For the volume charge density ρ, the divergence of the electric field intensity will be equal to Select one:a. ε0 ρv  b. ρv / ε0   c.  ε02ρv d. ρv  / ε02

To solve the problem, we will use Gauss's law and the relationship between the electric field intensity $\mathbf{E}$ and volume charge density $\rho$.

1. ### Break Down the Problem

We need to analyze the divergence of the electric field intensity and how it relates to the volume charge density. This can be described mathematically using Maxwell's equations, specifically Gauss's law.

2. ### Relevant Concepts

According to Gauss's law in differential form, the divergence of the electric field $\mathbf{E}$ is given by: $\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$ Where:

• $\nabla \cdot \mathbf{E}$ is the divergence of the electric field.
• $\rho$ is the volume charge density.
• $\varepsilon_0$ is the permittivity of free space.

3. ### Analysis and Detail

From the equation above, we can see that the divergence of the electric field is directly proportional to the volume charge density divided by the permittivity of free space. Thus, it can be explicitly written as: $\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$

4. ### Verify and Summarize

Now, we can analyze the options given:

• a. $\varepsilon_0 \rho v$
• b. $\frac{\rho v}{\varepsilon_0}$
• c. $\varepsilon_0^2 \rho v$
• d. $\frac{\rho v}{\varepsilon_0^2}$

The correct relationship matches option b: $\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$

Final Answer

The divergence of the electric field intensity is equal to: $\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \quad \text{(Option b)} \quad \frac{\rho v}{\varepsilon_0}$