Question

In free space, a volume charge of constant density \(\rho_{v}=\rho_{0}\) exists within the region \(-\infty

Short answer

Answer: The electric field \(\mathbf{E}\) everywhere in space is given by: \(\mathbf{E} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\)

Step-by-step solution

Choose a Gaussian surface

To apply Gauss's law, we will choose a Gaussian surface that will make the calculations easier. Since the volume is infinite in x and y directions and finite in the z direction, we can choose a cuboid Gaussian surface with a width \(w\) along the z-axis, height \(h\) along the y-axis, and length \(l\) along the x-axis, centered at the origin.

Apply Gauss's law for \(\mathbf{D}\)

Gauss's law states that \(\oint_{S} \mathbf{D} \cdot d\mathbf{a} = Q_{free}\), where \(S\) is the Gaussian surface and \(Q_{free}\) is the charge contained within the surface. For our cuboid Gaussian surface, we have six faces. We can see that the x and y components of \(\mathbf{D}\) should be zero, as there is no preferred direction in these dimensions. Thus, we will focus on the z-component, and our Gaussian surface integral becomes: \(\oint_{S} D_z d\mathbf{a} = Q_{free}\) Now we find the total charge enclosed by the Gaussian surface: \(Q_{free} = \rho_0 \cdot V = \rho_0 \cdot (w \cdot h \cdot l)\) where \(V\) is the volume of the Gaussian surface.

Calculate the electric displacement vector \(\mathbf{D}\)

We will now calculate the electric displacement \(\mathbf{D}\) through the Gaussian surface. Since \(D_x\) and \(D_y\) are zero, we only need to consider \(D_z\): \(\oint_{S} D_z d\mathbf{a} = D_z (\text{top face} - \text{bottom face})\) The top face has an area of \(l \cdot h\) and is parallel to the z-axis, so its contribution is \(D_z\cdot(l\cdot h)\). Similarly, the bottom face gives \(-D_z\cdot(l\cdot h)\). Plugging these into Gauss's law: \(D_z(l\cdot h) - D_z(l\cdot h) = \rho_0 \cdot (w \cdot h \cdot l)\) We see that \(D_z = \rho_0 \cdot w\). This gives us the electric displacement vector: \(\mathbf{D} = (0, 0, \rho_0 \cdot w)\)

Find the electric field \(\mathbf{E}\)

We have the electric displacement vector \(\mathbf{D}\), and we know its relationship with the electric field \(\mathbf{E}\): \(\mathbf{D} = \epsilon_0 \mathbf{E}\), where \(\epsilon_0\) is the vacuum permittivity. Hence, the electric field is given by: \(\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0} = \frac{(0, 0, \rho_0 \cdot w)}{\epsilon_0} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\) Finally, we have found the electric field \(\mathbf{E}\) everywhere in space: \(\mathbf{E} = (0, 0, \frac{\rho_0 \cdot w}{\epsilon_0})\)

Key concepts

Electric field

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

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:

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