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Chapter 24: Q. 39 (page 684)

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-4.5×105C.

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,

Φe=EA=Qinϵo

Qin=ϵoEA

localid="1648750984200" Qin=4πϵoEr2...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=6.37×106m

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

E=-100N/C

Substitute values in equation localid="1648752090427" 1,

we get,

Qin=4πϵoEr2

=4π8.85×10-12C2/N·m2(-100N/C)6.37×106m2

=-4.5×105C

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Most popular questions from this chapter

The square and circle in FIGURE Q24.3 are in the same uniform field. The diameter of the circle equals the edge length of the square. Is Φsquarelarger than, smaller than, or equal to Φcircle? Explain.

The cube in FIGURE EX24.8 contains no net charge. The electric field is constant over each face of the cube. Does the missing electric field vector on the front face point in or out? What is the field strength?

A very long, uniformly charged cylinder has radius Rand linear charge densityλ. Find the cylinder's electric field strength (a) outside the cylinder, r≥R, and (b) inside the cylinder, r≤R. (c) Show that your answers to parts a and b match at the boundary, r=R

aA uniformly charged ball of radiusaand charge -Qis at the center of a hollow metal shell with inner radius band outer radius c. The hollow sphere has net charge+2Q. Determine the electric field strength in the four regionsr≤a,a<r<b,b≤r≤c,andr>c.

The sphere and ellipsoid in FIGURE Q24.9 surround equal charges. Four students are discussing the situation. Student 1: The fluxes through A and B are equal because the average radius is the same. Student 2: I agree that the fluxes are equal, but that’s because they enclose equal charges. Student 3: The electric field is not perpendicular to the surface for B, and that makes the flux through B less than the flux through A. Student 4: I don’t think that Gauss’s law even applies to a situation like B, so we can’t compare the fluxes through A and B. Which of these students, if any, do you agree with? Explain
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