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Chapter 18: Q5CQ (page 661)

When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative鈥攜et both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain.

Short Answer

Expert verified

No, the dust particle does not have a third type of charge; it is attracted to both because the dust particle molecules become polarized in the direction of glass rod or silk.

Step by step solution

01

Charging by rubbing

When silk is rubbed on a glass rod, the rod loses electrons and becomes positively charged. On the other hand, silk gains electrons becoming negatively charged.

02

Attraction of the dust particle

When a positively charged glass rod is placed near the dust particle, the dust particle gets polarized. The neutral dust particles become negatively charged due to polarization and get attracted toward the rod. Becoming negatively charged while the opposite face acquires an equal positive charge, ensuring the neutrality of dust particles while still being attracted to the rod.

When a negatively charged silk is placed near the dust, the positive charges in the neutral dust particles get accumulated towards the silk, becoming positively charged, while the opposite face acquires an equal negative charge, ensuring the neutrality of dust particles while still being attracted to the silk.

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Most popular questions from this chapter

What can you say about two charges \({q_1}\) and \({q_2}\), if the electric field one-fourth of the way from \({q_1}\) to \({q_2}\) is zero?

(a) What magnitude point charge creates a \({\rm{10,000 N/C}}\) electric field at a distance of \(0.{\bf{250}}{\rm{ }}{\bf{m}}\)? (b) How large is the field at \({\bf{10}}.{\bf{0}}{\rm{ }}{\bf{m}}\)?

(a) Find the electric field at\(x = 5.00{\rm{ cm}}\)in Figure 18.52 (a), given that\(q = 1.00{\rm{ }}\mu C\). (b) At what position between\(3.00\)and\(8.00{\rm{ cm}}\)is the total electric field the same as that for\( - 2q\)alone? (c) Can the electric field be zero anywhere between\(0.00\)and\(8.00{\rm{ cm}}\)? (d) At very large positive or negative values of\(x\), the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of\(11.0{\rm{ cm}}\)is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)

Figure 18.52 (a) Point charges located at\[{\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\]along the x-axis. (b) Point charges located at\[{\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\]along the x-axis

Compare and contrast the Coulomb force field and the electric field. To do this, make a list of five properties for the Coulomb force field analogous to the five properties listed for electric field lines. Compare each item in your list of Coulomb force field properties with those of the electric field鈥攁re they the same or different? (For example, electric field lines cannot cross. Is the same true for Coulomb field lines?)

A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is\(2.50 \times {10^4}{\rm{ N/C}}\). (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole.

Figure 18.55 Parallel conducting plates with opposite charges on them create a relatively uniform electric field used to accelerate electrons to the right. Those that go through the hole can be used to make a TV or computer screen glow or to produce X-rays.
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