Question

A solid nonconducting sphere has a volume charge distribution given by \(\rho(r)=(\beta / r) \sin (\pi r / 2 R) .\) Find the total charge contained in the spherical volume and the electric field in the regions \(rR\). Show that the two expressions for the electric field equal each other at \(r=R\).

Short answer

Question: Find the total charge contained in a solid nonconducting sphere of radius R with volume charge distribution given by: \(\rho(r) = (\beta / r) \sin (\pi r / 2 R)\). Determine the electric field in the region \(rR\). Verify that they are equal at the boundary \(r=R\). Answer: 1. The total charge contained in the sphere is \(Q = 8 \pi \beta\). 2. The electric field for the region \(rR\) is \(E = \frac{2 \beta}{\epsilon_{0} r^{2}}\). 4. The expressions for the electric field are equal at \(r=R\).

Step-by-step solution

Total Charge within the Sphere

To calculate the total charge contained in the spherical volume, let's integrate the given charge density function. We have the charge density: \(\rho(r) = (\beta / r) \sin (\pi r / 2 R)\). The volume element in spherical coordinates is \(dV = r^2 \sin{\theta} dr d\theta d\phi\). We will integrate from r = 0 to R, \(\theta\) = 0 to \(\pi\), and \(\phi\) = 0 to \(2\pi\): \(Q = \int \rho (r) dV = \int_{0}^{R} \int_{0}^{\pi} \int_{0}^{2\pi} \big(\frac{\beta}{r}\big) \sin(\frac{\pi r}{2R}) r^2 \sin{\theta} dr d\theta d\phi\) Now, we compute the integral as three separate integrals: \(Q = \big(\int_{0}^{R} \sin(\frac{\pi r}{2R}) r dr\big) \big(\int_{0}^{\pi} \sin(\theta) d\theta\big) \big(\int_{0 }^{2 \pi} d \phi\big) \beta\) After integrating and simplifying, we get the total charge: \(Q = 8 \pi \beta\).

Electric Field in the Region \(r < R\)

For the region within the sphere (r < R), we will use Gauss's Law to calculate the electric field: \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\epsilon_0}\) We will consider a Gaussian surface of radius r inside the sphere. Since the electric field and area vector are parallel to each other, the dot product is simplified: \(E \oint dA = \frac{Q_{encl}}{\epsilon_0}\) Since the area of a sphere is \(A = 4\pi r^2\), we have \(E 4\pi r^2 = \frac{Q_{encl}}{\epsilon_0}\) Now, let's find the enclosed charge within the Gaussian surface: \(Q_{encl} = \int_{0}^{r} \int_{0}^{\pi} \int_{0}^{2\pi} (\frac{\beta}{r'}) (\sin(\frac{\pi r'}{2R})) r'^2 \sin{\theta} dr' d\theta d\phi\) Using the same procedure as in Step 1, we find \(Q_{encl} = 8 \pi \beta \frac{r^{3}}{R^3}\) Plugging this into Gauss's Law, we obtain: \(E 4\pi r^2 = \frac{8 \pi \beta \frac{r^{3}}{R^3}}{\epsilon_0}\) Solving for E, we get: \(E = \frac{2 \beta r}{\epsilon_{0} R^{3}}\) for \(r < R\).

Electric Field in the Region \(r > R\)

For the region outside the sphere (r > R), we can treat the sphere as a point charge. Therefore, we can use the formula for the electric field due to a point charge: \(E = \frac{kQ}{r^{2}} = \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{r^{2}}\) Substituting the total charge \(Q = 8\pi \beta\): \(E = \frac{1}{4 \pi \epsilon_{0}}\frac{8 \pi \beta}{r^{2}}\) Simplifying, we get: \(E = \frac{2 \beta}{\epsilon_{0} r^{2}}\) for \(r > R\)

Verify Equality of Electric Fields at \(r=R\)

To show that the expressions for the electric field are equal at \(r=R\), we equate the expressions for the two regions at \(r=R\): \(\frac{2 \beta R}{\epsilon_{0} R^{3}} = \frac{2 \beta}{\epsilon_{0} R^{2}}\) Simplifying, we get: \(\frac{2 \beta}{\epsilon_{0} R^{2}} = \frac{2 \beta}{\epsilon_{0} R^{2}}\) The expressions are equal at \(r=R\).

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle that relates the distribution of electric charge to the resulting electric field. In simple terms, it states that the electric field times the area of a closed surface, known as a Gaussian surface, is proportional to the total charge enclosed within that surface. The formula for Gauss's Law is expressed as: \[\begin{equation}\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{encl}}}{\epsilon_0}\end{equation}\]where \(\vec{E}\) is the electric field, \(d\vec{A}\) is a vector representing an infinitesimal area on the Gaussian surface, \(Q_{\text{encl}}\) is the total charge enclosed, and \(\epsilon_0\) is the permittivity of free space. This law is particularly useful for symmetrical charge distributions, such as spherical or cylindrical geometries, as it simplifies the computation of the electric field. In the context of our exercise, Gauss's Law is used to find the electrical field both inside and outside a sphere with a given charge distribution.

Spherical Charge Distribution

A spherical charge distribution is a charge distribution that exhibits spherical symmetry. This means that the charge density at any point within the distribution depends only on the distance from the center of the sphere and is independent of the direction. In the exercise given, we encounter a nonconducting sphere with a volume charge density given by \(\rho(r) = (\beta / r) \sin (\pi r / 2 R)\).

For a uniform spherical charge distribution, Gauss's Law allows us to compute the electric field by considering a spherical Gaussian surface concentric with the charge distribution. However, in our case, the charge distribution is non-uniform and the charge density varies with radius. Despite this, due to the symmetry of the problem, the electric field at a given radius r can still be determined by evaluating the total charge within a sphere of that radius and using Gauss's Law.

Volume Charge Density

Volume charge density, denoted as \(\rho(r)\), represents the amount of charge per unit volume at a particular location within a three-dimensional charge distribution. It quantifies how densely electric charge is packed in space. In our exercise, the volume charge density of the sphere varies with radius r and is described by the function \(\rho(r) = (\beta / r) \sin (\pi r / 2 R)\).

To find the total charge within the entire spherical volume, or within any spherical region thereof, one has to integrate the charge density over the respective volume. The volume charge density can be a complex function of the position within the charge distribution, as is the case in the given exercise. The concept of volume charge density is essential for understanding how charge is spread out in space and impacts the resulting electric field.

Electric Field Regions

The electric field generated by a charge distribution can have different expressions depending on the region of space being considered. For a spherically symmetric charge distribution, like the one in our problem, there are typically two regions of interest: inside and outside the spherical distribution.

Inside the sphere, marked as the region where \(r < R\), the electric field can vary with position and depends on the amount of charge contained within the radius r. For calculation, we use a Gaussian surface that lies within the sphere. Outside the sphere, in the region where \(r > R\), all the charges in the sphere can be treated as if they were at the center, similar to a point charge. Here, the electric field only depends on the distance from the center of the sphere and the total charge. Our exercise demonstrates how Gauss's Law can be used to determine these fields in the respective regions and confirms their continuity at the boundary by showing that the expressions for the electric field are the same at the surface of the sphere, \(r = R\).