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sics.com/myct/itemview?offset next&assignmentProblemID; 74969685 secure https Homework for Giancoli, Physics: Principles with Applicati.. in an Electric Field Speed of an Electron Speed of an Electron in an Electric Field Part A Two stationary positive point charges, charge 1 of What is the speed unral of the electron when it is 10.0 cm from charge 1? magnitude 305 nCand charge 2 of magnitude 1.85 nC Express your answer in meters per second. are separated by a distance of 48.0 cm.Anelectron is reeased from rest at the point midway between the two changes, and it moves along the line connecting the two m/s Hints yAnswers Gave Up Review PartExplanation / Answer
let, the 1 st point charge q1= 3.05 nC is placed at the origin, then accordingly
2 nd point charge q2 = 1.85 nC is placed on the x-axis at x=+ 48 cm = + 0.48 m as they are separated by 48 cm in a straight line.
when, 3 rd point charge electron ,q3 = 1.6*10^-19 C is placed at x=+ 25.0 cm= + 0.25 m from the origin.then, potential energy at that time is
Potential energy, PE 1 = U1= 9*10 ^ 9 ( q1q2/ 0.5 - q1q3/ 0.25 - q2q3/ 0.25 )
= 9*10^ 9 ( 2*3.05 *1.85*10^-18 - 4*3.05*1.6*10^-28 - 4*1.85*1.6*10^-28 )
Potential energy =PE1= 9*10^ -9[11.28 - 31.36*10^-10 ] J
When third point charge electron= q3 = 1.6*10^-19 C is placed at x=+10 cm= + 0.1m from origin i.e point charge 1, PE at that time equals to
Potential energy = PE2= 9*10 ^ 9 ( q1q2/ 0.5 - q1q3/ 0.1 - q2q3/ 0.4 )
= 9*10^ 9 ( 2*3.05 *1.85*10^-18 - 10*3.05*1.6*10^-28 - 2.5*1.85*1.6*10^-28 )
Potential energy =PE2= 9*10^ 9( 11.28*10^-18 - 56.2*10^-28 )
Potential energy =PE2= 9*10^- 9(11.28 - 56.2 *10^-10 ) J
Change in PE = PE1 - PE2= 9*10^- 9 *24.84*10^-10 J
=> PE1 - PE2=223.56 * 10^ -19 J
Kinetic energy =Change in PE = 223.56 *10^-19
=> KE = PE1 - PE2
(1/2)mv^2=223.56*10^-19
v = sq rt[(447.12 / 9.1)*10^12 ]
= sq rt[(49.134)*10^12 ]
v = 7 *10^6 m/s
The speed of the electron is 7 *10 ^ 6 m/s when it is 10.0 cm from the + 3.05-nC charge
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