/*! 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} Q18.8-68PE Consider identical spherical con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.

Short Answer

Expert verified

(a) In order to counteract the gravitational attraction a ship must have\(4.3 \times {10^{ - 5}}{\rm{ C}}\) charge on them. (b) No, the charge on the ship is unaffected by the distance between the ships' centres.

Step by step solution

01

Gravitational force of attraction

According to the universal law of gravitational, every object attracts another object towards its center with a force known as gravitational force. The gravitational force exists due to the mass of the object.

02

Construction of problem

The two identical spaceships of mass \(5 \times {10^5}{\rm{ kg}}\) are \(50{\rm{ m}}\) from each other. (a) How much charge should each ship carry to counteract the gravitational attraction between them? (b) Is the charge on the ship affected by the distance between the ships' centres?

03

(a) Charge on the spaceship

The gravitational force of attraction between two identical space ships of mass \(m\) and separated by a distance \(r\) is,

\({F_g} = \frac{{G{m^2}}}{{{r^2}}}\)

Here, \(G\) is the universal gravitational constant.

The electrostatic force between two identical space ships having same magnitude of charge \(q\) and separated by a distance \(r\) is,

\({F_e} = \frac{{K{q^2}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant.

Since, the gravitational force of attraction is balanced by the electrostatic force. Therefore,

\(\begin{array}{c}{F_e} = {F_g}\\\frac{{K{q^2}}}{{{r^2}}} = \frac{{G{m^2}}}{{{r^2}}}\end{array}\)

Rearranging the above expression in order to get an expression for the charge,

\(q = \sqrt {\frac{{G{m^2}}}{K}} \ldots \left( {1.1} \right)\)

Substitute \(6.67 \times {10^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}\) for \(G\), \(5 \times {10^5}{\rm{ kg}}\) for \(m\), and \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\)for\(K\),

\(\begin{array}{c}q = \sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right) \times {{\left( {5 \times {{10}^5}{\rm{ kg}}} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)}}} \\ = 4.3 \times {10^{ - 5}}{\rm{ C}}\end{array}\)

Hence, the charge on the ship is \(4.3 \times {10^{ - 5}}{\rm{ C}}\).

04

(b) Effect of distance

From equation \(\left( {1.1} \right)\), it is clear that the charge is independent of the distance between the centers of space ship.

Hence, the charge on the ship does not depend on the distance between the centers of the ships.

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

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.

What can you say about two charges \({q_1}\) and \({q_2}\), if the electric field one-fourth of the way from \({q_1}\) to \({q_2}\) is zero?

(a) Sketch the electric field lines near a point charge +q (b) Do the same for a point charge -3.00q.

A test charge of \({\rm{ + 2 \mu C}}\) is placed halfway between a charge of \({\rm{ + 6 \mu C}}\) and another of \({\rm{ + 4 \mu C}}\) separated by \(10{\rm{ cm}}\). (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the \({\rm{ + 6 \mu C}}\)charge)?

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.