Question

In a region in free space, electric flux density is found to be $$ \mathbf{D}=\left\\{\begin{array}{lr} \rho_{0}(z+2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (-2 d \leq z \leq 0) \\ -\rho_{0}(z-2 d) \mathbf{a}_{z} \mathrm{C} / \mathrm{m}^{2} & (0 \leq z \leq 2 d) \end{array}\right. $$ Everywhere else, \(\mathbf{D}=0 .\left(\right.\) a) Using \(\nabla \cdot \mathbf{D}=\rho_{v}\), find the volume charge density as a function of position everywhere. (b) Determine the electric flux that passes through the surface defined by \(z=0,-a \leq x \leq a,-b \leq y \leq b\). (c) Determine the total charge contained within the region \(-a \leq x \leq a\), \(-b \leq y \leq b,-d \leq z \leq d .(d)\) Determine the total charge contained within the region \(-a \leq x \leq a,-b \leq y \leq b, 0 \leq z \leq 2 d\).

Short answer

b) What is the electric flux, \(\Psi\), through the surface at \(z=0\)? c) What is the total charge, \(Q_c\), within the region \(-a \leq x \leq a\), \(-b \leq y \leq b\), \(-d \leq z \leq d\)? d) What is the total charge, \(Q_d\), within the region \(-a \leq x \leq a\), \(-b \leq y \leq b\), \(0 \leq z \leq 2d\)? a) The volume charge density, \(\rho_v\), is the same for both regions and equals \(\rho_0\). b) The electric flux, \(\Psi\), through the surface at \(z=0\) is equal to \(-4ab\rho_0d\). c) The total charge, \(Q_c\), within the given region is equal to \(8abd\rho_0\). d) The total charge, \(Q_d\), within the given region is also equal to \(8abd\rho_0\).

Step-by-step solution

Calculate the divergence of \(\mathbf{D}\)

To find the volume charge density, use the divergence of the electric flux density vector, \(\nabla \cdot \mathbf{D} = \rho_v\). Since \(\mathbf{D}\) only has a \(z\) component, the divergence can be simplified to: $$ \rho_v = \frac{\partial D_z}{\partial z} $$

Differentiate \(\mathbf{D}\) with respect to \(z\) within the given ranges

For the given ranges of \(z\), take the partial derivative of \(D_z\) with respect to \(z\): For \(-2d \leq z \leq 0\): $$ \frac{\partial}{\partial z}\left( \rho_0(z+2d) \right) = \rho_0 $$ For \(0 \leq z \leq 2d\): $$ \frac{\partial}{\partial z}\left(-\rho_0(z-2d) \right) = \rho_0 $$ Thus, the volume charge density \(\rho_v\) is the same for both regions and equals \(\rho_0\). #b) Determine the electric flux through the surface defined by \(z=0\)

Calculate the electric flux through the surface

Since \(\mathbf{D}\) has only a \(z\) component, and at the surface \(z=0\), the component of \(\mathbf{D}\) is parallel to the normal vector of the surface. Therefore, the electric flux \(\Psi\) through the surface can be calculated by integrating \(\mathbf{D} \cdot \mathbf{n}\) over the surface, where \(\mathbf{n}\) is the normal vector. $$ \Psi =\int_{-a}^a \int_{-b}^b D_z(0, x, y) dy dx $$

Evaluate the integral

As determined in part a, the \(Dz\) component at \(z=0\) is \(\rho_0(-2d)\). Thus, we can evaluate the integral: $$ \Psi =\int_{-a}^a \int_{-b}^b -2d\rho_0 dy dx = -4ab\rho_0d $$ #c) Determine the total charge within the region \(-a \leq x \leq a\), \(-b \leq y \leq b, -d \leq z \leq d\)

Integrate the volume charge density over the specified region

To find the total charge, integrate the volume charge density, \(\rho_v\), over the specified region: $$ Q_c = \int_{-a}^a \int_{-b}^b \int_{-d}^d \rho_v dx dy dz $$

Evaluate the integral

Since \(\rho_v = \rho_0\), the integral can be evaluated: $$ Q_c = \int_{-a}^a \int_{-b}^b \int_{-d}^d \rho_0 dx dy dz = 8abd\rho_0 $$ #d) Determine the total charge within the region \(-a \leq x \leq a\), \(-b \leq y \leq b, 0 \leq z \leq 2d\)

Integrate the volume charge density over the specified region

To find the total charge, integrate the volume charge density, \(\rho_v\), over the specified region: $$ Q_d = \int_{-a}^a \int_{-b}^b \int_{0}^{2d} \rho_v dx dy dz $$

Evaluate the integral

Since \(\rho_v = \rho_0\), the integral can be evaluated: $$ Q_d = \int_{-a}^a \int_{-b}^b \int_{0}^{2d} \rho_0 dx dy dz = 8abd\rho_0 $$ The total charge in the specified region is the same as in part c.

Key concepts

Volume Charge Density

When studying electric fields, understanding the distribution of charge in space is crucial, and this is where the concept of volume charge density, denoted by \(\rho_v\), becomes important. Think of volume charge density as a measure of how much electric charge is distributed within a unit volume of space.

Using the divergence of the electric flux density vector, which in mathematical terms is expressed as \(abla \cdot \mathbf{D} = \rho_v\), allows us to calculate this density when the flux density \(\mathbf{D}\) is known. In our example, by differentiating \(\mathbf{D}\) with respect to \(z\), we find that no matter where you are within the specified range, the volume charge density remains constant and equal to \(\rho_0\).

This information is essential for understanding the behavior of electric fields and can be used for further electrical engineering calculations, such as finding the total charge within a certain volume.

Divergence Theorem

The divergence theorem is a powerful mathematical tool that helps us relate the flow of a vector field through a surface to the behavior of the field within a volume. In the language of calculus, it equates the volume integral of the divergence of a vector field to the surface integral of that field.

In the context of electric flux density \(\mathbf{D}\), if we wanted to know how much of this field is emanating from a given volume, we can integrate the divergence of \(\mathbf{D}\) over that volume. This theorem simplifies complex three-dimensional problems by turning them into easier surface integral problems, thus being a cornerstone in electrical engineering, especially when combined with Gauss's Law, as we'll see in the following sections.

Electric Flux

Electric flux, represented by the symbol \(\Psi\), is a measure that quantifies the amount of electric field that passes through a given area. Imagine it like the number of field lines crossing a surface. The more lines that pass through, the greater the electric flux.

In regards to our exercise, to determine the electric flux through a surface at \(z=0\), we calculate it by integrating the product of \(\mathbf{D}\) and the normal vector \(\mathbf{n}\) over the area of the surface. Given that \(\mathbf{D}\) and \(\mathbf{n}\) are parallel at \(z=0\), the calculated flux ends with the value of \(\Psi = -4ab\rho_0d\), illustrating how charge distribution affects the field in a specific area.

Gauss's Law

Gauss's Law is a fundamental principle that links the electric flux emanating from a closed surface to the charge enclosed by that surface. Mathematically, it's represented by \(abla \cdot \mathbf{D} = \rho_v\), indicating that the divergence of the electric flux density is equal to the volume charge density.

In our scenario, Gauss's Law aids in understanding how charge is spread out within different regions by computing the total charge within a closed surface. We apply the law by integrating the volume charge density over the volume enclosed by the surface. This concept is crucial when designing and analyzing electric fields and forces in various practical applications, such as capacitors and insulators.