Problem 61 - College Physics

Question

\(\bullet$$\bullet\) A total charge of magnitude \(Q\) is distributed uniformly within a thick spherical shell of inner radius \(a\) and outer radius b. (a) Use Gauss's law to find the electric field within the cavity \((r \leq a)\) . (b) Use Gauss's law to prove that the electric field outside the shell \((r \geq b)\) is exactly the same as if all the charge were concentrated as a point charge \(Q\) at the center of the sphere. (c) Explain why the result in part (a) for a thick shell is the same as that found in Example 17.10 for a thin shell. A thick shell can be viewed as infinitely many thin shells.)

Step-by-Step Solution

Answer

(a) \(E = 0\) for \(r \leq a\); (b) \(E = \frac{Q}{4\pi \epsilon_0 r^2}\) for \(r \geq b\); (c) Fields inside both shells are zero.

Step 1: Understand the Sphere Geometry

We have a thick spherical shell with an inner radius \(a\) and an outer radius \(b\), and it carries a total charge \(Q\) that is uniformly distributed.

Step 2: Apply Gauss's Law for the Region \(r \leq a\)

For the region within the cavity \(r \leq a\), we consider a Gaussian surface with radius \(r\) (where \(r < a\)). There is no charge enclosed by this Gaussian surface since all the charges reside in the shell area between \(a\) and \(b\). By Gauss's law, the electric field \(E\) is given by \(E \cdot A = \frac{q_{enc}}{\epsilon_0}\). Here, \(q_{enc} = 0\), so we have \(E \cdot (4\pi r^2) = 0\). Thus, \(E = 0\) within the cavity \(r \leq a\).

Step 3: Apply Gauss's Law for the Region \(r \geq b\)

For the region outside the shell \(r \geq b\), consider a spherical Gaussian surface of radius \(r\) where \(r \geq b\). The charge enclosed by this Gaussian surface is the total charge \(Q\). By Gauss's law, \(E \cdot A = \frac{Q}{\epsilon_0}\), where \(A = 4\pi r^2\). Therefore, the electric field is \(E = \frac{Q}{4\pi \epsilon_0 r^2}\). This is the same as the electric field produced by a point charge \(Q\) at the center of the sphere.

Step 4: Compare with Thin Shell Result

In Example 17.10, the electric field inside a thin shell was also zero because a thin shell can be thought of as having no charge within its sphere region. Similarly, for a thick shell, any point \(r \leq a\) is still within a region with no enclosed charge, leading to an electric field of zero. The analogy to infinitely many thin shells supports the same result.

Key Concepts

Electric Field

The concept of an electric field is essential when discussing charged objects and their effects on their surroundings. Essentially, an electric field represents the force per unit charge experienced by a small test charge placed in its vicinity. It indicates the direction and magnitude of the force that a positive charge would experience.

To express this mathematically when considering a point charge, the electric field is given by the formula:

• \( E = \frac{kQ}{r^2} \), where
  • \( E \) is the electric field intensity at a distance \( r \),
  • \( k \) is Coulomb's constant,
  • \( Q \) is the charge causing the field, and
  • \( r \) is the distance from the charge.

For the scenario of a spherical shell, the understanding of electric field inside and outside the shell is crucial to applying Gauss's law effectively. Inside a charged spherical shell, if the point is within the cavity, then by symmetry, the electric field is zero. Outside the spherical shell, the field behaves as if all the charge is concentrated at the center.

Spherical Shell

A spherical shell is a three-dimensional structure resembling a hollow ball, with a certain amount of space between its inner and outer surfaces. In physics, we often deal with thin or thick spherical shells.

- **Thin Spherical Shell:** It is considered having an insignificant thickness compared to its radius.- **Thick Spherical Shell:** It includes any shell that has a significant difference between its inner and outer radii.
To solve problems involving spherical shells, we can use Gauss’s law. In the case of a charged spherical shell, charges reside on the surface. Thus, inside the hollow part (\( r \leq a \) for a thick shell), no charge is present, and hence, the electric field is zero there. This results from symmetry and the properties of electrical conductors, where charges move to the outer surface.

Gauss's law allows us to treat the spherical shell’s electric field problems as if the entire charge were at its center. Outside the shell, the shell's entire charge seems to act as a point charge located at its center.

Charge Distribution

Charge distribution refers to how electrical charge is distributed over an object. In the context of Gauss's law and our spherical shell problem, the distribution of charge affects how we calculate the electric field.

For a uniformly charged thick spherical shell:

• All the charge resides between the inner radius \( a \) and the outer radius \( b \).
• This results in the cavity (region where \( r \leq a \)) having a charge of \( q_{enc} = 0 \).

Because no charge is within this cavity, the electric field in this region is zero. For the outside region (\( r \geq b \)), irrespective of the shell's thickness, the entire charge seems centralized, resulting in the field pattern that a point charge \( Q \) would create. Here Gauss's law leads to the equation \( E = \frac{Q}{4\pi \epsilon_0 r^2}\), depicting how field intensity decays with distance squared.