5 questions Q1 Q2 Q3 Q4 Q5 MasteringPhysics: HW3-Google Chrome Secure l https://

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MasteringPhysics: HW3-Google Chrome Secure l https://session.masteringphysics.com/myct/itemView?assignmentProblemID=75502841&offset-prev; Phys 2200 Helo | Close HW3Exercise 23.5 Resources previous | 2 of 18 next » Exercise 23.5 Part A A small metal sphere, carrying a net charge of 91 =-310 uC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of g2 =-7.20 0 and mass 1.40 g , is projected toward g1. When the two spheres are 0.800 m apart, g2 is moving toward g with speed 22.0 m/s (Fiqure 1) Assume that the two spheres can be treated as point charges. You can ignore the force of gravity What is the speed of g2 when the spheres are 0.400 m apart? u= mis Submit My Answers Give Up Part B How close does g get to g1? Figure 1 ' of 1 Submit My Answers Give Up Continue m/S 1 -0.800 m

Explanation / Answer

as there is no friction,

total energy is conserved.

initial total energy

=electrical potential energy+kinetic energy of q2

=(k*q1*q2/distance)+0.5*m*v^2

where k=coloumb's constant

initial total energy=(9*10^9*(-3.1*10^(-6))*(-7.2*10^(-6))/0.8)+0.5*1.4*0.001*22^2

=0.5899 J

let speed of q2 when both are 0.4 m apart is v m/s

then final total energy

=(9*10^9*(-3.1*10^(-6))*(-7.2*10^(-6))/0.4)+0.5*1.4*0.001*v^2

=0.5022+7*10^(-4)*v^2

equating initial and final energy values:

0.5022+7*10^(-4)*v^2=0.5899

==>v=sqrt((0.5899-0.5022)/(7*10^(-4)))=11.1931 m/s

part B;

let the closest distance be d meters.

at this point, speed of q2 will be zero.

then 9*10^9*(-3.1*10^(-6))*(-7.2*10^(-6))/d=0.5899

==>d=9*10^9*(-3.1*10^(-6))*(-7.2*10^(-6))/0.5899=0.3405 m

so q2 can go at most till 0.3405 m distance of q1.

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