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Chapter 16: Problem 5

A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)

Short Answer

Expert verified
Answer: The sign of the net charge on Sphere 1 in diagram b is negative. In comparison with the charge on Sphere 1, Sphere 2 has a net positive charge with the same magnitude as the net negative charge on Sphere 1.

Step by step solution

01

Identify charge distribution when spheres are touching

When a positively charged rod is brought near two uncharged conducting spheres that are initially touching, the negatively charged electrons on the surface of the spheres will be attracted to the positively charged rod. Consequently, the side closer to the rod will have more negatively charged electrons while the side further away will have accumulated positive charges. However, since the spheres are touching, the charges will be distributed evenly between them.
02

Determine the charge distribution when spheres are separated

With the charged rod still near, if we now separate the two spheres, the charges on each sphere will remain constant as the rod is the source of the electric field, which caused the redistribution. Sphere 1 will have a net negative charge due to its proximity to the positively charged rod, while Sphere 2 will have a net positive charge.
03

Determine the charge on Sphere 1 after removing the charged rod

When we remove the positively charged rod, the redistribution of charges on Sphere 1 is no longer influenced by the external electric field. Therefore, Sphere 1 will maintain its net negative charge. The sign of the net charge on Sphere 1 in diagram b is negative.
04

Determine the charge and sign on Sphere 2 compared to Sphere 1

Since the charges on the spheres were initially uniform when they were touching each other and both spheres have the same size, the amount of negative charge attracted to Sphere 1 must equal the positive charge separated on Sphere 2. Thus, the charge on Sphere 2 is equal in magnitude but opposite in sign to the charge on Sphere 1. Sphere 2 has a net positive charge with the same magnitude as the net negative charge on Sphere 1. So, (a) the sign of the net charge on sphere 1 in diagram b is negative, and (b) in comparison with the charge on sphere 1, Sphere 2 has a net positive charge with the same magnitude as the net negative charge on Sphere 1.

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Most popular questions from this chapter

A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
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