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

Actually, after double checking, the equation should be:y = (\frac{kq\lambda}{2m})\frac{1}{x^{2}} + r_{1}So the final expression relating x and y with the elimination of time would be:y = (\frac{kq\lambda}{2m})\frac{1}{x^{2}} + r_{1}In summary, the conversation discusses finding an expression for the trajectory of a negative charge with an initial velocity placed near an infinite line of positive charge density. The charge will undergo projectile motion due to the force from the electric field. To eliminate time from the equations, the value of E is substituted with \frac{kq\lambda}{r_{1}}, resulting
  • #1
Void123
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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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  • #2
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.
 
  • #3
Isn't the value of [tex]E[/tex] a function of [tex]y[/tex] (and therefore [tex]t[/tex])?
 
  • #4
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.
 

1. What is a line of infinite charge?

A line of infinite charge is an imaginary line that extends infinitely in both directions and has a uniform distribution of electric charge along its length. It is often used in physics and engineering calculations as a simplified model for real-life systems.

2. How is the electric field calculated for a line of infinite charge?

The electric field at any point on a line of infinite charge is calculated using the equation E = λ/2πε0r, where λ is the charge density (charge per unit length) of the line, r is the distance from the point to the line, and ε0 is the permittivity of free space.

3. What is the direction of the electric field for a line of infinite charge?

The electric field for a line of infinite charge points radially outward from the line. This means that at any point, the electric field vector is perpendicular to the line and points away from it.

4. How does the electric field vary along a line of infinite charge?

The electric field along a line of infinite charge is inversely proportional to the distance from the line. This means that the electric field strength decreases as the distance from the line increases.

5. How is the potential calculated for a line of infinite charge?

The potential at any point on a line of infinite charge is calculated using the equation V = λ/2πε0ln(r), where V is the potential, λ is the charge density, r is the distance from the point to the line, and ε0 is the permittivity of free space. This equation assumes that the potential at infinity is zero.

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