/*! 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} Q. 39 The earth has a vertical electri... [FREE SOLUTION] | 91影视

91影视

The earth has a vertical electric field at the surface, pointing down, that averages100N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?

Short Answer

Expert verified

The excess charge on the surface of the earthQinis-4.5105C.

Step by step solution

01

Formula for electric flux and charge

The electric flux is that the amount of electrical field that flows through a closed surface.

Gauss' law states that the electrical field passing through a surface is proportional to the charge within the surface.

Electric flux is,

e=EA=Qino

Qin=oEA

localid="1648750984200" Qin=4蟺系oEr2...1

02

Calculation for excess charge on the surface of the world

The radius of the gaussian surface is up to the radius of the earth's outer surface,

r=6.37106m

As it enters the world, the electrical field is negative.

E=-100N/C

Substitute values in equation localid="1648752090427" 1,

we get,

Qin=4蟺系oEr2

=48.8510-12C2/Nm2(-100N/C)6.37106m2

=-4.5105C

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

FIGURE shows three Gaussian surfaces and the electric flux through each. What are the three charges q1,q2andq3?

A hollow metal sphere has6cmand 10cminner and outer radii, respectively. The surface charge density on the inside surface is -100nC/m2. The surface charge density on the exterior surface is +100nC/m2. What are the strength and direction of the electric field at points 4,8and12cm from the center?

Newton鈥檚 law of gravity and Coulomb鈥檚 law are both inversesquare laws. Consequently, there should be a 鈥淕auss鈥檚 law for gravity.鈥 a. The electric field was defined as E u = F u on q /q, and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g u of a point mass?

Write your answer using the unit vector nr, but be careful with signs; the gravitational force between two 鈥渓ike masses鈥 is attractive, not repulsive. b. What is Gauss鈥檚 law for gravity, the gravitational equivalent of Equation 24.18? Use 桅G for the gravitational flux, g u for the gravitational field, and Min for the enclosed mass. c. A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as r = r011 - r/2R2, where r0 is the density at the center. Determine r0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses. d. Find an expression for the gravitational field strength inside the planet at distance r 6 R.

What is the net electric flux through the torus (i.e., doughnut shape) of FIGURE ?

FIGURE EX24.2 shows a cross section of two concentric spheres. The inner sphere has a negative charge. The outer sphere has a positive charge larger in magnitude than the charge on the inner sphere. Draw this figure on your paper, then draw electric field vectors showing the shape of the electric field.

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.