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A solid conducting sphere with radius \(R\) that carries positive charge \(Q\) is concentric with a very thin insulating shell of radius \(2R\) 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 \(0 < r < R, R < r < 2R\), and \(r > 2R\). (b) Graph the electric-field magnitude as a function of \(r\).

Short Answer

Expert verified
(a) Electric fields: \( 0 < r < R, E = 0 \); \( R < r < 2R, E = \frac{Q}{4\pi \varepsilon_0 r^2} \); \( r > 2R, E = \frac{Q}{2\pi \varepsilon_0 r^2} \). (b) Graph plot E versus r shows step decrease at r=2R.

Step by step solution

01

Understand the Problem Setup

You have two objects: a solid conducting sphere with radius \( R \) and a charge \( Q \), and an insulating spherical shell with radius \( 2R \) also carrying a charge \( Q \). The task is to find the electric field in the regions: \( 0 < r < R \), \( R < r < 2R \), and \( r > 2R \).
02

Inside the conducting sphere \( 0 < r < R \)

In a conductor, the electric field inside the material is zero. Hence, in the region \( 0 < r < R \), the electric field is \( E = 0 \).
03

Outside the conducting sphere but inside the insulating shell \( R < r < 2R \)

Applying Gauss's Law to this region, consider a Gaussian surface with radius \( r \) where \( R < r < 2R \). The enclosed charge is just the charge on the conducting sphere, which is \( Q \).By Gauss's Law:\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0} \]\[ E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \]Thus:\[ E = \frac{Q}{4\pi \varepsilon_0 r^2} \] The direction of the electric field is radial and outward.
04

Outside both the sphere and shell \( r > 2R \)

Consider a Gaussian surface with radius \( r > 2R \). The enclosed charge within this surface will be the sum of the charges on both the sphere and the shell, totaling \( 2Q \).By Gauss's Law:\[ E \cdot 4\pi r^2 = \frac{2Q}{\varepsilon_0} \]Thus:\[ E = \frac{2Q}{4\pi \varepsilon_0 r^2} = \frac{Q}{2\pi \varepsilon_0 r^2} \] The direction of the electric field is radial and outward.
05

Summary of Electric Fields

- For \( 0 < r < R \): \( E = 0 \)- For \( R < r < 2R \): \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \)- For \( r > 2R \): \( E = \frac{Q}{2\pi \varepsilon_0 r^2} \)
06

Graphing the Electric Field Magnitude

Create a graph with \( r \) on the x-axis and \( |E| \) on the y-axis. The graph will show:- \( E = 0 \) for \( 0 < r < R \).- A decreasing curve \( \sim \frac{1}{r^2} \) between \( R < r < 2R \).- A lower decreasing curve \( \sim \frac{2}{r^2} \) beyond \( r > 2R \). The curve will step down at \( r = 2R \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauss's Law
Gauss's Law is a powerful tool for calculating electric fields, especially in cases involving symmetric charge distributions. It states that the total electric flux through a closed surface is equal to the charge enclosed divided by the electric constant, \[ \varepsilon_0 \]. This can be expressed mathematically as:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0} \]
Where:
  • \( \oint \vec{E} \cdot d\vec{A} \) is the electric flux through the closed surface.
  • \( Q_{enclosed} \) is the total charge enclosed within the surface.
Gauss's Law is particularly useful for spherically symmetric charge distributions, such as a conducting sphere or an insulating shell. Using a Gaussian surface that matches the symmetry of the problem, we can simplify the calculation of the electric field. This law helps us avoid complex integrations, making it easier to find electric fields in various regions around charged objects.
Conducting Sphere
A conducting sphere is an object where the conductive material allows the free flow of electric charge throughout its volume. When a conducting sphere is charged, the charges redistribute themselves on the surface of the sphere in such a way that the electric field inside the sphere is zero. This is due to electrostatic equilibrium.
For a spherical conductor of radius \( R \) carrying a total charge \( Q \), the charge resides on the outer surface.Here are a few important points about conducting spheres:
  • Inside the conductor (i.e., for any radius \( r < R \)), the electric field \( E \) is zero, because the charges have redistributed to the surface, canceling any internal electric fields.
  • Outside the conductor (i.e., for any radius \( r > R \)), the electric field behaves as if all the charge were concentrated at the center of the sphere.
These characteristics simplify the application of Gauss's Law in determining the electric fields associated with conducting spheres.
Insulating Shell
An insulating shell is quite different from a conducting sphere because it does not allow charges to move freely. Instead, the charge can be distributed throughout the volume or on the surface of the insulator.
In the case of a spherical insulator, it's common to have the charge uniformly distributed over the surface, as specified in this exercise with radius \( 2R \). The challenge here is to find the electric field at different distances from the center:
  • Inside the hollow region of the shell (for \( r < 2R \) but greater than \( R \)), the charge outside the Gaussian surface does not influence the field inside it due to symmetrical distribution. The field is determined by any enclosed charge, which would be zero if there is no inner charge.
  • Beyond the shell (for \( r > 2R \)), the shell's charge can be treated as a point charge at its center, as determined by using Gauss's Law.
This behavior is fundamentally different from that of the conducting sphere, owing to the charge retention abilities of insulating materials.
Electric Field Intensity
Electric field intensity is a measure of the strength of an electric field at a particular point. It is denoted by \( E \) and is defined as the force experienced per unit positive charge placed at that point, measured in Newton per Coulomb (N/C).
The intensity of the electric field varies depending on the distance from the charge source:
  • For \( 0 < r < R \): Inside the conducting sphere, \( E \) is zero due to the absence of an electric field inside a conductor.
  • For \( R < r < 2R \): The electric field intensity is inversely proportional to the square of the distance, \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \), resulting from the charge \( Q \) enclosed by the Gaussian surface centered at the sphere.
  • For \( r > 2R \): Both charges are now effectively enclosed, and \( E = \frac{Q}{2\pi \varepsilon_0 r^2} \), again reducing with distance to the power of two, but now with an increased magnitude reflecting the total charge \( 2Q \).
Understanding electric field intensity and how it varies is crucial to mastering concepts of electrostatics and analyzing behaviors of charges under different configurations.

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Most popular questions from this chapter

A very small object with mass 8.20 \(\times\) 10\(^{-9}\) kg and positive charge 6.50 \(\times\) 10\(^{-9}\) C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \(\times\) 10\(^{-8}\) C/m2. The object is initially 0.400 m from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

In one experiment the electric field is measured for points at distances \(r\) from a uniform line of charge that has charge per unit length \(\lambda\) and length \(l\), where \(l \gg r\). In a second experiment the electric field is measured for points at distances \(r\) from the center of a uniformly charged insulating sphere that has volume charge density \(\rho\) and radius \(R =\) 8.00 mm, where \(r > R\). The results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment: For each set of data, draw two graphs: one for \(Er^2\) versus r and one for \(Er\) versus \(r\). (a) Use these graphs to determine which data set, A or B, is for the uniform line of charge and which set is for the uniformly charged sphere. Explain your reasoning. (b) Use the graphs in part (a) to calculate \(\lambda\) for the uniform line of charge and \(\rho\) for the uniformly charged sphere.

In a region of space there is an electric field \(\overrightarrow{E}\) that is in the z-direction and that has magnitude \(E =\) [964 N/(C \(\cdot\) m)]\(x\). Find the flux for this field through a square in the \(xy\)-plane at \(z =\) 0 and with side length 0.350 m. One side of the square is along the \(+x\)-axis and another side is along the \(+y\)-axis.

A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 4.50 nC of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

A solid metal sphere with radius 0.450 m carries a net charge of 0.250 nC. Find the magnitude of the electric field (a) at a point 0.100 m outside the surface of the sphere and (b) at a point inside the sphere, 0.100 m below the surface.

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