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

[Void123]

Homework Statement

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

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

The Attempt at a Solution

Using the Gaussian integral, I calculated the electric field

E\propto\frac{\lambda}{r}

I found that x=v_{0}t

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.

 

[Void123]

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

Hence:

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

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

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

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

So,

x = v_{0}t (1)

y = \frac{qEt^{2}}{2m} + r_{1} (2)

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

I just need someone to corroborate this.

 

[Donaldos]

Isn't the value of E a function of y (and therefore t)?

 

[Void123]

Isn't the value of E a function of y (and therefore t)?

I think you're right.