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

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Homework Help Overview

The problem involves calculating the closest distance a proton can get to an infinitely long line of charge with a specified linear charge density. The proton is initially at a distance of 17.5 cm from the line and is moving towards it with a given speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the electric field generated by the line of charge and the forces acting on the proton. There is an attempt to apply the linear acceleration formula, but concerns are raised about the validity of this approach due to the assumption of constant acceleration.
  • Some participants suggest using energy conservation principles instead, while others express confusion about the electric field's dependence on distance from the line of charge.
  • Questions arise regarding the integration of the electric field and the distinction between initial and final electric field strengths.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the use of energy conservation and the need to consider changing electric field strengths, but there is still significant confusion among participants about the application of these concepts.

Contextual Notes

Participants note the challenge of not having a clear final state for the proton's distance from the line of charge, which complicates their calculations. There is also mention of constraints from a homework platform that limits attempts at solving the problem.

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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