Problem 42 - University Physics with Modern Physics

Question

22.42. A Sphere in a Sphere. A solid conducting sphere carry- ing charge \(q\) has radius a. It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\) . The hollow sphere has no net charge. (a) Derive expressions for the electric-field magnitude in terms of the distance \(r\) from the center for the regions \(rc .\) (b) Graph the magnitude of the electric field as a function of \(r\) from \(r=0\) to \(r=2 c\) (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2 c.

Step-by-Step Solution

Answer

(a) Electric field: 0 for \( r < a \) and \( b < r < c \), \( \frac{kq}{r^2} \) for \( a < r < b \) and \( r > c \). (c) Inner charge: \(-q\). (d) Outer charge: \(+q\). (e) Field lines radial outward.

Step 1: Understanding the Problem

We have a solid conducting sphere with a charge \( q \) and radius \( a \), placed inside a hollow conducting sphere with inner radius \( b \) and outer radius \( c \). The hollow sphere has no net charge. We need to find the electric field magnitude for different regions of \( r \) and address questions about charges and field lines.

Step 2: Electric Field for \( r < a \)

In this region, the electric field inside a conductor is zero. Hence, for \( r < a \), the electric field \( E \) is \( 0 \).

Step 3: Electric Field for \( a < r < b \)

Here, we apply Gauss's Law. The charge enclosed by a Gaussian surface of radius \( r \) (with \( a < r < b \)) is \( q \). By Gauss's Law, the electric field is given by \[ E = \frac{k q}{r^2} \] where \( k = \frac{1}{4 \pi \varepsilon_0} \).

Step 4: Electric Field for \( b < r < c \)

In this region, the hollow conducting sphere shields any internal electric fields. Since it's a conductor with no charge inside and overall charge neutrality, the electric field \( E = 0 \).

Step 5: Electric Field for \( r > c \)

In this region, applying Gauss's Law again, the charge enclosed is still \( q \), because the hollow sphere itself has no net charge. Thus, the electric field is given by \[ E = \frac{k q}{r^2} \].

Step 6: Graphing Electric Field

The graph of \( E \) versus \( r \) will show \( E = 0 \) for \( r < a \) and \( b < r < c \), and \( E = \frac{k q}{r^2} \) for \( a < r < b \) and \( r > c \). This results in a graph with peaks at \( a < r < b \) and after \( r = c \), with flat regions at \( r < a \) and \( b < r < c \).

Step 7: Charge on Inner Surface of Hollow Sphere

The inner surface of the hollow sphere will have a charge of \(-q\) to neutralize the field inside \( b < r < c \).

Step 8: Charge on Outer Surface of Hollow Sphere

Because the hollow sphere has no net charge, the outer surface must have a charge of \(+q\) to balance the \(-q\) on the inner surface.

Step 9: Sketching Electric Field Lines

The field lines emerge radially outward from the solid sphere for \( a < r < b \) and continue similarly for \( r > c \), illustrating the field around a charge distributed uniformly over a sphere.

Key Concepts

Conducting Spheres

In the concept of conducting spheres, it's crucial to understand how charges distribute themselves in materials that conduct electricity. Conductors have free electrons that can move about easily, and because of this, charges reside only on the surface of a conducting sphere. This happens because like charges repel, prompting the electrons to spread as far apart as possible, which means accumulating at the surface.

When there's an external electric field or when the conductor itself is charged, these electrons rearrange to cancel any internal electric fields. As a result, inside a conductor (like our solid sphere with radius less than a), the electric field remains zero.

• Conductors allow surface charge distribution.
• No internal electric field inside a conductor.
• Charges respond dynamically to external fields or charges.

Gauss's Law

Gauss's Law is a powerful tool for analyzing electric fields. It states that the net electric flux through a closed surface is proportional to the enclosed charge. Mathematically, it is represented as \[ \Phi_E = \oint \mathbf{E} \cdot \mathrm{d} \mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field vector, and \( Q_{\text{enc}} \) is the charge enclosed by the surface.

In our scenario of concentric spheres, Gauss's Law helps us determine the electric field at different radial distances from the center. By using a Gaussian surface at certain intervals, such as \( r > a \) or \( r > c \), we can derive the electric field magnitude accurately. This method simplifies calculations significantly, especially with symmetric charge distributions.

• Works well with symmetric charge distributions.
• Simplifies calculation of fields through surfaces.
• Key for understanding effects in and out of conductors.

Electric Field Magnitude

The magnitude of the electric field, denoted \( E \), depends on the distance from the charge source and the amount of charge. For our concentric spheres:

• For \( r < a \), \( E = 0 \) since the field inside the conductor is zero.
• For \( a < r < b \), the electric field is given by the formula \( E = \frac{k q}{r^2} \), where \( k \) is Coulomb's constant \( \left( k = \frac{1}{4 \pi \varepsilon_0} \right) \).
• For \( b < r < c \), the field is again zero, shielded by the conducting shell.
• For \( r > c \), the outer region captures the overall charge so the field returns to \( E = \frac{k q}{r^2} \).

The field decreases with distance, showing a characteristic \( 1/r^2 \) dependence, highlighting how electric influences diminish over space.

Charge Distribution

Charge distribution determines how charges are spread over the surfaces of conductors. In our double-sphere setup:

• Inner solid sphere carries a charge \( q \), which is distributed over its surface.
• Inner surface of the hollow sphere tries to neutralize interior fields, so it holds \(-q\). This results in zero net charge in the hollow region.
• Outer surface of the hollow sphere has to balance the entire sphere system's net zero charge. Therefore, it carries \( +q \).

Understanding these distributions helps in predicting field behavior. Visualizing charge and electric field lines around conductors can provide deeper insights into how fields interact with materials. These distributions ensure the electric neutrality and field continuity across the system.