Deriving the Yukawa potential from the field of a screened charge

Question

I am trying to derive the Yukawa potential from the electric field of a screened positive point charge, which is

$$ \vec{E}(\vec{r}) = \frac{q}{4\pi\epsilon_0}\frac{e^{-kr}(kr+1)}{r^2}\hat{r}. $$

The simplest way to do it I think is to use

$$ V(\vec{r}) = -\int_{\mathcal{O}}^{\vec{r}}\vec{E}\cdot d\vec{r} = -\int_{\infty}^r\frac{q}{4\pi\epsilon_0}\frac{e^{-kr'}(kr'+1)}{r'^2}dr' \\ = -\frac{q}{4\pi\epsilon_0}\left(\int_{\infty}^r\frac{ke^{-kr'}}{r'}dr' + \int_{\infty}^r\frac{e^{-kr'}}{r'^2}dr' \right) $$

but these integrals are advanced and I don't know how to solve them. I have also computed the corresponding charge distribution $\rho$ from the field through Gauss' law but I don't know if I can compute the potential from

$$ V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\int\int_{volume}\frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}d\tau' $$ because I think $\rho$ extends to infinity meaning the integral would diverge.

This is an exercise related to the second chapter of Griffiths (not in the book itself though), so I am looking for a way to solve it that doesn't use Poisson's equation and the methods of solving it.

Answer 1

Consider the integral $$ \int_{\infty}^r\frac{ke^{-kr'}}{r'}dr'. $$ This integral doesn't have an elementary solution (I don't think), but you can still use elementary techniques to transform it as follows: Perform an integration by parts, with $u = 1/r'$ and $dv = k e^{-kr'}$. We then have $v = -e^{-kr'}$ and $du = - dr'/{r'}^2$, and so $$ \int_{\infty}^r\frac{ke^{-kr'}}{r'}dr' = \left[ -\frac{e^{-kr'}}{r'} \right]_{\infty}^r - \int_{\infty}^r\frac{e^{-kr'}}{{r'}^2}dr', $$ and so $$ \int_{\infty}^r\frac{ke^{-kr'}}{r'}dr' + \int_{\infty}^r\frac{e^{-kr'}}{r'^2}dr' = -\frac{e^{-kr}}{r} $$ as desired.

Answer 2

Let the radial component of the electric field $E_r(r)$ be given by

$$E_r(r)=\frac{q}{4\pi\varepsilon_0}\frac{(1+kr)e^{-kr}}{r^2}\tag1$$

The potential function, $\phi(r)$, satisfies the differnetial equation

$$\frac{\partial \phi(r)}{\partial r}=E_r(r)\tag2$$

Using $(1)$ in $(2)$ and integrating we find that

$$\begin{align} \phi(r)&=\int^r E_r(r')\,dr'+C\\\\ &=\frac{q}{4\pi\varepsilon_0} \int^r \frac{(1+kr')e^{-kr'}}{r'^2}\,dr'+C\\\\ &=\frac{q}{4\pi\varepsilon_0} \int^r \frac{e^{-kr'}}{r'^2}\,dr'+\frac{q}{4\pi\varepsilon_0} \int^r \frac{ke^{-kr'}}{r'}\,dr'+C\\\\ &\overbrace{=}^{IBP}-\frac{q}{4\pi\varepsilon_0}\frac{e^{-kr}}{r}+\frac{q}{4\pi\varepsilon_0}\left(-\int^r \frac{ke^{-kr'}}{r'}\,dr'+\int^r \frac{ke^{-kr'}}{r'}\,dr'\right) +C\\\\ &=-\frac{q}{4\pi\varepsilon_0}\frac{e^{-kr}}{r}+C \end{align}$$

Inasmuch as we set the potential at infinity to $0$, we find

$$\phi(r)=-\frac{q}{4\pi\varepsilon_0}\frac{e^{-kr}}{r}$$

as was to be shown!