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Chapter 24: Q. 43 (page 685)

Find the electric field inside and outside a hollow plastic ball of radius R that has chargeQ uniformly distributed on its outer surface.

Short Answer

Expert verified

Electric field inside a hollow plastic ball is 0nC.

Electric field outside a hollow plastic ball is14πϵoQr2.

Step by step solution

01

Figure for Electric field inside a hollow plastic ball  

Figure is,

02

Calculation for electric field inside a hollow plastic ball  

Electric flux

Φe=EA=Qinϵo

localid="1648763653778" E=QinϵoA...1

The charge is spread on the outside surface of a plastic sphere, which is not conductive.

As a result, the sphere's net charge is zero.

The quantity of charge in the bulk determines the electric field.

As a result of the net charge being zero.

Inside the sphere, the electric field is zero.

Ein=0N/C

03

Figure for Electric field outside a hollow plastic ball  

Figure is,

04

Calculation for electric field outside a hollow plastic ball  

The Gaussian surface enclosed net charge for a point outside the spher+Q.

The enclosed surface's area is

A=4πr2

This charge is enclosedlocalid="1648918575953" Qinin the gaussian surface is,

Qin=Q

Substitute all values in equation 1,

We get,

Eout=QinϵoA

=Qϵo4πr2

=14πϵoQr2

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Most popular questions from this chapter

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.

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

All examples of Gauss's law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we've claimed that the net Φe=Qin/ϵ0is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length Lcentered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral.

a. Consider the face parallel to the yz-plane. Define area dA→as a strip of width dyand height Lwith the vector pointing in the x-direction. One such strip is located at position localid="1648838849592" y. Use the known electric field of a wire to calculate the electric flux localid="1648838912266" dΦthrough this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

b. Now integrate dΦto find the total flux through this face.

c. Finally, show that the net flux through the cube is Φe=Qin/ϵ0.

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?

A tetrahedron has an equilateral triangle base with20-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 200N/C passes upward through the tetrahedron. a. What is the electric flux through the base? b. What is the electric flux through each of the three sides?
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