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A small segment of wire in FIGURE Q23.4contains 10nCof charge.

a. The segment is shrunk to one-third of its original length. What is the ratio of λf/λi, where λiandλf are the initial and final linear charge densities?

b. A proton is very far from the wire. What is the ratio Ff /Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

c. Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to keep the linear charge density unchanged?

Short Answer

Expert verified

(a) The ratio of final to initial linear charge densities when the length shrunk to one-third of its original length=3

(b) The ratio of final to initial linear forces when the length shrunk to one-third of its original length=3.

(c) The amount of final charge required to keep linear charge densities constant when its final length is stretched 10 times of initial length=100nC

Step by step solution

01

Given information (part a)

Case1:lengthli=lChargeQi=10nCCase2:lengthlf=l3ChargeQf=10nCRatioofchargedensities=λfλi

02

Explanation (part a)

Linear charge densityλ=chargelengthRatioofLinearchargedensities=λfλi=QflfQili1λfλi=lilf(2)2

Substituting the values of given data in eqn(2)i.e.,λfλi=lilf

λfλi=l13λfλi=3

03

Given information (part b)

Case1:lengthli=lChargeQi=10nCChargeofprotonQp=QsayCase2:lengthlf=l3ChargeQf=10nCChargeofprotonQp=QsayRatioofforces=FfFi

04

Explanation (part b)

Substituting the values of given data ineqn(2)i.e.,λfλi=lilfWhere,Linearchargedensityλ=chargelength2FfFi=ll3FfFi=3

05

Given information (part c)

Case1:lengthli=lChargeQi=10nCCase2:lengthlf=10lRatioofchargedensities=λfλi=1

06

Explanation (part c)

Substituting the values of given data in eqn(2)

Linearchargedensityλ=chargelengthRatiooflinearchargedensities=λfλi=1=flfQili1Qf=l2×Qil12Qf=10l×10nClQf=100nC

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Most popular questions from this chapter

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?

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.4? Specify the direction as an angle above or below horizontal

What are the strength and direction of the electric field at the position indicated by the dot in FIGUREP23.37? Give your answer (a)in component form and (b)as a magnitude and angle measured cwor ccw(specify which) from the positive x-axis.

An electric dipole is formed from ±1.0nCcharges spaced apart. The dipole is at the 2.0mmorigin, oriented along the x-axis. What is the electric field strength at the points (a) (x,y)=(10cm,0cm)and(b)(x,y)=(0cm,10cm)?

A 12-cmlong thin rod has the nonuniform charge density λ(x)=(2.0nC/cm)e-|x|/(6.0cm), where localid="1648623923714" xis measured from the center of the rod. What is the total charge on the rod?

Hint: This exercise requires an integration. Think about how to handle the absolute value sign.

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