How Does a Negative Charge Move Near an Infinite Line of Positive Charge?

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Homework Help Overview

The discussion revolves around the motion of a negative charge near an infinite line of positive charge, specifically focusing on deriving an expression for the trajectory of the charge. The problem involves concepts from electrostatics and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field and its dependence on distance. There are attempts to express the trajectory in terms of distance traveled parallel and perpendicular to the line of charge, with considerations on eliminating time from the equations. Questions arise about the dependence of the electric field on the charge's position.

Discussion Status

Some participants are making progress in their calculations and are seeking validation of their work. There is an ongoing exploration of the relationship between the electric field and the charge's position, indicating a productive direction in the discussion.

Contextual Notes

Participants are considering the implications of the electric field's dependence on the charge's distance from the line, which may affect their calculations. There is also a mention of initial conditions and the setup of the problem that may influence the trajectory analysis.

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Homework Statement



Consider a negative charge placed initially at a distance [tex]r_{0}[/tex] from an infinite line with positive charge density [tex]\lambda[/tex]. The charge is given an initial velocity of magnitude [tex]v_{0}[/tex] parallel to the line of charge.

Find an expression for the trajectory of the particle in which time has been eliminated. This expression relates the distance traveled by the charge parallel to the line and the distance traveled perpendicular to the line.

Homework Equations



[tex]m\stackrel{\rightarrow}{\ddot{r}}(x, y) = q\stackrel{\rightarrow}{E}[/tex]

The Attempt at a Solution



Using the Gaussian integral, I calculated the electric field

[tex]E\propto\frac{\lambda}{r}[/tex]

I found that [tex]x=v_{0}t[/tex]

Since the charge has an initial velocity in the horizontal direction and a force acting on it in the vertical direction due to the electric field, I know that it will follow a projectile motion as it travels parallel to the line.

How would I eliminate time from my equations? And would my final expression be some r(x, y) of the parabolic type?

Thanks.
 
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I have mad some progress, I think. I just need someone to check my work.

Hence:

[tex]\vec{V} = \frac{d\vec{r}}{dt} = \frac{qEt}{m}\hat{\bold{j}} + v_{0}\hat{\bold{i}}[/tex]

[tex]\vec{r} = \frac{qEt^{2}}{2m}\hat{\bold{j}} + v_{0}t\hat{\bold{i}} + r_{1}\hat{\bold{j}}[/tex]

[tex]\vec{r} = (\frac{qEt^{2}}{2m} + r_{1})\hat{\bold{j}} + v_{0}t\hat{\bold{i}}[/tex]

Where [tex]r_{1}[/tex] is just the initial distance of the charge from the infinite line.

So,

[tex]x = v_{0}t (1)[/tex]

[tex]y = \frac{qEt^{2}}{2m} + r_{1} (2)[/tex]

Substituting t from (1) into (2) and replacing E with [tex]\frac{k\lambda}{r_{1}}[/tex], I should get a final formula relating [tex]y[/tex], the distance perpendicular to the line of charge, to [tex]x[/tex] the distance traveled parallel to the line of charge, with the elimination of time.

I just need someone to corroborate this.
 
Isn't the value of [tex]E[/tex] a function of [tex]y[/tex] (and therefore [tex]t[/tex])?
 
Donaldos said:
Isn't the value of [tex]E[/tex] a function of [tex]y[/tex] (and therefore [tex]t[/tex])?

I think you're right.
 

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