Verifying an Electron's Final Speed After Releasing from Rest

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SUMMARY

The final speed of the electron, when it is 10.0 cm from charge 1, is calculated to be 7.53 x 106 m/s. This result is derived using the conservation of energy principle, specifically the equations for kinetic energy (Ek) and electric potential energy (U). The calculations involve the magnitudes of the two positive point charges, 3.45 nC and 1.85 nC, and their respective distances from the electron. The solution confirms the accuracy of the final speed, addressing concerns about its magnitude.

PREREQUISITES
  • Understanding of electric potential energy (U) and kinetic energy (Ek)
  • Familiarity with Coulomb's law and point charge interactions
  • Basic knowledge of conservation of energy principles in physics
  • Ability to manipulate equations involving scientific notation
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  • Review the principles of conservation of energy in electrostatics
  • Learn about the implications of charge interactions on particle motion
  • Explore advanced topics in electric fields and potential energy calculations
  • Study the effects of varying charge magnitudes and distances on particle speed
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Homework Statement



Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed v(final) of the electron when it is 10.0 cm from charge 1?

Homework Equations



Ek= (mv^2)/2
U= (k(q1q2))/r

The Attempt at a Solution

I used the following equation:

Ek(f)+U(f) = Ek(i)+U(i)

(mv(f)^2)/2 + (k(q1q2))/r = (mv(i)^2)/2 + (k(q1q2))/r

(9.1x10^-31)(vf)^2)/2 + (((9x10^9)(3.45nC)(-1.6x10^-19))/0.10m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.40m) = (((9x10^9)(3.45nC)(-1.6x10^-19))/0.25m) + (9x10^9)(1.85nC)(-1.6x10^-19))/0.25m)

v(f) = 7.53 x 10^6 m/s

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So that's my answer. I lose marks if I submit an incorrect answer. That said, I just want to double check my answer before submitting it. A simple yes or no will suffice. Thanks in advance.

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That seems unreasonably high. Maybe I'm just ignorant but, guess we'll see.
 
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Took a stab at it. Was correct. Thanks everyone.
 

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