Problem 49 - University Physics with Modern Physics

Question

An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho\) (\(r\)) \(= \alpha/r\), where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a\), what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a < r < b\)? (b) A point charge \(q\) is placed at the center of the hollow space, at \(r =\) 0. In terms of \(\alpha\) and \(a\), what value must \(q\) have (sign and magnitude) in order for the electric field to be constant in the region \(a < r < b\), and what then is the value of the constant field in this region?

Step-by-Step Solution

Answer

a) \( E = \frac{\alpha(r^2 - a^2)}{2\varepsilon_0 r^2} \); b) \( q = -2\pi\alpha a^2 \), \( E_0 = \frac{\alpha (b^2 - 2a^2)}{2\varepsilon_0 b^2} \)

Step 1: Understand the Problem

We need to find the electric field due to a charge density inside a hollow insulating sphere. We are also given that a point charge is placed at the center of the hollow sphere, and we are tasked with finding conditions under which the electric field remains constant between the inner and outer radii.

Step 2: Apply Gauss's Law

To find the electric field, we apply Gauss's Law, which can be expressed as: \[ \oint E \, dA = \frac{Q_{enc}}{\varepsilon_0} \]where \(Q_{enc}\) is the charge enclosed by the Gaussian surface. We will use a spherical Gaussian surface of radius \(r\) such that \(a < r < b\).

Step 3: Calculate the Enclosed Charge

The charge density is given by \(\rho(r) = \frac{\alpha}{r}\). The volume element in spherical coordinates is \(dV = 4\pi r^2 \, dr\), and thus the total charge enclosed is:\[ Q_{enc} = \int_{a}^{r} \rho(r) \, dV = \alpha \int_{a}^{r} \frac{1}{r} \, 4\pi r^2 \, dr = 4\pi\alpha \int_{a}^{r} r \, dr \]Solving the integral gives:\[ Q_{enc} = 4\pi\alpha \left( \frac{r^2}{2} - \frac{a^2}{2} \right) = 2\pi\alpha (r^2 - a^2) \]

Step 4: Derive the Electric Field Expression

Substituting \(Q_{enc}\) into the Gauss's Law equation:\[ E(4\pi r^2) = \frac{2\pi \alpha (r^2 - a^2)}{\varepsilon_0} \]Solving for \(E\), we find:\[ E = \frac{\alpha(r^2 - a^2)}{2\varepsilon_0 r^2} \]

Step 5: Introduce Point Charge Condition

Let's consider the condition with the point charge \(q\) at the center. We want the electric field to be constant for \(a < r < b\). For a constant electric field \(E_0\), the net charge enclosed at any \(r > a\) must result in the same field:\[ E_0 = \frac{q + 2\pi\alpha (r^2 - a^2)}{4\pi\varepsilon_0 r^2} \]This must be independent of \(r\), hence set the derivative to zero.

Step 6: Solve for the Point Charge

For the field expression to be independent of \(r\), the term involving \(r\) must cancel out. Setting up:\[ \frac{-q}{4\pi\varepsilon_0 r^2} = \frac{2\pi\alpha}{4\pi\varepsilon_0} \]Solving gives:\[ q = -2\pi\alpha a^2 \]

Step 7: Find the Constant Electric Field

With the charge \(q = -2\pi\alpha a^2\) calculated, substitute back to solve for \(E_0\):\[ E_0 = \frac{q + 2\pi\alpha (b^2 - a^2)}{4\pi\varepsilon_0 r^2} \]Substituting \(q\) causes cancellation of \(r\)-dependent terms, giving:\[ E_0 = \frac{2\pi\alpha (b^2 - a^2 - a^2)}{4\pi\varepsilon_0 b^2} = \frac{\alpha (b^2 - 2a^2)}{2\varepsilon_0 b^2}\]

Key Concepts

Electric Field in a Hollow Sphere

When you're dealing with electric fields, understanding how they behave in various configurations is key. One common setup is having a hollow sphere with charge distributed within a shell, but not inside the inner cavity.

In the given problem, the electric field arises due to a volume charge density distributed in the hollow shell of an insulating sphere. It is important to determine how the electric field behaves between the inner radius (\(a\)) and the outer radius (\(b\)) of the sphere.

To find the electric field, we use **Gauss's Law**, which is a powerful technique that relates the electric field over a spherical surface to the charge enclosed within it. Gauss's Law states:

• \( \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \)

Here, **\(Q_{enc}\)** is the charge enclosed by the Gaussian surface, and **\(\varepsilon_0\)** is the permittivity of free space.

By calculating the enclosed charge and applying Gauss's Law, we derive the expression for the electric field **\(E\)** at a point between the inner and outer surfaces of the sphere. This field depends on how the charge is distributed within the material of the shell.

Volume Charge Density of the Sphere

Volume charge density (\(\rho\)) of a sphere tells us how charge is distributed throughout the volume of the sphere. It is crucial to understand this distribution in cases where materials have non-uniform charge density.

In this setup, the volume charge density is a function of the radial distance from the center, given as:

• \( \rho(r) = \frac{\alpha}{r} \)

Here, **\(\alpha\)** represents a positive constant that describes the strength of the charge density's variation with radius.

The total charge enclosed, **\(Q_{enc}\)**, within a spherical region from **\(a\)** to any intermediate radius **\(r\)** can be calculated using integration:

• The integral of the product of the charge density and the volume element (in spherical coordinates) is used:
  • \( Q_{enc} = 4\pi\alpha \left( \frac{r^2}{2} - \frac{a^2}{2} \right) \)

This equation formed by integrating the given charge density illustrates how the charge accumulates as you move outward within the material of the shell.

Properties of a Hollow Sphere

A hollow sphere is a classic geometry encountered in physics, especially when discussing electrostatics and gravity.

A uniform hollow sphere means that the region inside the sphere is empty (or hollow), but the outer shell contains distributed charge. The real interest is what happens when there's a point charge right at the center. Understanding these principles is key to many applications.
**Key Features of Hollow Spheres**

• Inner and outer radii, denoted by \(a\) and \(b\).
• The hollow part ensures no charge exists in the inner cavity.
• The shell, however, can hold a distributed charge that affects the field outside.

When addressing electric fields, the hollow sphere presents a unique scenario. Inside the shell (between the radii **\(a\)** and **\(b\)**), the electric field is shaped by both the shell's distribution of charge and any point charge located at the center. This causes the electric field to be zero within the hollow part if there's no central charge. If a point charge is placed in the center, it affects the symmetry and can lead to a uniform field across the inner boundary.

By manipulating the amount and sign of this central charge, you can regulate the electric field in particular regions, achieving conditions like uniform fields between specific regions of the sphere.