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A charge inside a shell * Is the following reasoning correct or incorrect (if incorrect, state the error). A point charge \(q\) lies at an off-center position inside a conducting spherical shell. The surface of the conductor is at constant potential, so, by the uniqueness theorem, the potential is constant inside. The field inside is therefore zero, so the charge experiences no force.

Short Answer

Expert verified
The given reasoning is incorrect. The charge inside the shell would experience a force as the field inside a conductor with enclosed charge is not zero.

Step by step solution

01

Understanding of The Problem

The problem presents a point charge \(q\) that is off-center inside a conducting spherical shell. From the general properties of conductors in equilibrium, we know the electric potential is constant at any point within a conductor and on its surface.
02

Addressing The Reasoning

The reasoning in the problem states that since the surface of the conductor is at constant potential, the potential inside the conductor is also constant. This, in turn, implies the electric field inside the shell is zero, meaning the charge inside experiences no force.
03

Analyzing the Reasoning

The error in this reasoning lies in the assumption that a constant potential implies a zero electric field. A zero field inside a conductor is true only if there are no charges inside it. For a conductor with a charge inside, the field will not be zero. Moreover, the charge will experience an electrostatic force.

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Key Concepts

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

Conducting Spherical Shell
A conducting spherical shell is a fascinating structure in electrostatics. It acts as a hollow, symmetrical conductor, and when it is in electrostatic equilibrium, interesting properties emerge. Inside the conductor, electric charges rearrange themselves to ensure that the electric field is zero within the shell's material. Therefore, the electric field inside any hollow cavity within the conductor, such as the conducting spherical shell itself, is zero, as long as there is no charge present inside.
However, if a charge is placed inside the cavity, the situation changes. The conducting shell will react by redistributing its charges on its inner and outer surfaces to maintain equilibrium. While the electric field remains zero within the metallic part of the shell, the region inside the cavity (where the point charge is) can have a non-zero electric field due to the presence of the charge itself, countering the assumption made in the exercise.
Electric Potential
Electric potential is a crucial concept in understanding electrostatics and conducting shells. It is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the electric field, without any acceleration. In the context of a conducting spherical shell, the potential is constant on the shell's surface due to its conductive nature. This is a key fact about equipotential surfaces.
While the shell’s surface maintains this constant potential, what happens inside depends on the distribution of charges. Without an internal charge, the potential would indeed be uniform everywhere inside. However, when there is a charge inside, such as in the problem provided, the potential inside is affected by the charge’s electric field, challenging the idea that the inside must also maintain a constant potential throughout.
Electric Field
The electric field is a vector field around charged particles that represents the force exerted per unit charge at any point in space. Around a conducting spherical shell, when in equilibrium, there is no electric field within the material of the shell. However, this is not the case if there is a charge inside the cavity of the shell.
  • If there is no charge inside, the electric field inside the cavity is indeed zero.
  • With a point charge inside, the shell will have a non-zero electric field in the cavity because the charge creates its own electric field.
This non-zero electric field leads to the conclusion that the charge will not be free of force within the shell, as implied in the exercise reasoning. Hence, while a conductor screens the inside from external electric fields, it does not shield the inside from its fields, originating from the charges within.
Uniqueness Theorem
The Uniqueness Theorem is a powerful tool in electrostatics used to demonstrate that a particular solution to an electrostatic problem is the only possible solution given specific boundary conditions. It asserts that the solution to Poisson's or Laplace's equations in a region is uniquely determined if the boundary conditions of that region are specified.
In the context of conducting spherical shells, this theorem explains how once the potential on the surface is known, it determines the potential everywhere else, assuming there are no other charges. The key detail is that the theorem requires that the boundary conditions remain unchanged, but in cases where a charge is added inside, as in our problem, the boundary conditions within the shell are altered by the presence of this charge. Thus, while the theorem holds true, it only holds true under the "correct" conditions, not assuming extraneous properties like constant inner potential without accounting for internal charges.

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Most popular questions from this chapter

Point charge near a corner Locate two charges \(q\) each and two charges \(-q\) each on the corners of a square, with like charges diagonally opposite one another. Show that there are two equipotential surfaces that are planes. In this way sketch qualitatively the field of the system where a single point charge is located symmetrically in the inside corner formed by bending a metal sheet through a right angle. Which configurations of conducting planes and point charges can be solved this way and which can't? How about a point charge located on the bisector of a \(120^{\circ}\) dihedral angle between two conducting planes?

Acquiring transverse momentum In the rest frame of a particle with charge \(q_{1}\), another particle with charge \(q_{2}\) is approaching, moving with velocity \(v\) not small compared with \(c\). If it continues to move in a straight line, it will pass a distance \(b\) from the position of the first particle. It is so massive that its displacement from the straight path during the encounter is small compared with \(b\). Likewise, the first particle is so massive that its displacement from its initial position while the other particle is nearby is also small compared with \(b\).(a) Show that the increment in momentum acquired by each particle as a result of the encounter is perpendicular to \(\mathbf{v}\) and has magnitude \(q_{1} q_{2} / 2 \pi \epsilon_{0} v b\). (Gauss's law can be useful here.) (b) Expressed in terms of the other quantities, how large (in order of magnitude) must the masses of the particles be to justify our assumptions?

A capacitor consists of two parallel rectangular plates with a vertical separation of \(2 \mathrm{~cm}\). The east-west dimension of the plates is \(20 \mathrm{~cm}\), the north-south dimension is \(10 \mathrm{~cm}\). The capacitor has been charged by connecting it temporarily to a battery of \(300 \mathrm{~V}\). What is the electric field strength between the plates? How many excess electrons are on the negative plate? Now give the following quantities as they would be measured in a frame of reference that is moving eastward, relative to the laboratory in which the plates are at rest, with speed \(0.6 c:\) the three dimensions of the capacitor; the number of excess electrons on the negative plate; the electric field strength between the plates. Answer the same questions for a frame of reference that is moving upward with speed \(0.6 c\).

Van de Graaff current ? In a Van de Graaff electrostatic generator, a rubberized belt \(0.3 \mathrm{~m}\) wide travels at a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of \(10^{6} \mathrm{~V} / \mathrm{m}\) on each side of the belt. What is the current in milliamps?

Compressing a sphere ** A spherical conducting shell has radius \(R\) and potential \(\phi\). If you want, you can consider it to be part of a capacitor with the other shell at infinity. You compress the shell down to essentially zero size (always keeping it spherical) while a battery holds the potential constant at \(\phi\). By calculating the initial and final energies stored in the system, and also the work done by (or on) you and the battery, verify that energy is conserved. (Be sure to specify clearly what your conservation-of-energy statement is, paying careful attention to the signs of the various quantities.)

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