Respuesta :
The frequency of the electron is :
[tex]$\omega = 3.34 \times 10^{10} \ Hz$[/tex]
Determining the frequency of the electron :
The magnitude of the electric field at a point on the [tex]z[/tex] a-x-is is given by :
[tex]$E=\frac{q|z|}{4 \pi \epsilon_0(z^2+R^2)^{3/2}}$[/tex]
Force on the electron is :
[tex]$F_z=(-e)E_z[/tex]
[tex]$=\frac{-eqz}{4 \pi \epsilon_0(z^2+R^2)^{3/2}}$[/tex]
Now if z is very small,
[tex]$(z^2+R^2)^{3/2} \sim (R^2)^{3/2}=R^3$[/tex]
[tex]$F_z = \frac{-eqz}{4 \pi \epsilon_0 R^3} = - \left(\frac{eq}{4 \pi \epsilon_0 R^3 \right)z}$[/tex]
Comparing the equation with
[tex]$F_z=-kz$[/tex]
Similarly using,
[tex]$k=\frac{eq}{4 \pi \epsilon_0 R^3}$[/tex]
[tex]$\omega =\sqrt{\frac{k}{m}} $[/tex]
[tex]$=\sqrt{\frac{eq}{4 \pi \epsilon_0 R^3 m}}$[/tex]
[tex]$=\sqrt{\frac{1.6 \times 10^{-19}\times 1}{4 \pi \epsilon_0 \times 1^3 \times 9.1 \times 10^{-31}}}$[/tex]
[tex]$= 3.977 \times 10^{10} \ Hz$[/tex]
The electric field on the a-x-is of a square is given by
[tex]$E_z=\frac{\lambda z a}{4 \pi \epsilon_0 k^2 \sqrt{k^2+\frac{a^2}{4}}}$[/tex]
[tex]$=\frac{4\lambda z a}{4 \pi \epsilon_0 \left(z^2+\frac{a^2}{4}\right) \sqrt{z^2+\frac{a^2}{2}}}$[/tex]
If we as-sumed that [tex]$ z << a$[/tex]
[tex]$E_z=\frac{4 \lambda a \times 4 \sqrt z}{4 \pi \epsilon_0 a^3}z$[/tex]
[tex]$\omega = \sqrt{\frac{k}{m}}$[/tex]
[tex]$=\sqrt{\frac{16 \sqrt z \lambda q}{4 \pi \epsilon_0 a^2 m}}$[/tex]
here,
[tex]$\lambda = 1/8, a = 2$[/tex]
[tex]$\omega = 3.34 \times 10^{10} \ Hz$[/tex]
Learn more about "frequency" here :
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