I Work to move a point charge from infinity to the centre of a charge distribution

[LeoJakob]

Electrostatic potential $$ \Phi(\vec{r})=k \int \mathrm{d}^{3} r \frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} (i) $$ with $$ k=\frac{1}{4\pi\epsilon_{0}} $$ in SI units.
What work is required to move a point charge q from infinity to the center of the through $$ \rho(\vec{r})=\rho_{0}\mathrm{e}^{-a r} (ii) $$ given charge distribution, where $$ a \text{ and } \rho_{0} $$ are constants? Work in Gaussian units.

To solve the problem I would use spherical coordiantes. The potential only depends on the radial difference between the charge q with position
$$\vec{r} = r\hat{e}_r \text{ and the location vector of the charge distribution element } \vec{r'} = r' \hat{e}_r \text{ such that }\\
\left|\vec{r}-\vec{r'}\right| = \sqrt{(r-r')^2}
$$

$$\text{Define } \Phi(\infty)=0 \text{ then the work is given by } W= q \Phi(\vec r)=q \Phi(r)$$

$$W= q\Phi(\vec{r})=k \int \mathrm{d}^{3} r' \frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}= \int \mathrm{d}^{3} r' \frac{\rho(r^{\prime})}{\sqrt{(r-r')^2}}=4\pi\int \limits_\infty^0 (r')^3 \frac{\rho(r^{\prime})}{\sqrt{(r-r')^2}}dr^{\prime}$$

The factor 4pi comes from the integration in spherical coordinates and k=1 in Gaussian units.

Is my approach right?

 

[BvU]

I assume the 3 in $$4\pi\int \limits_\infty^0 (r')^3 \frac{\rho(r^{\prime})}{\sqrt{(r-r')^2}}dr^{\prime}$$is a typo ?

I don't agree with

the potential only depends on the radial difference between the charge q with position

And certainly not with $\left |\vec{r}-\vec{r}^{\prime}\right| = {\sqrt{(r-r')^2}}$

However -- fortunately -- you are only interested in $\vec r=\vec 0$, so if you fill that in in $(i)$ you're allright.

$\$

 

[LeoJakob]

I assume the 3 in $$4\pi\int \limits_\infty^0 (r')^3 \frac{\rho(r^{\prime})}{\sqrt{(r-r')^2}}dr^{\prime}$$is a typo ?

I don't agree withAnd certainly not with $\left |\vec{r}-\vec{r}^{\prime}\right| = {\sqrt{(r-r')^2}}$

However -- fortunately -- you are only interested in $\vec r=\vec 0$, so if you fill that in in $(i)$ you're allright.

$\$

First of all: Thanks for taking the time to answer me :)

the 3 in $$4\pi\int \limits_\infty^0 (r')^3 \frac{\rho(r^{\prime})}{\sqrt{(r-r')^2}}dr^{\prime}$$is a typo because:

$$W= q\Phi(\vec{r})=k \int \mathrm{d}^{3} r' \frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}=\int \limits_{0}^{2 \pi} \int \limits_{0}^{\pi} \int \limits_{\infty}^{0} \frac{p\left(r^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}(\sin \theta) \cdot\left(r^{\prime}\right)^{2} d r^{\prime} d \theta d \phi \\ =\int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{\pi} \sin \theta d \theta \int \limits_{\infty}^{0}\left(r^{\prime}\right)^{2} \frac{\rho \left(r^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d r^{\prime} =4\pi\int \limits_\infty^0 (r')^2 \frac{\rho(r^{\prime})}{\left|\vec{r}-\vec{r}^{\prime}\right|}dr^{\prime} \\ \Rightarrow $$

Would you agree with(?):

$$q\Phi(0)=4\pi\int \limits_\infty^0 (r')^2 \frac{\rho_{0}\mathrm{e}^{-a r^{\prime}})}{\left|\vec{r}^{\prime}\right|}dr^{\prime} $$

Why do you think the following is wrong?

$$\left|\vec{r}-\vec{r}^{\prime}\right|=\left|\left(r-r^{\prime}\right) \vec{e}_{r}\right|=\left|r-r^{\prime}\right| \cdot \underbrace{\| \vec{e}_{r}||}_{=1}=\sqrt{\left(r-r^{\prime}\right)^{2}} $$

 

[BvU]

First of all:

Ah, and I forgot:

$\qquad$!​

Why do you think the following is wrong?$$\left|\vec{r}-\vec{r}^{\prime}\right|=\left|\left(r-r^{\prime}\right) \vec{e}_{r}\right|=\left|r-r^{\prime}\right| \cdot \underbrace{\| \vec{e}_{r}||}_{=1}=\sqrt{\left(r-r^{\prime}\right)^{2}}$$

Which $\vec e_r$ would that be ? The one from $\vec r$ or the one from $\vec r'$ ?

If $|\vec r| = |\vec r'|$ then $|\vec r -\vec r'| =0$ only if $\theta=\theta'\ \&\ \phi=\phi'$ !

The $\hat e_r$ unit vector for $\vec r -\vec r'$ is not along $\hat e_r$ nor is it along $\hat e_{r'}$ !$\$

 

[LeoJakob]

Thanks for the welcome :) I now understand my mistake, thank you very much ! :)