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Chapter 23: Q. 72 (page 657)

A proton orbits a long charged wire, making 1.0×106 revolutions per second. The radius of the orbit is 1.0cm. What is the wire’s linear charge density?

Short Answer

Expert verified

The wire’s linear charge density is-2.28nC/m

Step by step solution

01

Given information and Formula used

Given :

Proton makes revolutions per second : 1.0×106

The radius of the orbit : 1.0cm

Theory used :

Electric field due to long charged wire at distance r

E=2λ4πε0r

(Electrostatic force is attractive, therefore λis negative.)

The proton receives centripetal force from electrostatic force, that isFc=Fe

02

Calculating the wire’s linear charge density 

Now, PE=mw2r

Using the above relation, we can deduce :

-2λ×e2πε0r=mw2r⇒-λ=1.67×10-27×(2π×106)2×(10-2)22×1.6×10-19×9×109=-2.28x109C/m⇒λ=-2.28nC/m

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Most popular questions from this chapter

FIGUREP23.41 is a cross section of two infinite lines of charge that extend out of the page. Both have linear charge density λ. Find an expression for the electric field strength E at height y above the midpoint between the lines.

A sphere of radius R has charge Q. The electric field strength at distancer>RisEi. What is the ratio Ef/Eiof the final to initial electric field strengths if

(a) Q is halved,

(b) R is halved, and

(c) r is halved (but is still>R)? Each part changes only one quantity; the other quantities have their initial values

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.1? Specify the direction as an angle above or below horizontal

FIGURE is a cross section of two infinite lines of charge that extend out of the page. The linear charge densities are ±λ. Find an expression for the electric field strength Eat height yabove the midpoint between the lines.

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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