PLEASE HELP ! THANK YOU Chater 18 sreed of an Electron in an Electric Field Reso

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PLEASE HELP ! THANK YOU

Chater 18 sreed of an Electron in an Electric Field Resources previous 2 of 15 next Speed of an Electron in an Electric Field Part A Two stationary positive point charges, charge 1 of magnitude 3.35 nC and charge 2 of magnitude 1.65 nC, are separated by a distance of 56.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the ling connecting the two charg6s What is the peed nl of the electran when it is 10.0 em from charge 1? Express your answer in meters per second. Hints m/s Submit My Answers Give Up

Explanation / Answer

Vi = kQ1/r + kQ2/r
Vi = k(Q1 + Q2)/r
Vi = 8.998 * 10^9 * (3.35nC + 1.65nc)/ 0.28
Vi = 160.7 V

Now find the beginning potential energy of the electron:

Ui = q * Vi
Ui = 1.602*10^-19 * 160.7
Ui = 2.57*10^-17

Vf = kQ1/r1 + kQ2/r2
Vf = 8.998 * 10^9 * 3.35nC / 0.1 + 8.998 * 10^9 * 1.65nC / 0.46
Vf = 333.7 V

Now find the final potential energy of the electron:

Uf = q * Vf
Uf = 1.602*10^-19 * 333.7
Uf = 5.34 * 10^-17

Now solve for kinetic energy (use the absolute value of the difference in potential energies):

KE = abs(Ui – Uf)
KE = abs((2.57*10^-17) – (5.34* 10^-17))
KE = 2.77 * 10^-17

Convert to velocity (the mass of an electron is 9.1094 * 10^-31 kg):

KE = 1/2mv^2
2.77 * 10^-17 = 1/2*9.1094 * 10^-31*v^2
v = 7.8*10^6 m/s

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