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Chapter 23: Q. 11 (page 652)

Rank in order, from largest to smallest, the electric field strengths E1 to E5 at the five points inFIGURE Q23.11. Explain.

Short Answer

Expert verified

The rank in order, from largest to smallest, the electric field strengthsE1 to E5 at the five points is E1=E2=E3=E4=E5

Step by step solution

01

Given information

Given the electric field strengths

02

Explanation

Since an electric field E→is a vector quantity ⇒E→has both magnitude and direction.

For the present scenario, E→is constant everywhere between the two plates.

This is specified by E→, which are all of the same length and in the same direction.

∴E1=E2=E3=E4=E5

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Most popular questions from this chapter

A ring of radius Rhas total chargeQ.

aAt what distance along the z-axis is the electric field strength a maximum?

bWhat is the electric field strength at this point?

The two parallel plates in FIGURE P23.53are 2.0cmapart and the electric field strength between them is 1.0×104N/C. An electron is launched at a 45∘ angle from the positive plate. What is the maximum initial speed v0 the electron can have without hitting the negative plate?

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

The combustion of fossil fuels produces micron-sized particles of soot, one of the major components of air pollution. The terminal speeds of these particles are extremely small, so they remain suspended in air for very long periods of time. Furthermore, very small particles almost always acquire small amounts of charge from cosmic rays and various atmospheric effects, so their motion is influenced not only by gravity but also by the earth's weak electric field. Consider a small spherical particle of radius r, density ρ, and charge q. A small sphere moving with speed v experiences a drag force Fdrag=6πηrv, where η is the viscosity of the air. (This differs from the drag force you learned in Chapter 6 because there we considered macroscopic rather than microscopic objects.)

a. A particle falling at its terminal speed vtermis in equilibrium with no net force. Write Newton's first law for this particle falling in the presence of a downward electric field of strength E, then solve to find an expression for vterm.

b. Soot is primarily carbon, and carbon in the form of graphite has a density of 2200kg/m3. In the absence of an electric field, what is the terminal speed in mm/s of a 1.0-μm-diameter graphite particle? The viscosity of air at 20°C is 1.8×10-5kg/ms.

c. The earth's electric field is typically (150 N/C , downward). In this field, what is the terminal speed in mm/s of a 1.0 μm-diameter graphite particle that has acquired 250 extra electrons?

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