The potential \(\phi(\boldsymbol{r})=\left(q / 4 \pi \epsilon_{0} r\right) \mathrm{e}^{-\mu r}\) may be regarded as representing the effect of screening of a charge \(q\) at the origin by mobile charges in a plasma. Calculate the charge density \(\rho\) (at points where \(r \neq 0\) ) and find the total charge throughout space, excluding the origin.

To find the charge density from the given potential function, we will use Poisson's equation: \(\nabla^2 \phi(\boldsymbol{r}) = -\frac{\rho(\boldsymbol{r})}{\epsilon_0}\).

Now we need to compute the Laplacian of the given potential function in spherical coordinates. Remembering that in spherical coordinates, the Laplacian operator is: \(\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \cdots\) (We're skipping the angular part since the potential function only depends on the \(r\)-coordinate) The given potential function is: \(\phi(\boldsymbol{r})=\left(\frac{q}{4 \pi \epsilon_{0} r}\right)\mathrm{e}^{-\mu r}\)

Now, we must calculate the first derivative of the potential function with respect to r: \(\frac{\partial \phi}{\partial r} = \frac{\partial}{\partial r}\left(\frac{q}{4\pi\epsilon_0r}\mathrm{e}^{-\mu r}\right) = \frac{-q}{4\pi\epsilon_0 r^2}\mathrm{e}^{-\mu r} + \frac{q\mu}{4\pi\epsilon_0 r}\mathrm{e}^{-\mu r}\)

Now, we must calculate the second derivative of the potential function with respect to r: \(\frac{\partial^2 \phi}{\partial r^2} = \frac{\partial}{\partial r}\left(\frac{-q}{4\pi\epsilon_0 r^2}\mathrm{e}^{-\mu r} + \frac{q\mu}{4\pi\epsilon_0 r}\mathrm{e}^{-\mu r}\right)\) By taking the derivative, we have: \(\frac{\partial^2 \phi}{\partial r^2} = \frac{2q}{4\pi\epsilon_0 r^3}\mathrm{e}^{-\mu r} - \frac{q\mu}{4\pi\epsilon_0 r^2}\mathrm{e}^{-\mu r} - \frac{q\mu}{4\pi\epsilon_0 r^2}\mathrm{e}^{-\mu r} + \frac{q\mu^2}{4\pi\epsilon_0 r}\mathrm{e}^{-\mu r}\)

Now that we have obtained the second derivative, we will plug it into the Laplacian operator: \(\nabla^2\phi(r) = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \phi}{\partial r}\right) = \frac{1}{r^2}\left(4\pi r^2\frac{\partial^2 \phi}{\partial r^2}\right)\) Simplifying, we get: \(\nabla^2\phi(r) = \frac{2q}{\epsilon_0}\mathrm{e}^{-\mu r} - \frac{q\mu^2}{\epsilon_0}\mathrm{e}^{-\mu r}\)

Now we can use Poisson's equation to find the charge density: \(-\frac{\rho(r)}{\epsilon_0} = \frac{2q}{\epsilon_0}\mathrm{e}^{-\mu r} - \frac{q\mu^2}{\epsilon_0}\mathrm{e}^{-\mu r}\) Solving for \(\rho(r)\), we get: \(\rho(r) = -2q\mathrm{e}^{-\mu r} + q\mu^2\mathrm{e}^{-\mu r}\)

To find the total charge throughout space, excluding the origin, we will integrate the charge density over the volume of the space: \(Q = \int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\rho(r)r^2\sin(\theta)drd\theta d\phi\) Here, we're using the fact that \(Q = \int_{V}\rho(r)dV\) where V is the volume of space. Now integrating, we let the theta and phi integrals go from 0 to pi and 0 to 2pi respectively. This results in \(4\pi\) as the factor for those integrals. The integration with respect to r would give us the remaining part: \(Q = 4\pi\int_{0}^{\infty}(-2q\mathrm{e}^{-\mu r} + q\mu^2\mathrm{e}^{-\mu r}) r^2 dr\) Now, we need to integrate the two parts separately: \(Q = 4\pi\left(-2q\int_{0}^{\infty}r^2\mathrm{e}^{-\mu r} dr + q\mu^2\int_{0}^{\infty}r^2\mathrm{e}^{-\mu r} dr\right)\) Using integration by parts and evaluating the integrals, we find that they diverge. Thus, the total charge in space excluding the origin is infinite.