Electron Between Infinite Line Charges and SHM

[roam]

Homework Statement

Two infinite line charges, of charge density λ Cm^-1, are aligned parallel to one another as shown in the diagram below. The line charges are a distance 2a apart. An electron is placed at location (x, 0, 0), x << a and released. Show that it will execute simple harmonic motion about the origin if the sign of the charge density is chosen correctly. Find an expression for the frequency of the motion.

http://img15.imageshack.us/img15/1507/45768328.jpg

Homework Equations

The integral form of Gauss’s law: [itex]\oint \vec{E} \ . \ \hat{n} da = \frac{q_{enc}}{\epsilon_0}[/itex]

SHM equation: [itex]\frac{d^2x}{dt^2}=- \frac{k}{m} x[/itex]

The relationships: [itex]f=\frac{1}{T} = \frac{kv}{2 \pi} = \frac{v}{\lambda}[/itex]

The Attempt at a Solution

So using Gauss's law the electric field due to the two line charges at any point would be:

[itex]E= \frac{\lambda}{2 \pi \epsilon_0 s} \hat{s}[/itex]​

Since the electron has negative charge and it is closer to the left-hand line charge, I think the left line charge must also be negative so they repel and the electron stays somewhere in the middle. Is this right?

So, how can I show that the motion would be SHM? I know that the solution to the SHM equation is given by x(t) = A cos (ωt-ϕ). But how do I relate this to the electric fields and the motion of the electron?

Any help would be greatly appreciated.

 

[voko]

1. Can we assume the electron stays on the axis x at all times? Why?

2. Assuming that, what is the combined force acting on it?

 

[bigerst]

try to write down a differential equation for the electron, simplify it using approximations and compare it with the differential equation for SHM

 

[roam]

1. Can we assume the electron stays on the axis x at all times? Why?

2. Assuming that, what is the combined force acting on it?

The problem statement doesn't make this assumption. But if we do, the force acting on it is

F=QE

Where Q=e (charge of the electron), E is the total electric field due to the two line charges. But how does velocity factor into the equation?

Lorentz force law that contains velocity F=Q[E+(vxB)] only applies in the presence of both electric and magnetic fields.

try to write down a differential equation for the electron, simplify it using approximations and compare it with the differential equation for SHM

What kind of DE do I have to write? Does it have to be in terms of velocity and distance?

 

[vela]

If the electron is to undergo simple harmonic motion, its equation of motion should look like the SHM one you wrote down as a relevant equation. Note that the first derivative does not appear in that equation, so you don't want the electron's velocity to appear in your equation.

Start by applying F=ma to the electron.

 

[TSny]

So, how can I show that the motion would be SHM?

You know the motion will be SHM if the net force on the electron obeys Hooke's law: F = -kx, for some constant k.

Thus, derive an expression for the net force when the electron is at (x, 0 ,0). Then try to find a simplified, approximate form of this expression under the assumption that x<<a.

 

[roam]

Thank you very much for your responses.

If the electron is to undergo simple harmonic motion, its equation of motion should look like the SHM one you wrote down as a relevant equation. Note that the first derivative does not appear in that equation, so you don't want the electron's velocity to appear in your equation.

Start by applying F=ma to the electron.

But shouldn't we work out the force by considering the electric force F=QE due to the line charges (in this case Q=e=1.602x10^-19)?

If the two line charges are of opposite sign, and assuming that the electron is closer to -a than it is to a, then we have

[itex]F=e \left( \frac{\lambda}{2 \pi \epsilon_0 r} - \frac{\lambda}{2 \pi \epsilon_0 R} \right)[/itex]

where R>r.

You know the motion will be SHM if the net force on the electron obeys Hooke's law: F = -kx, for some constant k.

Thus, derive an expression for the net force when the electron is at (x, 0 ,0). Then try to find a simplified, approximate form of this expression under the assumption that x<<a.

I'm not sure if I understood that correctly. So I have the following equation:

[itex]e \left( \frac{\lambda}{2 \pi \epsilon_0 r} - \frac{\lambda}{2 \pi \epsilon_0 R} \right) = - k x[/itex]

Do I have to then write this in terms of the spring constant k and substitute in the SHM differential equation?

 

[voko]

You need to think of the physical significance of the symmetries in your problem. Will there be any cause for the electron to leave the x axis?

If yes, what?

If no, does that help you to obtain the total force acting on the electron?

 

[TSny]

I'm not sure if I understood that correctly. So I have the following equation:

[itex]e \left( \frac{\lambda}{2 \pi \epsilon_0 r} - \frac{\lambda}{2 \pi \epsilon_0 R} \right) = - k x[/itex]

Do I have to then write this in terms of the spring constant k and substitute in the SHM differential equation?

First of all, you mentioned that the wires should have opposite sign. Is that right? Keep in mind that we want x = 0 to be a stable equilibrium point so that small displacements from x = 0 (along the x axis) yield a net force back toward x = 0. Nevertheless, I think the left hand side of your equation is correct in that it represents the net force assuming the same sign of charge for the two wires.

You want to show that, to a good approximation, the left side of your equation above reduces to the right side when x << a. To do this you'll need to express r and R in terms of x and a. In finding a good approximation recall that 1/(1+z) [itex]\approx[/itex] 1-z when z<<1. You should be able to get the left hand side to take the form of the right hand side: -kx. In the process you will discover an expression for k.

 

[vela]

As I said above, you want to apply F=ma to the problem. The righthand side supplies the second derivative of x with respect to time that appears in the SHM differential equation. The lefthand side, the F, is where the linear x term appears from when you follow TSny's advice.