5. Consider an electric field $$ \vec{E}=C \frac{\vec{r}}{r^{2}} $$ (c:constant) (a) Calculate the divergence and curl of this $$ \vec{E} $$ field. (b) What is the unit of constant $$ C $$ ? (c) What charge density $$ \rho $$ ( $$ \vec{r} $$ ) would produce this $$ \vec{E} $$ feld? (d) What is the total charge for a sphere with radius $$ \vec{R} $$ ?

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To solve the problem step by step, we will address each part of the question regarding the electric field $$ \vec{E} = C \frac{\vec{r}}{r^2} $$.

### Part (a): Calculate the divergence and curl of the electric field $$ \vec{E} $$

1. **Divergence**: The divergence of a vector field $$ \vec{E} $$ is given by the formula:
   $$
   \nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}
   $$
   where $$ \vec{E} = (E_x, E_y, E_z) $$.

In spherical coordinates, the divergence of a radial field $$ \vec{E} = E_r \hat{r} $$ is given by:
   $$
   \nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r)
   $$
   Here, $$ E_r = C \frac{1}{r} $$.

2. **Curl**: The curl of a vector field $$ \vec{E} $$ is given by the formula:
   $$
   \nabla \times \vec{E} = \left( \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}, \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}, \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right)
   $$
   For a radial field, the curl is generally zero.

Now, let's calculate the divergence and curl of $$ \vec{E} $$.

#### Divergence Calculation
$$
E_r = C \frac{1}{r}
$$
$$
\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r) = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \cdot C \frac{1}{r}) = \frac{1}{r^2} \frac{\partial}{\partial r} (C r) = \frac{C}{r^2}
$$

#### Curl Calculation
Since $$ \vec{E} $$ is a radial field, we can conclude:
$$
\nabla \times \vec{E} = 0
$$

### Part (b): What is the unit of constant $$ C $$?

The electric field $$ \vec{E} $$ has units of volts per meter (V/m) or equivalently, newtons per coulomb (N/C). The term $$ \frac{\vec{r}}{r^2} $$ has units of $$ \frac{\text{m}}{\text{m}^2} = \frac{1}{\text{m}} $$. Therefore, the units of $$ C $$ must be such that:
$$
[C] \cdot \frac{1}{\text{m}} = \text{N/C}
$$
Thus, the units of $$ C $$ are:
$$
[C] = \text{N} \cdot \text{m} = \text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{kg} \cdot \text{m}^2/\text{s}^2
$$

### Part (c): What charge density $$ \rho(\vec{r}) $$ would produce this $$ \vec{E} $$ field?

Using Gauss's law, we know:
$$
\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}
$$
From part (a), we found:
$$
\nabla \cdot \vec{E} = \frac{C}{r^2}
$$
Thus,
$$
\frac{\rho}{\epsilon_0} = \frac{C}{r^2} \implies \rho = \epsilon_0 \frac{C}{r^2}
$$

### Part (d): What is the total charge for a sphere with radius $$ R $$?

To find the total charge $$ Q $$ within a sphere of radius $$ R $$, we integrate the charge density over the volume of the sphere:
$$
Q = \int_V \rho \, dV
$$
In spherical coordinates, the volume element $$ dV = r^2 \sin \theta \, dr \, d\theta \, d\phi $$. Thus,
$$
Q = \int_0^R \int_0^\pi \int_0^{2\pi} \epsilon_0 \frac{C}{r^2} r^2 \sin \theta \, d\phi \, d\theta \, dr
$$
This simplifies to:
$$
Q = \epsilon_0 C \int_0^R \int_0^\pi \int_0^{2\pi} \sin \theta \, d\phi \, d\theta \, dr
$$
Calculating the integrals:
$$
\int_0^{2\pi} d\phi = 2\pi, \quad \int_0^\pi \sin \theta \, d\theta = 2
$$
Thus,
$$
Q = \epsilon_0 C \cdot 2\pi \cdot 2 \cdot R = 4\pi \epsilon_0 C R
$$

### Summary of Results
(a) $$ \nabla \cdot \vec{E} = \frac{C}{r^2}, \quad \nabla \times \vec{E} = 0 $$

(b) $$ [C] = \text{kg} \cdot \text{m}^2/\text{s}^2 $$

(c) $$ \rho(\vec{r}) = \epsilon_0 \frac{C}{r^2} $$

(d) $$ Q = 4\pi \epsilon_0 C R $$
(a) Divergence: $$ \frac{C}{r^2} $$, Curl: 0 (b) $$ C $$ has units of $$ \text{kg} \cdot \text{m}^2/\text{s}^2 $$ (c) Charge density $$ \rho(\vec{r}) = \epsilon_0 \frac{C}{r^2} $$ (d) Total charge $$ Q = 4\pi \epsilon_0 C R $$

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