Electric Field Velocities from Point A to Point B

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SUMMARY

The discussion focuses on calculating the final speed of a proton as it moves from point A at 30V to point B at -10V, given its initial speed of 4.90×104 m/s. Using the equation for potential energy change, deltaV = deltaK/q, the final speed at point B is determined to be 8.75×104 m/s. The relevant constants include the charge of the proton (1.6×10-19 C) and its mass (1.67×10-27 kg). The calculation simplifies the energy equation to find the final velocity effectively.

PREREQUISITES
  • Understanding of electric potential and kinetic energy concepts
  • Familiarity with the equations of motion in physics
  • Knowledge of the properties of protons, including charge and mass
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the concept of electric potential energy and its relation to kinetic energy
  • Learn about conservation of energy in electric fields
  • Explore the effects of electric fields on charged particles
  • Investigate advanced topics in particle physics, such as the behavior of protons in varying electric fields
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of charged particles in electric fields.

drprevet
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Homework Statement



A proton's speed as it passes point A is 4.90×104 . It follows the trajectory: Point A @ 30V through Point B @-10V. What is the proton's speed at point B?

Homework Equations



deltaV=deltaK/q
V=40
vi=4.9*10^4
q=1.6*10^-19
m=1.67*10^-27
deltaK=1/2mvf^2-1/2mvi^2

The Attempt at a Solution



40=(0.5)(1.67*10^-27)(vf^2)-=(0.5)(1.67*10^-27)(4.9*10^4)/1.6*10^-19

velocity final=8.75*10^4
 
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The calculation is easier if you factor the right side of the equation:

[tex]40=\frac{0.5*1.67\cdot 10^{-27}}{1.6\cdot 10^{-19}}(v^2-(4.9\cdot 10^4)^2)[/tex]

and then simplify.

ehild
 

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