You have a summer intern position with a company that designs and builds. An eng

Question

You have a summer intern position with a company that designs and builds. An engineer with the company is designed a microscopic oscillator to help keep time, and you've been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small positively charged metal loop. His claim is that the negative charge will undergo simple harmonic motion at a frequency determine by the amount of charge on the loop. Consider a negative charge near the center of a positively charged ring. Find the frequency of small oscillations, with amplitude

Explanation / Answer

A) It is not mentioned how the circle is positioned with respect to the z-axis, but I'd assume the z-axis is the straight line that runs through the center of the circle parallel to its normal. In this case you don't need to show much at all. Draw a diagram of the loop (circle) and z-axis. Place a charge -q somewhere along the z-axis and draw the forces that act between the positive loop and the negative charge. Since opposite attract, and since the charge travels along the z-axis, the charge is pulled towards the ring when its position z 0, and NO forces act on the charge when it is at z = 0. If the charge is now placed a distance d 0 along the z-axis it will be pulled towards the circle, continue past z = 0 until it starts heading back again. It will oscillate!


Let us start by finding F(z), that is, the force between the charge and ring as a function of the distance between the center of the loop and the charge.

F(z) = F = dF

dF is the force between a very small part of the loop and the charge -q. Let us find the expression for dF:

dF = -(qdq) / (4×r²) = -(qdq) / (4×(z² + R²)²)

Here dq is the charge of the small part of the loop, when I say small I mean infinitesimally small! r is the distance between the small part and the charge -q. This distance is given by (z² + R²), where z is the distance between the charge -q and the center of the loop.

Let the loop bet positioned such that the z-axis points to the right. When studying the loop and its many small parts dq we notice that pairs of dq's located at opposite sides of the loop will have corresponding dF's that cancel each other leaving only the z component of the dF's.

The z component of dF is given by:

sin * dF

where is the angle between R and r. This angle is given by:

sin = z / (z² + R²)

the z component of dF now becomes:

[z / (z² + R²)] × dF = [z / (z² + R²)] × -(qdq) / (4×(z² + R²)²)

this simplifies to:

-zqdq / [4×(z² + R²)^(3/2)]

since z << R

(z² + R²)^(3/2) = (R²)^(3/2) = R³

so the z component of dF ends up being:

-zqdq / [4×R³]

Now:

F(z) = dF = -zqdq / [4×R³] = -zqQ / [4×R³]

Try comparing this to Hooks law!

F(x) = -kx

now, the frequency at which a mass m on a perfect spring oscillates is given by:

f = (1/2)(k/m)

Our Force is given by:

F(z) = -zqQ / [4×R³]

but if we let k = qQ / [4×R³]

we get

F(z) = -zk

now this looks familiar eh ;)

we now have from the example with hooks law and simple harmonic motion with a spring that:

f = (1/2)(k/m) = (1/2)((qQ / [4×R³]) / m)

this simplifies to:

f = (1/2)(qQ / 4m×R³)

part B

f = (1/2)(qQ / 4m×R³) = (1/2)(1.6*10^-19*1.3*10^-13*9*10^9/9.1*10^-31*(1.3*10^-6)^3)

= 0.154 *10^13 Hz

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