Problem 43 - University Physics with Modern Physics

Question

22.43. A solid conducting sphere with radius \(R\) that carries poositive charge \(Q\) is concentric with a very thin insulating shell of radius 2\(R\) that also carries charge \(Q\) . The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions \(02 R\) . (b) Graph the electric- field magnitude as a function of \(r .\)

Step-by-Step Solution

Answer

(a) Inside sphere, \(E=0\); between sphere and shell, \(E=\frac{Q}{4\pi\varepsilon_0 r^2}\); outside shell, \(E=\frac{Q}{2\pi\varepsilon_0 r^2}\). (b) E-field is zero inside, decreases between sphere and shell, and continues to decrease outside.

Step 1: Understanding the Problem

We need to find the electric field in three distinct regions: inside the conducting sphere (\(02R\)). Given that the sphere and shell carry charges \(+Q\), we apply the principles of electric fields concerning conductors and shells.

Step 1: Region \(0<r<R\)

In this region, the point is inside a solid conducting sphere. In a conducting sphere, any excess charge resides on its surface, and the electric field inside a charged conductor is zero. Therefore, the electric field \(E(r) = 0\).

Step 2: Region \(R<r<2R\)

Here, we're outside the conductor but inside the insulating shell.The electric field at any point is due to the charge \(Q\) on the conducting sphere's surface.Using Gauss's law, take a Gaussian surface of radius \(r\) with \(R

Step 3: Region \(r>2R\)

In this region, past the insulating shell, the total charge enclosed by a Gaussian surface of radius \(r\) is \(Q + Q = 2Q\).Applying Gauss's law, we have:\[E(4\pi r^2) = \frac{2Q}{\varepsilon_0} \]Solving for \(E\):\[E(r) = \frac{2Q}{4\pi \varepsilon_0 r^2} = \frac{Q}{2\pi \varepsilon_0 r^2}\]

Step 4: Graphing the Electric Field Magnitude

To graph the electric field magnitude as a function of \(r\), plot the following points:- From \(02R\), \(E(r) = \frac{Q}{2\pi \varepsilon_0 r^2}\), which continues to decrease, aligning with a classical \(\frac{1}{r^2}\) decay for an enclosed charge of \(2Q\).

Key Concepts

Gauss's law

Gauss's law is a fundamental principle in electromagnetism that helps us understand electric fields. It relates the electric flux through a closed surface to the charge enclosed by that surface. The law is described mathematically as: \[\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\]where \(\Phi\) is the electric flux, \(\vec{E}\) is the electric field, \(d\vec{A}\) is a differential area on the closed surface, \(Q_{enc}\) is the charge enclosed, and \(\varepsilon_0\) is the vacuum permittivity.
This law simplifies the calculation of electric fields around symmetric charge distributions, such as spheres and cylinders.

• It allows us to find electric fields by using a hypothetical closed surface known as a Gaussian surface.
• This approach is especially useful in cases with high symmetry, like spherical or cylindrical shapes, making complex calculations much simpler.

Understanding Gauss's law helps in predicting how electric fields behave when enclosed charges change or when they are distributed evenly over a surface.

conducting sphere

A conducting sphere is a perfect example of applying Gauss's law due to its symmetrical properties. Conductors have a unique feature: any excess charge resides exclusively on the surface. Therefore, the electric field inside a conducting sphere is zero.
In our exercise, the sphere is charged with \(+Q\), and analysis of a point inside the sphere (region \(0

The electric field is zero inside because charges redistribute themselves on the surface to maintain equipotential surfaces.

Only when you move outside the surface, the sphere acts like a point charge, and hence we can use Gauss's law effectively there.

This characteristic denotes why the electric field inside a conducting body under static conditions is nil; it is an essential aspect of conductors and how they distribute charges.

insulating shell

The insulating shell in the problem has its charge \(+Q\) distributed uniformly over its surface. Unlike conductors, insulators allow charge distribution to remain fixed in place.
When examining a region between the sphere and the shell (\(R

The electric field between the shell and sphere is associated with the charge on the sphere.

The insulating shell's charge affects the field only when considering a Gaussian surface that extends beyond the shell.

This discrete distribution results in an electric field that varies differently than if the shell were conducting; key to this distinction is understanding how insulating properties affect charge behavior and thereby the surrounding electric field.

charge distribution

Charge distribution is crucial to determine the electric field in any system. In our scenario, both the conducting sphere and the insulating shell have a charge \(+Q\). However, the way these charges are allocated affects the resulting field.

• In conductors, charges move to the outer surface, creating a zero field inside.
• For the insulating shell, the charges remain where placed, uniformly spreading over the surface.

This understanding of charge distribution enables the determination of electric fields in and around these objects using Gauss's law.
By comparing the two regions of interest, \(R2R\), we see distinct behaviors:

• Between the surfaces, the electric influence is only from the sphere, as the shell's charges have no electric field effect until we cross beyond it.
• Beyond the shell, the entire charge system, \(2Q\), must be acknowledged according to Gauss's law, yielding an electric field that exhibits an inverse square decay due to the total enclosed charge of \(2Q\).

These charge characteristics are fundamental when mapping out electric field magnitudes across different regions in problem scenarios.

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Problem 42

Next

Problem 45

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Problem 48

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Key Concepts

Gauss's law

Gauss's law is a fundamental principle in electromagnetism that helps us understand electric fields. It relates the electric flux through a closed surface to the charge enclosed by that surface. The law is described mathematically as: \[\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\]where \(\Phi\) is the electric flux, \(\vec{E}\) is the electric field, \(d\vec{A}\) is a differential area on the closed surface, \(Q_{enc}\) is the charge enclosed, and \(\varepsilon_0\) is the vacuum permittivity.
This law simplifies the calculation of electric fields around symmetric charge distributions, such as spheres and cylinders.

• It allows us to find electric fields by using a hypothetical closed surface known as a Gaussian surface.
• This approach is especially useful in cases with high symmetry, like spherical or cylindrical shapes, making complex calculations much simpler.

Understanding Gauss's law helps in predicting how electric fields behave when enclosed charges change or when they are distributed evenly over a surface.

conducting sphere

A conducting sphere is a perfect example of applying Gauss's law due to its symmetrical properties. Conductors have a unique feature: any excess charge resides exclusively on the surface. Therefore, the electric field inside a conducting sphere is zero.
In our exercise, the sphere is charged with \(+Q\), and analysis of a point inside the sphere (region \(0

The electric field is zero inside because charges redistribute themselves on the surface to maintain equipotential surfaces.

Only when you move outside the surface, the sphere acts like a point charge, and hence we can use Gauss's law effectively there.

This characteristic denotes why the electric field inside a conducting body under static conditions is nil; it is an essential aspect of conductors and how they distribute charges.

insulating shell

The insulating shell in the problem has its charge \(+Q\) distributed uniformly over its surface. Unlike conductors, insulators allow charge distribution to remain fixed in place.
When examining a region between the sphere and the shell (\(R

The electric field between the shell and sphere is associated with the charge on the sphere.

The insulating shell's charge affects the field only when considering a Gaussian surface that extends beyond the shell.

This discrete distribution results in an electric field that varies differently than if the shell were conducting; key to this distinction is understanding how insulating properties affect charge behavior and thereby the surrounding electric field.

charge distribution

Charge distribution is crucial to determine the electric field in any system. In our scenario, both the conducting sphere and the insulating shell have a charge \(+Q\). However, the way these charges are allocated affects the resulting field.

• In conductors, charges move to the outer surface, creating a zero field inside.
• For the insulating shell, the charges remain where placed, uniformly spreading over the surface.

This understanding of charge distribution enables the determination of electric fields in and around these objects using Gauss's law.
By comparing the two regions of interest, \(R2R\), we see distinct behaviors:

• Between the surfaces, the electric influence is only from the sphere, as the shell's charges have no electric field effect until we cross beyond it.
• Beyond the shell, the entire charge system, \(2Q\), must be acknowledged according to Gauss's law, yielding an electric field that exhibits an inverse square decay due to the total enclosed charge of \(2Q\).

These charge characteristics are fundamental when mapping out electric field magnitudes across different regions in problem scenarios.