Question

Volume charge density is located as follows: \(\rho_{v}=0\) for \(\rho<1 \mathrm{~mm}\) and for \(\rho>2 \mathrm{~mm}, \rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3}\) for \(1<\rho<2 \mathrm{~mm} .\) (a) Calculate the total charge in the region \(0<\rho<\rho_{1}, 0

Short answer

To summarize the solutions of each part: a) The total charge in the region is given by: $$ Q = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}. $$ b) The electric displacement \(D_\rho\) at \(\rho = \rho_1\) is given by: $$ D_\rho = \frac{4\mu C/m^3(\rho_1^3 - 1)}{3\rho_1}. $$ c) The values of \(D_\rho\) at the specified values of \(\rho\) are: - \(\rho = 0.8\) mm: \(D_\rho = 0 \mu C/m^2\) - \(\rho = 1.6\) mm: \(D_\rho \approx 1.409 \mu C/m^2\) - \(\rho = 2.4\) mm: \(D_\rho \approx 1.667 \mu C/m^2\).

Step-by-step solution

Identify volume charge density function

The volume charge density \(\rho_v\) is given as follows: $$\rho_{v}=0 \text{ for } \rho<1 \mathrm{~mm} \text{ and for } \rho>2 \mathrm{~mm},$$ $$\rho_{v}=4 \rho \mu \mathrm{C} / \mathrm{m}^{3} \text{ for } 1<\rho<2 \mathrm{~mm}.$$

Calculate the total charge

The total charge in the region is given by integrating the volume charge density over the volume under consideration: $$Q = \int_{0}^{2\pi} \int_{0}^{L} \int_{0}^{\rho_1} \rho_v \cdot \rho d\rho dz d\phi. $$ Since \(\rho_v = 0\) for \(\rho < 1\) mm, we can consider only the region where \(\rho_v = 4\rho \mu C / m^3\). Thus, the integral will give: $$Q = \int_{0}^{2\pi} d\phi \int_{0}^{L} dz \int_{1}^{\rho_1} (4 \rho \mu C/m^3) \rho d\rho.$$ We can split the multi-variable integral to make calculations easier: $$Q= [\int_{0}^{2\pi} d\phi][\int_{0}^{L} dz][\int_{1}^{\rho_1} (4 \rho^2 \mu C/m^3) d\rho].$$

Evaluate the integrals

Compute the integrals separately: $$\int_{0}^{2\pi} d\phi = 2\pi,$$ $$\int_{0}^{L} dz = L,$$ $$\int_{1}^{\rho_1} (4\rho^2\mu C/m^3) d\rho = 4 \mu C/m^3 \int_{1}^{\rho_1} \rho^2 d\rho.$$ Evaluate the last integral: $$4\mu C/m^3 [(\rho_1^3 / 3) - (1^3 / 3)] = 4\mu C/m^3 (\rho_1^3 - 1)/3.$$

Multiply the results

Now, combine the results of the three integrals: $$Q = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}.$$ This gives the total charge in the region of interest. #b) Determine \(D_{\rho}\) at \(\rho=\rho_{1}\)#

Apply Gauss's Law

We can use Gauss's law to determine the electric displacement \(D_\rho\): $$\oint \vec{D} \cdot \mathrm{d}\vec{A} = Q_{enc},$$ where \(Q_{enc}\) is the charge enclosed within the Gaussian surface.

Choose appropriate Gaussian surface

Select a cylindrical Gaussian surface with height \(L\), radius \(\rho_1\), and a base parallel to the \(xy\) plane. Only \(D_\rho\) contributes to the integral, as \(D_z\) is perpendicular to the curved surface area and \(D_\phi\) is perpendicular to the top and bottom surfaces.

Integrating over the surface

The integral becomes: $$\oint \vec{D} \cdot \mathrm{d}\vec{A} = D_\rho \int_{0}^{2\pi} \int_{0}^{L} \rho_1 dL d\phi = 2\pi L \cdot \rho_1 D_\rho.$$ Since we have found the total charge in part (a), we can substitute it into Gauss's Law: $$2\pi L \cdot \rho_1 D_\rho = 2\pi L\cdot \frac{4\mu C/m^3(\rho_1^3 - 1)}{3}.$$

Solve for \(D_\rho\)

Divide both sides by \(2\pi L \cdot \rho_1\): $$D_\rho = \frac{4\mu C/m^3(\rho_1^3 - 1)}{3\rho_1}.$$ This expression provides the value of \(D_\rho\) at \(\rho = \rho_1\). #c) Evaluate \(D_{\rho}\) at specified values of \(\rho\)# Using the result from part (b), we can now evaluate \(D_{\rho}\) for the three values of \(\rho\) given in the problem:

\(\rho = 0.8\) mm

The value of \(\rho\) is less than 1 mm, so no charge is enclosed. Therefore, \(D_\rho = 0\).

\(\rho = 1.6\) mm

Using the obtained formula: $$D_\rho = \frac{4\mu C/m^3(1.6^3 - 1)}{3(1.6)} \approx 1.409\mu C/m^2.$$

\(\rho = 2.4\) mm

Since the volume charge density is zero for \(\rho > 2\) mm, the enclosed charge will be the same as when \(\rho = 2\) mm. Using the formula: $$D_\rho = \frac{4\mu C/m^3(2^3 - 1)}{3(2.4)} \approx 1.667\mu C/m^2.$$

Key concepts

Volume Charge Density

The volume charge density is a measurement that tells us how much electric charge is packed into a given volume of space. In this exercise, the volume charge density, denoted by \( \rho_v \), is defined piecewise, meaning it changes depending on the location \( \rho \), a radial coordinate in cylindrical coordinates.

• For \( \rho < 1 \) mm, there is no charge, so \( \rho_v = 0 \).
• For \( \rho > 2 \) mm, the charge density is also zero.
• But for \( 1 < \rho < 2 \) mm, the density is \( \rho_v = 4\rho \mu \text{C/m}^3 \).

This variation means charges are present only within a specific cylindrical shell, where their density increases linearly with \( \rho \). Understanding this helps to visualize where and how the charges are distributed within the region, which is crucial for calculations involving electric fields and forces.

Gauss's Law

Gauss's Law is a fundamental principle used in electromagnetism, which provides a relation between the electric displacement field \( \vec{D} \) and the charge enclosed within a given surface. Gauss's Law is expressed mathematically as:\[ \oint \vec{D} \cdot d\vec{A} = Q_{enc}, \]where:

• \( \oint \vec{D} \cdot d\vec{A} \) is the integral of the electric displacement over the closed surface.
• \( Q_{enc} \) is the total charge enclosed by the surface.

In our exercise, we use this law to determine the electric displacement, \( D_\rho \), at specific radial distances \( \rho \). By selecting an appropriate Gaussian surface, such as a cylindrical surface co-axial with our charge distribution, we can simplify the computation to find \( D_\rho \), given the symmetric distribution of charges.The utility of Gauss's Law lies in its ability to calculate the effects of complex charge distributions in a straightforward manner, especially when symmetry is involved.

Electric Displacement

The electric displacement field, represented by \( \vec{D} \), is an important concept in understanding how electric fields are organized in materials. It accounts for both free and bound charges within a medium.In cylindrical coordinates, \( D_\rho \) specifically refers to the electric displacement component in the radial direction. It takes into account the enclosed charge quantified through Gauss's Law. In our solution, \( D_\rho \) at various values of \( \rho \) is computed using the formula derived in part (b) of the exercise:\[ D_{\rho} = \frac{4\mu C/m^3(\rho_1^3 - 1)}{3\rho_1} \]This formula gives \( D_\rho \) by taking into account that the volume charge density is zero outside of the 1 to 2 mm range. Thus, it helps in predicting the influence of the enclosed charge at varying radial distances.

Cylindrical Coordinates

Cylindrical coordinates are particularly useful when dealing with problems that have cylindrical symmetry, like the one in this exercise. These coordinates consist of \( \rho \), the radial distance from the z-axis; \( \phi \), the angular coordinate; and \( z \), the height along the axis.

• \( \rho \) is analogous to the radial distance in polar coordinates and is vital when considering radial distributions like volume charge densities.
• \( \phi \) represents the angular position and stays constant in this problem because our charge distribution does not depend on angle.
• \( z \) adds a third dimension, allowing us to consider volumes, not just areas.

The choice of cylindrical coordinates simplifies the integration process. Key aspects of the volume, like height \( L \) and angle symmetry, let us easily compute the total charge and analyze the electric fields using Gauss's Law. Thus, understanding cylindrical coordinates enhances comprehension of problems involving round or tubular regions and symmetrical charge distributions.