91Ó°ÊÓ

In physics, the electric field is a vector field surrounding electric charges. It represents the force that a electric charge would experience per unit of charge placed within the field. The electric field, denoted as \(\mathbf{E}\), shows both the direction and magnitude of the force. It plays a crucial role in understanding how charges interact in different scenarios.
An electric field has units of volts per meter (V/m) and its direction is defined as the direction of the force that a positive test charge would experience. For a uniform electric field as in our case, where the charge distribution is along the z-axis, the electric field can be derived from the electric displacement \(\mathbf{D}\) and vacuum permittivity \(\epsilon_0\) as:

• \(\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}\)
• Given \(\mathbf{D} = (0, 0, \rho_0 \cdot w)\), then \(\mathbf{E} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\)

This shows that the electric field has only a z-component and remains constant in the x and y directions.

Electric displacement is a vector field often denoted as \(\mathbf{D}\), used to account for free and bound charge distributions in materials. In free space, it directly correlates to the amount of free charge enclosed within a given region. Gauss's law for electric displacement states:

• \(\oint_{S} \mathbf{D} \cdot d\mathbf{a} = Q_{free}\)

In our problem, since the volume charge density is distributed along the z-axis, only the z-component \(D_z\) plays a role. By choosing an appropriate Gaussian surface, we simplified the problem because:

• The net contribution from x and y components of \(\mathbf{D}\) is null.

Thus, you can determine the electric displacement by evaluating the charges within the Gaussian surface. As calculated:

• \(\mathbf{D} = (0, 0, \rho_0 \cdot w)\)