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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 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 \(0 < r < R, R < r < 2 R,\) and \(r > 2 R .(\) b) Graph the electric-field magnitude as a function of \(r .\)

Short Answer

Expert verified
The electric field is zero for \(0 < r < R\), \(\frac{kQ}{r^2}\) for \(R < r < 2R\), and \(\frac{k(2Q)}{r^2}\) for \(r > 2R\).

Step by step solution

01

Understand the Problem Setup

We have a conducting sphere at the center with radius \( R \) and total charge \( Q \). Around it, an insulating shell with radius \( 2R \) also carries a total charge \( Q \). Charges are uniformly distributed on both.
02

Region 1 - Electric Field for \(0 < r < R\)

Within a conductor, the electric field inside is zero. Therefore, for \( 0 < r < R \), the electric field is \( E = 0 \).
03

Region 2 - Electric Field for \(R < r < 2R\)

The electric field at a distance \( r \) from the center, where \( R < r < 2R \), is determined only by the charge \( Q \) on the conducting sphere. Using Gauss's Law, the electric field is given by:\[ E = \frac{kQ}{r^2} \]where \( k \) is Coulomb's constant.
04

Region 3 - Electric Field for \(r > 2R\)

For \( r > 2R \), the system can be thought of as a single point charge made up of the charges on the sphere and the shell. Thus, the total charge here is \( 2Q \). Using Gauss's Law, the electric field is:\[ E = \frac{k(2Q)}{r^2} \].
05

Graph the Electric Field as a Function of Distance

We construct a graph to show how the electric field magnitude changes with distance \( r \):- For \( 0 < r < R \), the electric field is zero, so the graph is a flat line at zero.- For \( R < r < 2R \), the electric field decreases with \( \frac{1}{r^2} \), reaching its minimum right before \( 2R \).- For \( r > 2R \), the electric field again decreases with \( \frac{1}{r^2} \) but has twice the field magnitudes compared to the \( R < r < 2R \) region because of the doubled charge.

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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 vital concept in electromagnetism. It relates the electric field produced by charged objects to the charge enclosed within a given surface. Mathematically, Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In formula terms, it is:\[ \Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \( \Phi_E \) is the electric flux, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.
For symmetrical systems like a sphere or shell, this law makes it simpler to calculate electric fields. It allows us to consider an imaginary Gaussian surface around our charged objects.
  • If we can determine the total charge within this Gaussian surface,
  • then we can deduce the electric field.
This simplifies complex scenarios with difficult charge distributions.
Understanding Gauss's Law can vastly reduce the complexity of electric field calculations, as seen in systems like concentric spheres and shells.
Conducting Sphere
A conducting sphere is a classical component in electrostatics. It holds particular rules about charge behavior. In a conductor, charges can move freely. As a result:
  • Any excess charge resides on the sphere's surface.
  • The electric field inside, at any point, is zero.

In our case with the conducting sphere of radius \(R\) carrying a charge \(Q\), the charge is uniformly distributed over its surface. Gauss's Law tells us that inside the conducting sphere, the electric field is zero. This is due to the conductive nature ensuring that any electric field would cause a charge movement until neutrality is achieved inside. Hence,:\[ E = 0 \; \text{for} \; 0 < r < R \]
When moving outside the sphere but still within the insulating shell, the electric field depends on charges inside the Gaussian surface formed at distance \( r \). Here:\[ E = \frac{kQ}{r^2} \; \text{for} \; R < r < 2R \]
Understanding conducting spheres helps in comprehending more complex multi-layered systems.
Insulating Shell
An insulating shell characterizes a non-conducting material covered over a sphere. While the charge in a conductor relocates to its surface, charges in an insulator remain fixed in position. This causes different influences on the electric field calculations.
In this scenario, the insulating shell has a charge\(Q\) distributed uniformly across it, creating an additional source of electric field effect at greater distances:
  • Inside the shell, no charge is enclosed till you reach its inner radius.
  • The charges only begin to affect as you cross the inner boundary, influencing the total field outside.
For distances beyond the shell (\( r > 2R \)), the system is treated as a collection of charges. With both the sphere and shell combined spatially into a point charge of \(2Q\), the field is given by:\[ E = \frac{k(2Q)}{r^2} \; \text{for} \; r > 2R \]
This highlights the insulating role, setting it apart from the conductive behavior, and showing how it influences multi-layered systems like the one in the exercise.
Electric Field Graph
An electric field graph visually represents how the electric field varies with distance. It's a powerful tool for understanding field behavior over different regions. In our example:
  • For \( 0 < r < R \), the electric field is zero inside the conducting sphere.
  • From \( R < r < 2R \), the field starts from its maximum value at \( R \) and decreases with distance, following the inverse square law \( \frac{1}{r^2} \).
  • For \( r > 2R \), the field continues to decrease with distance, as it now accounts for charge \(2Q\), still following the \( \frac{1}{r^2} \) rule.

The graph typically starts with a flat line at zero, a decline in the middle, and another declining trend at greater distances with higher magnitudes for greater distances.
Graphs like these aid in comparing theoretical predictions to real measurements, helping in grasping field behavior analytically and visually. They encapsulate the entire spatial electric field description in a straightforward curve, reinforcing insights into charge distribution effects on the electric field.

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Most popular questions from this chapter

The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c .\) The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \lambda. Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) . (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1150 \mathrm{~N} / \mathrm{C}\) just outside the surface?

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length \(+5.20 \mu \mathrm{C} / \mathrm{m} .\) What magnitude of force does one line of charge exert on a \(0.0500-\mathrm{m}\) section of the other line of charge?

A very long conducting tube (hollow cylinder) has inner radius \(a\) and outer radius \(b\) . It carries charge per unit length \(+\alpha\) where \(\alpha\) is a positive constant with units of \(\mathrm{C} / \mathrm{m} .\) A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+\alpha .\) (a) Calculate the electric field in terms of \(\alpha\) and the distance \(r\) from the axis of the tube for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(r > b .\) Show your results in a graph of \(E\) as a function of \(r .\) (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho .\) (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\) . What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r=R .\) (d) Graph the electric-field magnitude as a function of \(r\) from \(r=0\) to \(r=3 R .\)

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