q}{\eta}\tag{9}$$
(I am not sure how to show this exactly, and have asked how in a separate question)
But moving on, if we accept that (9) is true, then because of superposition we have that for $n$ discrete charges
$$V(\vec{r})=k_e\sum_{i=1}^n\frac{q_i}{\eta_i}\tag{10}$$
and if we have a continuous distribution then
$$V(\vec{r})=k_e\int\frac{1}{\eta}dq\tag{11}$$
If the continuous distribution of charge is a volume charge, then
$$V(\vec{r})=k_e\iiint_V\frac{\rho(\vec{r_s})}{\eta}d\tau\tag{12}$$
The path to finding (12) above seems to be that you compute the electric potential generated by each infinitesimal charge $dq$ on the charged volume, and then use superposition principle to add the potentials up, which is an integral over the volume.
That is, if we think about the volume as a bunch of tiny charges $dq$ then Iwe have
$$d\vec{E}(\vec{r})=k_e\frac{dq}{\eta^2}\hat{\eta}\tag{15}$$
where $\vec{\eta}=\vec{r}-\vec{r}_{dq}$
and the potential at $\vec{r}$ due to this electric field is
$$V_{dq}(\vec{r})=k_e\frac{dq}{\eta}\tag{16}$$
$$=k_e\frac{\rho(\vec{r}_{dq})d\tau}{\eta}\tag{17}$$
And so if we sum all of these up over our volume, we are integrating over the volume of charge, and so we get (12).
My question is how to obtain (12) by first calculating the total electric field at $\vec{r}$ and then integrating to get the electric potential.
For a charged volume with charge density per unit volume $\rho(\vec{r_s})$ we have
$$\vec{E}(\vec{r})=\iiint_V k_e\frac{\rho(\vec{r}_s)d\tau}{\eta^2}\vec{\eta}\tag{13}$$
$$V(\vec{r})=\int_{\vec{r}_\infty}^{\vec{r}} \vec{E}(\vec{r})\cdot d\vec{r}\tag{14}$$
Is there a way to go from (13) to (12) by subbing in (13) into (14)?
Best Answer
We begin with the expression for the electric field
$$\vec E(\vec r')=k_e\int_{V}\frac{\rho(\vec r_s)(\vec r'-\vec r_s)}{|\vec r'-\vec r_s|^3}\,dV_s\tag1$$
Now, if we integrate $(1)$ from $\vec r_1$ to $\vec r$, and let $|\vec r_1|\to \infty$, we find that
$$\begin{align} -\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\vec E(\vec r')\cdot d\vec r'&=-\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\left(k_e\int_{V}\frac{\rho(\vec r_s)(\vec r'-\vec r_s)}{|\vec r'-\vec r_s|^3}\,dV_s\right)\cdot d\vec r'\\\\ &=-\lim_{|\vec r_1|\to \infty}\int_{\vec r_1}^{\vec r}\left(-k_e\int_{V}\rho(\vec r_s)\nabla '\frac{1}{|\vec r'-\vec r_s|}\,dV_s\right)\cdot d\vec r'\\\\ &=k_e \lim_{|\vec r_1|\to \infty}\int_{V} \rho(\vec r_s)\int_{\vec r_1}^{\vec r}\nabla '\frac{1}{|\vec r'-\vec r_s|}\cdot d\vec r'\,dV_s\\\\ &=k_e \lim_{|\vec r_1|\to \infty}\int_{V} \rho(\vec r_s)\left(\frac1{|\vec r-\vec r_s|}-\frac1{|\vec r_1-\vec r_s|}\right)\,dV_s\\\\ &=k_e \int_{V} \frac{\rho(\vec r_s)}{|\vec r-\vec r_s|}\,dV_s \end{align}$$
as was to be shown!