How Close Does a Proton Get to an Infinitely Long Line of Charge?

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SUMMARY

The discussion focuses on calculating how close a proton can approach an infinitely long line of charge with a linear charge density of 6.50 x 10^-12 C/m. The proton starts at a distance of 17.5 cm and moves towards the line with an initial speed of 2900 m/s. The kinetic energy of the proton will convert entirely into potential energy as it approaches the line. The relationship governing this conversion is expressed as q (V_{f} - V_{i}) = 1/2 m_{p} v^2, where potential can be determined through integration to find a general expression.

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  • Understanding of electrostatics, specifically electric potential due to line charges
  • Knowledge of energy conservation principles in physics
  • Familiarity with integration techniques for calculating potential energy
  • Basic concepts of particle motion and kinetic energy
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This discussion is beneficial for physics students, educators, and anyone interested in electrostatics and energy conservation principles, particularly in the context of charged particle dynamics.

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Q An infinitely long line of charge has a linear charge density of 6.50*10^-12 C/m. a proton is a distance of 17.5 cm from the line and moving directly toward the line with a speed of 2900 m/s. How close does the proton get to the line charge? I am lost and do not know how to approach the problem or set it up.
 
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from taht distance what is the potential due to this infinite rod?
When the proton gets as close as it can all of its kinetic energy will be converted to potential energy
q (V_{f} - V_{i}) = \frac{1}{2} m_{p} v^2

Cna you find the potential at the starting point using integration? Try using variables so you cna get a general expresion. It will be useful later on.
 

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