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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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