Calculating the Closest Distance of a Proton from an Infinite Line of Charge

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SUMMARY

The discussion focuses on calculating the closest distance a proton can approach an infinite line of charge with a linear charge density of 8.00×10-12 C/m. The proton starts 17.5 cm away and moves toward the line at 2700 m/s. The participant initially uses the electric field formula E = λ/(2πεr) and the force equation F = E*q to find acceleration, but encounters issues due to the non-constant acceleration. The correct approach involves energy conservation, leading to the conclusion that the closest distance the proton reaches is 0.134 m.

PREREQUISITES
  • Understanding of electric fields, specifically E = λ/(2πεr)
  • Knowledge of Newton's second law, F = ma
  • Familiarity with energy conservation principles in physics
  • Basic calculus for integrating electric fields (if applicable)
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  • Study energy conservation methods in electrostatics
  • Learn about electric fields generated by infinite line charges
  • Review the concept of non-constant acceleration in physics
  • Explore integration techniques for calculating work done by electric fields
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Students and educators in physics, particularly those studying electromagnetism and electric fields, as well as anyone seeking to understand the dynamics of charged particles in electric fields.

Ivegottheskill
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The question I've been attempting:

An infinitely long line of charge has a linear charge density of 8.00×10−12 C/m. A proton is a distance of 17.5 cm from the line and moving directly toward the line with a speed of 2700 m/s.

How close does the proton get to the line of charge?
Use 1.60×10−19 C for the magnitude of the charge on an electron, 1.67×10−27 kg for the mass of a proton, and 8.85×10−12 F/m for the permittivity of free space

From my notes and working etc. I've got:

E = lambda/2*pi*epsilon*r (where lambda = charge per unit length and epsilon = permittivity of free space)

F = E * q (where q = the charge of the proton, -1.60*10^-19, as apparently defined by the question)

I've used the F calculated to get acceleration by

a = F/m (where m = mass of proton)

I get a value of -7.88*10^7

I then use the linear acceleration formula v^2 = u^2 + 2*a*s to try and calculate s.

I get 0.04627... (4.63*10^-2)

Apparently this is incorrect however. Can anyone see where I'm messing up?
 
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The linear acceleration formula holds only when the acceleration is constant.

Try using energy conservation instead! :)
 
Hmm, still having trouble,

1/2*m*v^2 + E*q*y = 0 + E*q*(unknown y value)

The formula for E that I'm using doesn't make sense to me but appears right in my notes and textbook: E = lambda (charge per unit length)/2*pi*r*epsilon(permittivity of free space)

I wouldn't think E relies on r (for a straight line of charge). Aren't field lines parallel for a line of charge? Meaning E is constant at any point in the field?

I'm not given the r value for the final state of the particle, so I can't work out either E or the unknown y on the right hand side of that formula.
 
The "Electric Field Lines" are spread out more (1-dim), farther from the wire.
That means the E-field strength decreases as 1/r .
You need substripts to distinguish "E_final" from "E_initial" ... not equal!

You want to integrate E(r) from y_initial to y_final ...
or if this isn't for calc-based physics, use Potential.
 
No, still confused out of my brain. Was doing it on Mastering Physics.com, but exceeded attempts and failed the question. I tried a billion random different formulas. Worst thing is having no idea if they were even valid to use in an equation

The answer was apparently 0.134 m, but I still can't see how. I'll probably have to see tutorial teacher or something :confused::confused:
 

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