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An electron is released from rest at a distance of 9cm from a proton.

An electron is released from rest at a distance of 9 cm from a proton. How fast will the electron be moving when it is 3 cm from the proton?

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Answer #1
Initial potential energy U1 = Kq1q2/r = (9*10^9 * 1.6*10^-19 * 1.6*10^-19)/(0.09)= 2.56*10^-27 JFinal potential energy U2 = Kq1q2/r' = (9*10^9 * 1.6*10^-19 * 1.6*10^-19)/(0.03)=7.68*10^-27 JLet v be the speed of the electron.From conservation of energy(1/2)mv^2 = U2 - U10.5*9.1*10^-31*v^2 = 5.12*10^-27v^2 = 11252.747v = 106.07 m/s
answered by: virangergal
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