Question

A uniform volume charge density of \(80 \mu \mathrm{C} / \mathrm{m}^{3}\) is present throughout the region \(8 \mathrm{~mm}10 \mathrm{~mm}\), find \(D_{r}\) at \(r=20 \mathrm{~mm}\).

Short answer

Answer: The radial component of the electric displacement field is \(D_r(10 \text{ mm}) \approx 7.85 \text{ N/C}\) and \(D_r(20 \text{ mm}) \approx 1.96 \text{ N/C}\).

Step-by-step solution

Calculate total charge inside the sphere

To find the total charge inside the spherical surface with radius \(r=10\) mm, we need to integrate the volume charge density \(\rho_v\) over the given region (between radius \(r=8\) mm and \(r=10\) mm). Let \(Q\) be the total charge, \(r_1 = 8\) mm, \(r_2 = 10\) mm and \(\rho_v = 80 \times 10^{-6} \text{ C/m}^3\). The total charge \(Q\) can be found as: $$Q = \int_{V} \rho_v dV$$ Where \(dV\) is the volume element, given by \(dV = r^2 \sin \theta dr d\theta d\phi\) in spherical coordinates, with \(\theta\) and \(\phi\) representing the polar and azimuthal angles, respectively: $$Q = \int_{r_1}^{r_2} \int_{0}^{\pi} \int_{0}^{2\pi} (80 \times 10^{-6})r^2 \sin\theta dr d\theta d\phi$$

Perform the integration

Now, we will perform the integration: $$Q = \int_{r_1}^{r_2} (80 \times 10^{-6}) r^2 dr \cdot \int_{0}^{\pi}\sin\theta d\theta \cdot \int_{0}^{2\pi} d\phi$$ Solving each integral separately, we get: $$Q = (80 \times 10^{-6}) \left[\frac{r^3}{3}\right]_{r_1}^{r_2} \cdot [-\cos\theta]_{0}^{\pi} \cdot [\phi]_{0}^{2\pi}$$ $$Q = (80 \times 10^{-6}) \cdot \frac{(10^3 - 8^3)}{3} \cdot 2 \cdot 2\pi$$ Calculate the total charge: $$Q = 80 \times 10^{-6} \cdot \frac{936}{3} \cdot 4\pi \approx 3.14 \times 10^{-3} \text{ C}$$

Find \(D_r\) at \(r=10\) mm

To find the radial component of the electric displacement, \(D_r\), at \(r=10\) mm, we will use Gauss's law: $$\oint_{\Sigma} \vec{D} \cdot d\vec{\Sigma} = Q_{enclosed}$$ Since the problem has spherical symmetry, we can assume that the radial component of \(D_r\) is constant over the Gaussian surface. Thus, $$D_r \cdot \oint_{\Sigma} d\Sigma = Q_{enclosed}$$ Considering the Gaussian surface to be a sphere of radius \(r=10\) mm, the total surface area \(\Sigma\) is given by \(4\pi r^2\): $$D_r \cdot 4\pi (10 \times 10^{-3})^2 = 3.14 \times 10^{-3}$$ Solving for \(D_r\), $$D_r = \frac{3.14 \times 10^{-3}}{4\pi (10 \times 10^{-3})^2} = 7.85 \text{ N/C}$$

Find \(D_r\) at \(r=20\) mm

Now, let's find \(D_r\) at \(r=20\) mm, considering no charges for \(r>10\) mm. According to Gauss's law, $$D_r \cdot \oint_{\Sigma} d\Sigma = Q_{enclosed}$$ As the charge enclosed does not change, even though we have now doubled the radius, the enclosed charge is still \(Q = 3.14 \times 10^{-3}\) C. $$D_r \cdot 4\pi (20 \times 10^{-3})^2 = 3.14 \times 10^{-3}$$ Solving for \(D_r\) at \(r=20\) mm, $$D_r = \frac{3.14 \times 10^{-3}}{4\pi (20 \times 10^{-3})^2} = 1.96 \text{ N/C}$$ So, the radial component of the electric displacement field \(D_r\) is: (a) \(Q \approx 3.14 \times 10^{-3}\) C; (b) \(D_r(10 \text{ mm}) \approx 7.85 \text{ N/C}\); and (c) \(D_r(20 \text{ mm}) \approx 1.96 \text{ N/C}\).

Key concepts

Volume Charge Density

Volume charge density, often denoted by \(\rho_v\), is a measure of how much electric charge is distributed within a given volume. It tells us how much charge we have in one cubic meter of space. In many problems involving electrostatics, understanding the distribution of electric charge is essential to determine resultant electric fields and potential differences.

In our exercise, the volume charge density is given as \(80 \mu \mathrm{C}/\mathrm{m}^3\). This means that for every cubic meter of the volume we examine, there is 80 micro-coulombs of charge present. However, it's important to note that this charge density is not present everywhere. In the provided problem, the volume charge density applies only to a specific region where the radius \(r\) is between 8 mm and 10 mm. For all other regions, \(\rho_v = 0\).

Understanding volume charge density is crucial as it is the starting point for calculating the total charge within a specified volume. This sets the stage for applying further concepts like Gauss's Law to find electric fields.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux emerging from a closed surface to the charge enclosed by that surface. The mathematical representation of Gauss's Law is:

\[\oint_{\Sigma} \vec{E} \cdot d\vec{\Sigma} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\]

Where:

• \(\oint_{\Sigma} \vec{E} \cdot d\vec{\Sigma}\) is the electric flux through a closed surface \(\Sigma\).
• \(Q_{\text{enclosed}}\) is the total charge enclosed within surface \(\Sigma\).
• \(\varepsilon_0\) is the permittivity of free space.

In our problem, we simplified Gauss's Law to find the electric displacement field \(D\), using the relationship \(\vec{D} \cdot \vec{A} = Q_{\text{enclosed}}\) and considering symmetry.

For spherical symmetry, the magnitude of the electric field is uniform across the spherical surface. This allows us to solve for \(D_r\) (the radial component of the electric displacement field), given the total charge we calculated in the first step. Applying the law, we determine the electric displacement field both at the boundary at \(r = 10\, \text{mm}\) and further out at \(r = 20\, \text{mm}\).

Gauss's Law helps us resolve complex electric field calculations into more manageable forms by exploiting symmetry, as we see in solving for \(D_r\).

Electric Displacement Field

The electric displacement field, denoted **D**, is a vector field in electromagnetism that accounts for both the free charge and the bound charge in a material. It simplifies the relation between charge distributions and electric fields within materials. Specifically, it helps handle complex situations involving dielectric materials.

In our exercise, the problem simplifies due to the space being free of dielectric material effects. Here, the relationship between the electric displacement field \(D_r\) and the enclosed charge is given by:

\[D_r \cdot A = Q_{\text{enclosed}}\]

where \(A\) is the area of the spherical Gaussian surface. For a sphere with radius \(r\), \(A = 4\pi r^2\).

To find \(D_r\), we use the calculated total charge \(Q_{\text{enclosed}} = 3.14 \times 10^{-3} \text{ C}\) and solve accordingly for different radii. When the radius is 10 mm, we find the results to be \(D_r = 7.85 \text{ N/C}\), and when the radius is extended to 20 mm, \(D_r = 1.96 \text{ N/C}\).

Understanding **D** is imperative, as it simplifies calculations in the presence of charge distributions, especially in symmetric situations like spheres.

Spherical Coordinates

Spherical coordinates are a system of coordinates that fully describe a point in three-dimensional space using three values: the radial distance, the polar angle, and the azimuthal angle. In problems involving spherically symmetric charge distributions, spherical coordinates greatly simplify the mathematical process of integration.

In spherical coordinates:

• \(r\) is the radial distance from the origin to the point.
• \(\theta\) (theta) is the polar angle measured from the positive z-axis.
• \(\phi\) (phi) is the azimuthal angle measured in the x-y plane from the positive x-axis.

These coordinates are particularly useful for integrating over a spherical volume. In the given exercise, the volume element \(dV\) uses spherical coordinates, expressed as:

\[dV = r^2 \sin \theta dr d\theta d\phi\]

This representation aligns perfectly with the symmetry of the charge distribution and helps us calculate the total charge when integrated with the volume charge density. Using spherical coordinates eases the computation process by making the expressions for volume and surface area straightforward and manageable in the context of spherical symmetry.