The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.

We are given the electric potential \(V(r) = V_0 \exp \left(-r^2 / a^2 \right)\), where \(r = \sqrt{x^2 + y^2 + z^2}\). To find the electric field, we will compute the gradient of the potential, \(\vec{E}(\vec r) = -\vec{\nabla} V(\vec{r})\). We can write down the gradient in spherical coordinates: \(\vec{\nabla} V(r) = \frac{\partial V}{\partial r}\hat{r}\). Now, we need to find \(\frac{\partial V}{\partial r}\). \(\frac{\partial V}{\partial r} = \frac{\partial}{\partial r} \left(V_{0} \exp \left(-r^{2} / a^{2}\right)\right) = -2V_0 \frac{r}{a^2}\exp \left(-r^2 / a^2\right)\) So, the electric field is: \(\vec{E}(r) = -\left(-2V_0 \frac{r}{a^2}\exp \left(-r^2 / a^2\right)\right)\hat{r} = 2V_0 \frac{r}{a^2}\exp \left(-r^2 / a^2\right)\hat{r}\).

Now, we need to find the charge density. We will use Gauss' law, which states that \(\nabla \cdot \vec{E}(\vec r) = \frac{\rho(\vec r)}{\epsilon_0}\). Using spherical coordinates, the divergence is given by: \(\nabla \cdot \vec{E}(r) = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2E_r)\). Let's calculate the result of the differentiation: \(\frac{\partial}{\partial r}(r^2E_r) = \frac{\partial}{\partial r}\left(2V_0 \frac{r^3}{a^2}\exp \left(-r^2 / a^2\right)\right)\). Now, we will apply the product rule for differentiation: \(\frac{\partial}{\partial r}(r^2E_r) = 2V_0 \frac{1}{a^2} \left( 3r^2 \exp \left(-r^2 / a^2\right) - 2r^4 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right)\). Therefore, the divergence is: \(\nabla \cdot \vec{E}(r) = \frac{1}{r^2}\left[2V_0 \frac{1}{a^2} \left( 3r^2 \exp \left(-r^2 / a^2\right) - 2r^4 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right)\right]\). Now, we can find the charge density using Gauss' law: \(\rho(r) = \epsilon_0 \nabla \cdot \vec{E}(r) = 2\epsilon_0 V_0 \frac{1}{a^2} \left( 3 \exp \left(-r^2 / a^2\right) - 2r^2 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right)\).

To calculate the total charge, we will integrate the charge density over the entire volume: \(Q = \int \rho(\vec r) dV = \int \rho(r) r^2 \sin\theta dr d\theta d\phi\). Evaluating the integral: \(Q = 2\epsilon_0 V_0 \frac{1}{a^2} \int_0^\infty \int_0^\pi \int_0^{2\pi} \left( 3 \exp \left(-r^2 / a^2\right) - 2r^2 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right)r^2 \sin\theta dr d\theta d\phi\). Notice that the charge density is spherically symmetric, so we can simplify the integral: \(Q = 4\pi \epsilon_0 V_0 \frac{1}{a^2} \int_0^\infty \left( 3r^2 \exp \left(-r^2 / a^2\right) - 2r^4 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right) dr\). Calculating this integral, we get: \(Q = 4\pi \epsilon_0 V_0 a^3 \left( \frac{3\pi}{8} - \frac{3\pi}{4} + \frac{\pi}{4}\right) = 0\). The total charge of the region is 0.

The charge density is given by \(\rho(r) = 2\epsilon_0 V_0 \frac{1}{a^2} \left( 3 \exp \left(-r^2 / a^2\right) - 2r^2 \frac{1}{a^2}\exp \left(-r^2 / a^2\right)\right)\). Based on this expression, we can see that the charge density is spherically symmetric and centered at the origin. There is higher concentration of charge near the origin, and it decreases exponentially as we move away from the center. Practically, it will be like a Gaussian centered at the origin, having positive charges near the center and negative charges surrounding them to create a net charge of 0.