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A small, metal sphere hangs by an insulating thread within the larger, hollow conducting sphere of FIGURE Q24.10. A conducting wire extends from the small sphere through, but not touching, a small hole in the hollow sphere. A charged rod is used to transfer positive charge to the protruding wire. After the charged rod has touched the wire and been removed, are the following surfaces positive, negative, or not charged? Explain. a. The small sphere. b. The inner surface of the hollow sphere. c. The outer surface of the hollow sphere.

Short Answer

Expert verified

a. Positive

b. Negative

c. Positive.

Step by step solution

01

Step 1. Given information

The following diagram is provided.

A little metal sphere suspended on an insulating thread inside a bigger hollow conducting sphere is depicted in the illustration. The little sphere has a wire extending from it that does not contact the hollow sphere. A charged rod was used to transfer positive charge through the wire, then the rod was removed after touching the wire.

02

Part (a) Step 1. Find the positive, negative, or uncharged surfaces of the small spheres.

The usage of a charged rod to transfer positive charge to the wire is assumed. Because the wire is attached to the inner sphere and does not touch the outer hollow sphere, when the rod touches the wire, the positive charge will be transferred through the wire to the inner little sphere. As a result, the little sphere will have a positive charge.

03

Part (b) Step 1. Find the positive, negative, or uncharged surfaces of the Inner surface of hollow sphere.

Because the inner tiny sphere is positively charged as a result of the charge transfer from the rod, a negative charge will form on the inner surface of the hollow sphere to balance the positive charge on the small sphere. Because the net charge on the Gaussian surface through the conductor of the outer hollow sphere must be zero, the hollow sphere's inner surface develops a negative charge to balance the positive charge on the inner tiny sphere. With a negative charge magnitude equal to the positive charge on the inner sphere.

04

Part (c) Step 1. Find the positive, negative, or uncharged surfaces of the Outer surface of hollow sphere

The hollow sphere's inner surface develops a negative charge from its outer surface (all extra charge settles on a conductor's surface), resulting in a positive charge on the larger sphere's outside. Consider a Gaussian surface outside the larger sphere to illustrate what I mean. It has a net positive charge on it (i.e., the charge on the smaller sphere, since the larger sphere is overall neutral). Because the charge on the inner surface of the larger sphere is equal to and opposite that of the smaller sphere, the larger sphere's outer must be positively charged.

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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.

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ϵ0, with Qin/ϵ, where ϵ is the permittivity of the material. (Technically,ϵ0 is called the vacuum permittivity.) Suppose that a 50nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is2500N/C . What is the permittivity of rubber?

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.

A positive point chargeq sits at the center of a hollow spherical shell. The shell, with radius R and negligible thickness, has net charge -2q. Find an expression for the electric field strength (a) inside the sphere, r<K, and (b) outside the sphere, r>K. In what direction does the electric field point in each case?

A spherical shell has inner radius Rinand outer radius Rout. The shell contains total charge Q, uniformly distributed. The interior of the shell is empty of charge and matter.

a. Find the electric field strength outside the shell,r≥Rout .

b. Find the electric field strength in the interior of the shell, r≤Rin.

c. Find the electric field strength within the shell, Rin≤r≤Rout.

d. Show that your solutions match at both the inner and outer boundaries

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