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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 \(02 R\). (b) Graph the electric-field magnitude as a function of \(r\).

Short Answer

Expert verified
Inside sphere: \(E = 0\); between sphere and shell: \(E = \frac{kQ}{r^2}\); outside shell: \(E = \frac{k(2Q)}{r^2}\).

Step by step solution

01

Understand the Problem

We have a conducting sphere with charge \(Q\) and an insulating shell, also with charge \(Q\), both concentric. We need to find the electric field in three regions: inside the conducting sphere \(0 < r < R\), between the sphere and the shell \(R < r < 2R\), and outside the shell \(r > 2R\).
02

Electric Field Inside the Conducting Sphere

For radial distances \(0 < r < R\), inside the conducting sphere, the electric field \(E\) is zero because the charge resides on the surface of the conductor due to electrostatic equilibrium: \[ E = 0 \text{ for } 0 < r < R. \]
03

Electric Field Between the Sphere and Shell

For \(R < r < 2R\), we only consider the sphere's charge. By Gauss's Law, the electric field can be found as:\[ E = \frac{kQ}{r^2} \quad \text{for } R < r < 2R, \]where \(k\) is Coulomb's constant, due to the symmetry and spherical distribution of the charge on the solid sphere.
04

Electric Field Outside the Shell

For \(r > 2R\), use both charges as a single point at the center (superposition principle): \[ E = \frac{k(2Q)}{r^2} \quad \text{for } r > 2R, \]where the total charge is \(2Q\).
05

Graph the Electric Field

To graph \(E(r)\), plot a piecewise function:- \(E = 0\) from \(r = 0\) to \(r = R\).- Then, \(E\) rapidly decreases as \(\frac{kQ}{r^2}\) from \(r = R\) to \(r = 2R\).- Beyond \(r = 2R\), \(E\) drops as \(\frac{k(2Q)}{r^2}\), reflecting the combined charge effect.

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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 used to determine electric fields around symmetrical charge distributions. It states that the electric flux through a closed surface is proportional to the enclosed charge. In mathematical form, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]where \(\Phi_E\) is the electric flux, \(\vec{E}\) is the electric field, \(d\vec{A}\) is a differential area vector on the closed surface, \(Q_{\text{enc}}\) is the total charge enclosed in the surface, and \(\varepsilon_0\) is the permittivity of free space.
  • Gauss's Law is especially useful for spheres, cylinders, and planes due to their symmetry.
  • It simplifies calculations by allowing consideration of only the charges enclosed by a Gaussian surface.
In our exercise, Gauss's Law helps us find the electric fields at different radial points by considering the total charge enclosed within the chosen Gaussian surfaces.
Conducting Sphere
A conducting sphere is a solid object that can carry charge and is perfect for studying electrostatics because charges redistribute across its surface. In electrostatic equilibrium, a conducting sphere has these notable properties:
  • All excess charge resides on its outer surface.
  • Inside the conducting material, the electric field is zero. This is due to internal charges creating a canceling electric field that results in the absence of net electric forces.
In the region \(0 < r < R\), the electric field is zero because any charge on the sphere's surface does not influence points inside it. This explains why, for a conducting sphere in equilibrium, the electric field inside (up to its surface) is always zero.
Insulating Shell
An insulating shell is different from its conducting counterpart, as it carries charge in a uniform distribution over its volume or surface. It doesn't allow free movement of charges across its structure.
  • With a charge distributed on its outer surface, the field within an insulating shell can be zero due to this arrangement.
  • In the given problem, it doesn't impact the region between the sphere and the shell, thus maintaining the calculated field solely from the inner sphere’s charge.
This shell, in the problem, is concentric to the conducting sphere, expanding the analysis by adding to charge calculations for the external field (after the shell end) due to superposition of its charge.
Electrostatic Equilibrium
Electrostatic equilibrium represents a state where charges are at rest, and the net electric field within a conductor is zero. This is reached when charges stop experiencing forces pushing them through the conductor.
  • Once equilibrium is achieved, any additional charges added to a conductor are swiftly spread over the surface, ceasing internal electric field effects.
  • For a conductor, this means: within its surface, the electric field is nearly zero. The surface charge creates a symmetrical external field.
In our exercise, the conducting sphere is in electrostatic equilibrium, ensuring no internal electric field and a surface field only, partly explaining the zero field values in regions internally up to the sphere.

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Most popular questions from this chapter

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho(r)\) given as follows: $$ \begin{array}{ll} p(r)=\rho_{0}(1-4 r / 3 R) & \text { for } r \leq R \\ \rho(r)=0 & \text { for } r \geq R \end{array} $$ where \(\rho_{0}\) is a positive constant. (a) Find the total charge contained in the charge distribution. (b) Obtain an expression for the electric field in the region \(r \geq R\). (c) Obtain an expression for the electric field in the region \(r \leq R\). (d) Graph the electric-field magnitude \(E\) as a function of \(r .\) (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{~cm}\), giving it a charge of \(-36 \mu \mathrm{C}\). Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) \(6 \mathrm{~cm}\) outside the surface of the paint layer.

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 \(r\). (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.

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.

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