You have a summer intern position with a company that designs and builds nanomac

Question

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing 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 determined by the amount of charge on the loop.

Evaluate the oscillation frequency for an electron at the center of a 2.8 -diameter ring charged to 1.7×10-13 .

Explanation / Answer

Let us f(z) 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

Here dF is the force between a very small element of the loop and the charge -q. So
           df = -(qdq) / (4peor²)                 = -(qdq) / (4peo(z² + R²)²)

Here dq is the charge of the small element of the loop            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. The z- component of df is
             sin ? * df

      Here ? is the angle between R and r and
    sin ? = z / v(z² + R²)

Now, the z- compoenent of the df is
[z / v(z² + R²)] df = [z / v(z² + R²)] ( -(qdq) / (4peo(z² + R²)²))
                           = -zqdq / [4peo(z² + R²)^(3/2)]

since z << R

(z² + R²)^(3/2) = (R²)^(3/2) = R³

  
[z / v(z² + R²)] df = -zqdq / [4peoR³]


F(z) = ? dF       = ? -zqdq / [4peoR³]       = -zqQ / [4peoR³]

Accordig to Hook's law

f(x) = -kx                -----------------(1)

and

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



f(z) = -zqQ / [4peoR³]          --------------(2)

From the equations (1) and (2) ,wwe get k = qQ / [4peoR³],

f(z) = -zk


The frequency of the simple harmonic motion with a spring that:

f = (1/2p)v(k/m) = (1/2p)v((qQ / [4peoR³]) / m)
    = (1/2p)v(qQ / 4peomR³)
------------------------------------------------------------------------ Given that m = 9.11 x 10^-31 kg R = 1.35 x 10^-6 m q = 1.6 x 10^-19 C Q = 1.6 x 10^-13 C So the oscillation frequency for an electron is            f = (1/2p)v(qQ / 4peomR³)              = (1/2p)v((1.6 x 10^-19 C)(1.6 x 10^-13 C) / 4peo(9.11 x 10^-31 kg)(1.35 x 10^-6 m)³)             = 1.6136x10^+12 Hz

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