Speed of an Electron in an Electric Field

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SUMMARY

The discussion focuses on calculating the final speed of an electron released from rest between two stationary positive point charges, specifically 3.25 nC and 1.55 nC, separated by 57.0 cm. The problem utilizes the conservation of energy principle, where the initial potential energy is converted into kinetic energy as the electron moves. The relevant equations include the electric potential due to point charges and the kinetic energy formula, leading to the final speed calculation using energy conservation: 0 + Kqq1/(d/2) + Kqq2/(d/2) = mv^2/2 + Kqq1/a + Kqq2/(d-a).

PREREQUISITES
  • Understanding of electric potential due to point charges
  • Familiarity with conservation of energy principles
  • Knowledge of kinetic energy equations
  • Basic algebra for solving equations
NEXT STEPS
  • Review the concept of electric potential and its calculation for multiple point charges
  • Study conservation of energy in electrostatics
  • Learn how to derive the final speed of charged particles in electric fields
  • Practice problems involving kinetic and potential energy conversions
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Students in physics courses, particularly those studying electromagnetism, as well as educators looking for examples of energy conservation in electric fields.

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Homework Statement



Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 1.55nC , are separated by a distance of 57.0cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

find Vfinal of the electron when it is 10.0 cm from charge 1 the want the final answer in m/s

i don't get any of this my professor went to fast. can anyone help?
 
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This looks like a conservation-of-energy problem.

The two relevant forms of energy are kinetic and electric potential.
 
do you know of any equations that could help me
 
HELLO11 said:
do you know of any equations that could help me

Your book or class notes will have an equation for the electric potential (or voltage) due to a point charge.

To get the net electric potential of the two stationary charges, add up the individual potentials of each of the two charges.

To get the potential energy, multiply electric potential times the charge on the electron. Watch the +/- signs carefully.
 
can anyone help me
 
q1 = 3.25 nC, q2 = 1.55nC, d = 0.570 m. charge of electron = q, a = 0.100 m, find the final speed of the electron, v = ?
initial kinetic energy Eki = 0, potential energy Epi = Kqq1/(d/2) + Kqq2/(d/2)
final kinetic energy Ekf = mv^2/2, potential energy Epf = Kqq1/a + Kqq2/(d-a)
energy conservation:
0 + Kqq1/(d/2) + Kqq2/(d/2) = mv^2/2 + Kqq1/a + Kqq2/(d-a)
solve for v.
 

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