Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

One of the hazards facing humans in space is space radiation: high-energy charged particles emitted by the sun. During a solar flare, the intensity of this radiation can reach lethal levels. One proposed method of protection for astronauts on the surface of the moon or Mars is an aligned of large, electrically charged spheres placed high above areas where people live and work. The spheres would produce a strong electric field E S to deflect the charged particles that make up space radiation. The spheres would be similar in construction to a Mylar balloon, with a thin, electrically conducting layer on the outside surface on which a net positive or negative charge would be placed. A typical sphere might be 5 m in diameter. Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field E S of magnitude 1\( \times \) 106 N/C at 25 m from the center of the sphere. What is the magnitude of E S just outside the surface of such a sphere? (a) 0; (b) 106 N/C; (c) 107 N/C; (d) 108 N/C.

Short Answer

Expert verified

The electric field outside the surface of each sphere is \({10^8}\;{\rm{N}}/{\rm{C}}\) and the option (d) is correct.

Step by step solution

01

Identification of given data

The magnitude of electric field outside the surface of each sphere is\(E = 1 \times {10^6}\;{\rm{N}}/{\rm{C}}\)

The diameter of each sphere is\(d = 5\;{\rm{m}}\)

The distance for electric field from the centre of each sphere is \(D = 25\;{\rm{m}}\)

02

Conceptual Explanation

The number of excess electron is found by calculating the charge due to electric field on the outside of sphere.

03

Determination of electric field outside the surface of sphere

The electric field just outside the surface of each sphere is given below:

\(E = \frac{{kQ}}{{{D^2}}}\)

Here,\(k\)is Coulomb’s constant and its value is\(9 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\).

Substitute all the values in the above equation.

\(\begin{aligned}1 \times {10^6}\;{\rm{N}}/{\rm{C}} = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)Q}}{{{{\left( {25\;{\rm{m}}} \right)}^2}}}\\Q = 0.0694\;{\rm{C}}\end{aligned}\)

The electric field outside the surface of each sphere is given as:

\(\vec E = \frac{{kQ}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\)

Substitute all the values in the above equation.

\(\begin{aligned}\vec E = \frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {0.0694\;{\rm{C}}} \right)}}{{{{\left( {\frac{{5\;{\rm{m}}}}{2}} \right)}^2}}}\\\vec E = 0.0999 \times {10^9}\;{\rm{N}}/{\rm{C}}\\\vec E \approx {10^8}\;{\rm{N}}/{\rm{C}}\end{aligned}\)

Therefore, the electric field outside the surface of each sphere is \({10^8}\;{\rm{N}}/{\rm{C}}\) and the option (d) is correct.

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 Vaia!

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 particle of mass 0.195 g carries a charge of-2.50×10-8C. The particle is given an initial horizontal velocity that is due north and has magnitude4.00×104m/s. What are the magnitude and direction of the minimum magnetic field that will keepthe particle moving in the earth’s gravitational field in the samehorizontal, northward direction?

The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) and (b) the terminal voltage Vabof the 16.0-V battery.

Fig. E25.33

A particle with charge-5.60nCis moving in a uniform magnetic fieldrole="math" localid="1655717557369" B=-(1.25T)k^

The magnetic force on the particle is measured to berole="math" localid="1655717706597" F=-(3.40×10-7N)i^-(7.40×10-7N)j^ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar productv֏F. What is the angle between velocity and force?

A particle with mass1.81×10-3kgand a charge of has1.22×10-8C, at a given instant, a velocityV=(3.00×104m/s).What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic fieldB=(1.63T)i+(0.980T)j^?

The current in a wire varies with time according to the relationship

I=55A-(0.65As2)t2. (a) How many coulombs of charge pass a cross section of the wire in

the time interval between t=0and role="math" localid="1655721302619" t=8.0s? (b) What constant current would transport the

same charge in the same time interval?

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.

Sign-up for free