/*! 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} Q22-51P Using Thomson鈥檚 (outdated) mod... [FREE SOLUTION] | 91影视

91影视

Using Thomson鈥檚 (outdated) model of the atom described in Problem 22.50, consider an atom consisting of two electrons, each of charge -e, embedded in a sphere of charge +2e and radius R. In equilibrium, each electron is a distance d from the center of the atom (Fig. P22.51). Find the distance d in terms of the other properties of the atom.

Short Answer

Expert verified

The distance of each atom from the center of sphere is \(\frac{R}{2}\).

Step by step solution

01

Identification of given data

The charge of the sphere is\(Q = 2e\)

The charge of each atom in sphere is\(q = - e\)

The radius of sphere is\(R\)

The distance from the centre for each atom is \(d\)

02

Conceptual Explanation

The distance of each atom in sphere is calculated by equating the force on each atom by the charge of sphere.

03

Determination of distance of each atom from the center of sphere

The distance of each atom from center of sphere is given below:

\(\begin{aligned}\frac{{kQq}}{{{R^2}}} &= \frac{{kQq}}{{{{\left( {d + d} \right)}^2}}}\\2d &= R\\d &= \frac{R}{2}\end{aligned}\)

Therefore, the distance of each atom from the center of sphere is \(\frac{R}{2}\).

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 91影视!

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

In Europe the standard voltage in homes is 220 V instead of the 120 used in the United States. Therefore a 鈥100-W鈥 European bulb would be intended for use with a 220-V potential difference (see Problem 25.36). (a) If you bring a 鈥100-W鈥 European bulb home to the United States, what should be its U.S. power rating? (b) How much current will the 100-W European bulb draw in normal use in the United States?

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

A parallel-plate capacitor is charged by being connected To a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric Field change? The potential difference? The total energy? Explain.

You want to produce three 1.00-mm-diameter cylindrical wires,

each with a resistance of 1.00 鈩 at room temperature. One wire is gold, one

is copper, and one is aluminum. Refer to Table 25.1 for the resistivity

values. (a) What will be the length of each wire? (b) Gold has a density of1.9310-4kgm3.

What will be the mass of the gold wire? If you consider the current price of gold, is

this wire very expensive?

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00 nC is on the x-axis at x = 3.00 cm. Point P is on the y-axis at y = 4.00 cm. (a) Calculate the electric fieldsandat point P due to the charges q1 and q2. Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

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.