Problem 48 - University Physics with Modern Physics

Question

22.48. A solid conducting sphere with radius \(R\) carries a positive total charge \(Q .\) The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius 2\(R\) . The insulating shell has a uniform charge density \(\rho\) (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field (magnitude and direction) in each of the regions \(02 R\) . Show your results in a graph of the radial component of \(\overrightarrow{\boldsymbol{E}}\) as a function of \(\boldsymbol{r}\) . (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.

Step-by-Step Solution

Answer

(a) \(\rho = -\frac{3Q}{7\pi R^3}\), (b) E: 0 for \(r2R\); varies for \(R

Step 1: Determine charge density for neutral system

To ensure the net charge of the system is zero, the charge on the insulating shell must be equal and opposite to the charge of the sphere. The sphere has charge \(Q\). The volume of the insulating shell is \(\frac{4}{3}\pi((2R)^3 - R^3)\), and the charge of the shell is \(\rho\times \text{volume}\). Setting the shell's charge to \(-Q\):\[\rho \cdot \frac{4}{3}\pi((2R)^3 - R^3) = -Q\]. This simplifies to \[\rho = -\frac{3Q}{7\pi R^3}\].

Step 2: Find the electric field for the region r < R

Inside the conducting sphere, the electric field is zero because any excess charge resides on its surface. Thus, for \(0 < r < R\), \(E = 0\).

Step 3: Determine the electric field for region R < r < 2R

Use Gauss's Law to find the electric field in the region. A Gaussian surface with radius \(r\) where \(R < r < 2R\) encloses the charge \(Q\) on the sphere and part of the charge on the shell.\[Q_{enclosed} = Q + \rho \cdot \frac{4}{3}\pi(r^3 - R^3) = Q - Q (\frac{r^3}{R^3} - 1) = Q (2 - \frac{r^3}{R^3})\].\The electric field at this radius is: \[E = \frac{1}{4\pi\varepsilon_0}\frac{Q(2 - \frac{r^3}{R^3})}{r^2}\].

Step 4: Find the electric field for r > 2R

For this region, the total enclosed charge within a Gaussian surface of radius \(r > 2R\) is zero since the total system charge is zero. Thus, using Gauss's Law, the electric field \(E = 0\) for \(r > 2R\).

Step 5: Plot the electric field as a function of r

For \(0 < r < R\), \(E = 0\). For \(R < r < 2R\), the electric field decreases from a peak value at \(r = R\) until it reaches zero at \(r = 2R\), as described by \(E = \frac{1}{4\pi\varepsilon_0}\frac{Q(2 - \frac{r^3}{R^3})}{r^2}\). When \(r > 2R\), \(E = 0\) again. The graph is continuous at \(r = R\) and \(r = 2R\), confirming the rule about the discontinuity of electric fields.

Step 6: Discuss the continuity of electric field

The results agree with the rule since the electric field experiences no discontinuity at \(r = R\) or \(r = 2R\), where there are no thin charged surfaces, indicative of continuity at boundaries of distributed charges.

Key Concepts

Gaussian Surface

The Gaussian surface is a critical tool in evaluating electric fields, as used in the context of Gauss's Law. In the exercise, we use different Gaussian surfaces for different regions. A Gaussian surface is an imaginary closed surface used to apply Gauss's Law, allowing us to calculate electric fields easily.
It should enclose the charge distribution symmetrically. For instance, when evaluating charges within a spherical shell, a keen choice is a spherical Gaussian surface. In analyzing our conducting sphere and insulating shell, consider these helpful points:

• Gaussian surfaces help simplify symmetry-dependent electric field problems.
• In regions with constant charge density, enclosing more volume means enclosing more charge.
• Positioning the surface strategically results in effective calculation of enclosed charge, tailoring results for specific regions.

Understanding Gaussian surfaces is vital for effectively applying Gauss's Law and determining electric fields in symmetric distributions.

Charge Density

Charge density, denoted by \( \rho \), refers to the amount of charge per unit volume. In our exercise, it determines how much charge is distributed throughout the insulating shell.
The charge density calculation is pivotal to ensure that the entire system remains electrically neutral.Let's recall some essential points about charge density:

• Charge density \( \rho \) is calculated using the formula \( \rho = \frac{q}{V} \), where \( q \) is the charge and \( V \) is the volume.
• To ensure zero net charge in the system, the shell’s charge density must produce a charge equal in magnitude but opposite to the sphere’s charge.
• The expression \( \rho = -\frac{3Q}{7\pi R^3} \) results from equating the charge on the sphere with that of the shell.

Properly calculating charge density helps in avoiding errors in predicting electric field behavior and maintaining system neutrality.

Conducting Sphere

A conducting sphere is essential in electrostatics as it affects how charge is distributed. When a sphere is conducting, the charge resides on the surface, creating uniform charge distribution across this outer layer.
For the region inside the conducting sphere, the electric field is zero, as the charges spread out symmetrically over the surface. Important features include:

• Conductors allow charges to move freely until equilibrium is reached on the outer surface.
• The inner region maintains zero electric field due to cancellation of charge effects.
• An external electric field results solely from the surface charge.

This fundamental understanding of conducting spheres assists in establishing how charge distributions might influence surrounding electric fields and further informs the application of Gauss's Law.

Gauss's Law

Gauss’s Law is one of Maxwell’s equations dealing with electric charge. It connects electrical flux through a closed surface to the charge enclosed within it. In simple terms, it relates electric fields to the distribution of electric charge.
The law is expressed as:\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\]For practical applications like our exercise:

• Calculating the electric field within the insulating shell involves using Gauss’s Law to evaluate enclosed charges at varying radii.
• In regions where the net enclosed charge is zero, the electric field also becomes zero. This simplifies calculations significantly.
• It endorses using symmetry to simplify complex charge distributions into manageable shapes for integration over closed surfaces.

Applying Gauss's Law effectively simplifies complex electrostatic problems, providing insightful relationships between charge and electric fields.

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Problem 49

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Key Concepts

Gaussian Surface

The Gaussian surface is a critical tool in evaluating electric fields, as used in the context of Gauss's Law. In the exercise, we use different Gaussian surfaces for different regions. A Gaussian surface is an imaginary closed surface used to apply Gauss's Law, allowing us to calculate electric fields easily.
It should enclose the charge distribution symmetrically. For instance, when evaluating charges within a spherical shell, a keen choice is a spherical Gaussian surface. In analyzing our conducting sphere and insulating shell, consider these helpful points:

• Gaussian surfaces help simplify symmetry-dependent electric field problems.
• In regions with constant charge density, enclosing more volume means enclosing more charge.
• Positioning the surface strategically results in effective calculation of enclosed charge, tailoring results for specific regions.

Understanding Gaussian surfaces is vital for effectively applying Gauss's Law and determining electric fields in symmetric distributions.

Charge Density

Charge density, denoted by \( \rho \), refers to the amount of charge per unit volume. In our exercise, it determines how much charge is distributed throughout the insulating shell.
The charge density calculation is pivotal to ensure that the entire system remains electrically neutral.Let's recall some essential points about charge density:

• Charge density \( \rho \) is calculated using the formula \( \rho = \frac{q}{V} \), where \( q \) is the charge and \( V \) is the volume.
• To ensure zero net charge in the system, the shell’s charge density must produce a charge equal in magnitude but opposite to the sphere’s charge.
• The expression \( \rho = -\frac{3Q}{7\pi R^3} \) results from equating the charge on the sphere with that of the shell.

Properly calculating charge density helps in avoiding errors in predicting electric field behavior and maintaining system neutrality.

Conducting Sphere

A conducting sphere is essential in electrostatics as it affects how charge is distributed. When a sphere is conducting, the charge resides on the surface, creating uniform charge distribution across this outer layer.
For the region inside the conducting sphere, the electric field is zero, as the charges spread out symmetrically over the surface. Important features include:

• Conductors allow charges to move freely until equilibrium is reached on the outer surface.
• The inner region maintains zero electric field due to cancellation of charge effects.
• An external electric field results solely from the surface charge.

This fundamental understanding of conducting spheres assists in establishing how charge distributions might influence surrounding electric fields and further informs the application of Gauss's Law.

Gauss's Law

Gauss’s Law is one of Maxwell’s equations dealing with electric charge. It connects electrical flux through a closed surface to the charge enclosed within it. In simple terms, it relates electric fields to the distribution of electric charge.
The law is expressed as:\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\]For practical applications like our exercise:

• Calculating the electric field within the insulating shell involves using Gauss’s Law to evaluate enclosed charges at varying radii.
• In regions where the net enclosed charge is zero, the electric field also becomes zero. This simplifies calculations significantly.
• It endorses using symmetry to simplify complex charge distributions into manageable shapes for integration over closed surfaces.

Applying Gauss's Law effectively simplifies complex electrostatic problems, providing insightful relationships between charge and electric fields.