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Grounding a shell ** A conducting spherical shell has charge \(Q\) and radius \(R_{1}\). A larger concentric conducting spherical shell has charge \(-Q\) and radius \(R_{2}\). If the outer shell is grounded, explain why nothing happens to he charge on it. If instead the inner shell is grounded, find its final charge.

Short Answer

Expert verified
When the outer shell with charge \(-Q\) is grounded, no charges are transferred. But when the inner shell with charge \(Q\) is grounded, it loses all its charge to the ground due the grounding. Hence, the final charge on the inner shell becomes zero.

Step by step solution

01

Understanding the grounding effect

When an object or conductor is grounded, it means it's connected to the earth, which is a vast reservoir of charges. It can provide or absorb charges without changing its potential significantly. Now, any conductor in electrostatic equilibrium does not carry any net charge inside. All the excess charges reside on the surface.
02

Case 1: Grounding the Outer Shell

In the given task, the outer shell has an initial charge of \(-Q\). When it is grounded, no charges are transferred. Despite the inner shell carrying charge \(Q\), no charges migrate from the outer shell to the earth because the outer shell is already at a lower potential compared to the earth. Hence, the grounding of the outer shell does not affect the charges on it.
03

Case 2: Grounding the Inner Shell

When the inner shell is grounded, the behavior changes. While the inner shell initially has a charge of \(Q\), as soon as it's connected to the ground, it starts losing charge. The ground acts as a sink to neutralize the potential of the inner shell. So, it will drain as much charge necessary to ensure that the potential of the inner-shell, with respect to the infinitely far point, matches with that of the ground. Thus all the charge \(Q\) is drained from the inner shell to the ground to make the potential of the inner shell zero. Hence, the final charge on the inner shell is Zero.

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Key Concepts

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

Conducting Spherical Shell
Let's dive into the concept of a conducting spherical shell. A spherical shell, in simple terms, is a hollow sphere made from a conducting material. Conductors allow electric charge to move freely, due to their free electrons. In the context of electrostatics, if a shell is charged, the charges will distribute themselves on the surface.

This distribution happens because charges repel each other and will try to stay as far apart as possible. Within the conducting shell, there will be no net electric field if it's in electrostatic equilibrium (we will cover this in detail shortly). Because of the symmetric shell shape, the distribution of charge is uniform across its surface.

You might wonder: Why just the outer surface? This occurs because the charge inside a conductor cancels out when it's in electrostatic equilibrium, leaving the outermost surface to bear the entire charge. This attribute is fundamental in understanding electrostatic shielding, where the interior experience no electric field from outside charges.
Electrostatic Equilibrium
Electrostatic equilibrium refers to the state in which the electric field within a conductive material is zero. When a conductor is in this state, any excess charge resides entirely on the conductor's surface. The movement of charges stops once they are evenly distributed, as any further movement would create an electric field, contradicting the zero field condition.

In essence, electrostatic equilibrium ensures complete charge distribution balance and subsequently a zero electric field internally. This principle explains why charges remain on the outer surface of a conducting sphere. It also simplifies calculations involving electrostatic problems, as calculations pertaining to internal fields become unnecessary. Instead, actions and interactions with other charges need only be calculated based on surface behavior.
Charge Distribution
Charge distribution on conducting surfaces is influenced heavily by electrostatic principles. On a spherical conductor, charges uniformly distribute across the outer surface, as we've mentioned. The reason is rooted in electrostatic repulsion — like charges repel, trying to maximize distance between each other.

In the situation described in the exercise, a conducting spherical shell must consider the grounding and effects of neighboring charges. When the outer shell is grounded, its potential relative to the earth's (which is zero) doesn't change significantly due to its already negative charge. Hence, no charge movement occurs.

However, grounding the inner shell allows charge flow to and from the earth, changing its distribution to satisfy a zero potential condition. Therefore, the grounding alters how charges distribute themselves across the shells, showing the selective influence of grounding procedures.
Electric Potential
Electric potential is a concept that relates to the energy that a charge possesses due to its position in an electric field. It's a measure of the "potential energy per charge" at a point, usually referenced to some baseline potential value, often the ground with a potential of zero.

When it comes to conducting spheres, electric potential ties into grounding - connecting an object to earth affects its potential. A grounded conductor reaches the same potential as earth, and this might influence how charges distribute, as seen in the inner shell situation from the exercise.

Changes in the potential, when an object is grounded, drive the flow of charge until equilibrium is met. For instance, when the inner shell gets grounded, it loses all charge to ensure zero potential, highlighting the tangible implications of electric potential on charge behavior.

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Most popular questions from this chapter

Tapered rod Two graphite rods are of equal length. One is a cylinder of radius \(a\). The other is conical, tapering (or widening) linearly from radius \(a\) at one end to radius \(b\) at the other. Show that the end-to-end electrical resistance of the conical rod is \(a / b\) times that of the cylindrical rod. Hint: Consider the rod to be made up of thin, disk-like slices, all in series. (This result is actually only an approximate one, valid in the limit where the taper is slow. See Problem \(4.6\) for a discussion of this.)

Capacitance of a spheroid * Here is the exact formula for the capacitance \(C\) of a conductor in the form of a prolate spheroid of length \(2 a\) and diameter \(2 b\) : \(C=\frac{8 \pi \epsilon_{0} a \epsilon}{\ln \left(\frac{1+\epsilon}{1-\epsilon}\right)}, \quad\) where \(\quad \epsilon=\sqrt{1-\frac{b^{2}}{a^{2}}}\) First verify that the formula reduces to the correct expression for the capacitance of a sphere if \(b \rightarrow a\). Now imagine that the spheroid is a charged water drop. If this drop is deformed at constant volume and constant charge \(Q\) from a sphere to a prolate spheroid, will the energy stored in the electric field increase or decrease? (The volume of the spheroid is \((4 / 3) \pi a b^{2}\).)

Electron in an oscilloscope * The deflection plates in a high-voltage cathode ray oscilloscope ure two rectangular plates, \(4 \mathrm{~cm}\) long and \(1.5 \mathrm{~cm}\) wide, and spaced, \(.8 \mathrm{~cm}\) apart. There is a difference in potential of \(6000 \mathrm{~V}\) between he plates. An electron that has been accelerated through a potenial difference of \(250 \mathrm{kV}\) enters this deflector from the left, moving, of find the position of the electron and its direction of motion when t leaves the deflecting field at the other end of the plates. We shall lates is uniform right up to the end. The rest energy of the electron nay be taken as \(500 \mathrm{keV}\). (a) First carry out the analysis in the lab frame by answering the following questions: \- What are the values of \(\gamma\) and \(\beta ?\) \- What is \(p_{x}\) in units of \(m c ?\) \- How long does the electron spend between the plates? (Neglect the change in horizontal velocity discussed in Exercise 5.25.) \- What is the transverse momentum component acquired, in units of \(m c\) ? \- What is the transverse velocity at exit? \- What is the vertical position at exit? \- What is the direction of flight at exit?(b) Now describe this whole process as it would appear in an inertial frame that moved with the electron at the moment it entered the deflecting region. What do the plates look like? What is the field between them? What happens to the electron in this coordinate system? Your main object in this exercise is to convince yourself that the two descriptions are completely consistent.

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.

Field just outside a capacitor * A capacitor consists of two disks with radius \(R\), small separation \(s\), and surface charge densities \(\pm \sigma\). Find the electric field just outside the capacitor, an infinitesimal distance from the center of the positive disk.

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