/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q16DQ A negative charge -Q is placed i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?

Short Answer

Expert verified

An excess charge is there in inner surface and no charge on the surface due to the ground, therefore the electric field outside the sphere cannot be measured. and the gravity will not be enough to act as a ground.

Step by step solution

01

Step 1:

the electric field inside the conductor is zero because the charge in the cavity is negative charge. if the inner surface of the conductor must have the opposite charge of the cavity, then zero electric field inside the conductor is there. Thus, it has an excess of electrons

02

Step 2

the surface of the sphere is connected to the ground because the earth is a good conductor so the excess electrons on the surface of the sphere will flow to the ground. Thus, there is no excess charge on the outside surface of the metal. That is shown in the diagram.

the excess electrons on the surface are removed by the ground and the net charge on the sphere will be zero thus, no electric field is applied outside the sphere. Thus, outside the solid measure there is no electric field due to the charge.

the electron's mass is very small and the gravity depends on the mass of the elements thus it will be difficult for gravitational force to act as a ground or shield. Thus, the same thing cannot be done for gravity

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80

min. Silver containsfree electrons per cubic meter. (a) What is the

current in the wire? (b) What is the magnitude of thedrift velocity of the

electrons in the wire?

Consider the circuit of Fig. E25.30. (a)What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Fig. E25.30.

A 5.00-A current runs through a 12-gauge copper wire (diameter

2.05 mm) and through a light bulb. Copper has8.5×108free electrons per

cubic meter. (a) How many electrons pass through the light bulb each

second? (b) What is the current density in the wire? (c) At what speed does

a typical electron pass by any given point in the wire? (d) If you were to use

wire of twice the diameter, which of the above answers would change?

Would they increase or decrease?

The freshness of fish can be measured by placing a fish between the plates of a capacitor and measuring the capacitance. How does this work? (Hint: As time passes, the fish dries out. See Table )

In the circuit in Fig. E25.47, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.