Q72P - Fundamentals Of Physics

Question

An electron is constrained to the central axis of the ring of charge of radius R in Fig. 22-11, with.$\mathit{z}\mathbf{ }\mathbf{\ll }\mathbf{ }\mathit{R}$ Show that the electrostatic force on the electron can cause it to oscillate through the ring center with an angular frequency$\mathit{\omega }\mathbf{=}\sqrt{\frac{\mathbf{e}\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\in }}_{\mathbf{o}}\mathbf{m}{\mathbf{R}}^{\mathbf{3}}}}$where, q is the ring’s charge and m is the electron’s mass.

 

Step-by-Step Solution

Answer

The electrostatic force on the electron can cause it to oscillate with an angular frequency of.

 $\sqrt{\frac{eq}{4\pi {ϵ}_{o}m{R}^3}}$

Step 1: The given data

An electron is constrained to the central axis of the ring of charge of radius R, with.$z\ll R$

Step 2: Understanding the concept of electrostatic force

Using the concept of the electrostatic force, we can get the spring constant of a body by comparing the electrostatic force with the restoring force. Again, by substituting the value of the spring constant in the formula of angular frequency we can get the required value.

 

Formulae:

The electric field at a point on the axis of a uniformly charged ring, a distance  from the ring center,    $E=\frac{qz}{4\pi {ϵ}_o{|z^2+R^2|}^{3/2}}$                                                                 (i)

Where,  is the charge on the ring and R is the radius of the ring.

The restoring force of a spring,    $F=-kz$                                                                               (ii)

The angular frequency of a body,       $\omega =\sqrt{\frac{k}{m}}$                                                                       (iii)

The electrostatic force of a particle,     $F=qE$                                                    (iv)

Step 3: Calculation of the angular frequency of an oscillation

 

For q positive, the field points upward at points above the ring and downward at points below the ring. We take the positive direction to be upward. Then, the force acting on an electron on the axis using equation (i) in equation (iv) is given as:

$F=-\frac{eqz}{4\pi {ϵ}_o{|z^2+R^2|}^{3/2}}$

For small amplitude oscillations and  can be neglected in the denominator. Thus, the above equation of the force is given as:

$F=-\frac{eqz}{4\pi {ϵ}_o{R}^3}\mathrm{................}\left(a\right)$

The force is a restoring force: it pulls the electron toward the equilibrium point. $z=0$So, the value of the spring constant can be given by comparing equations (a) with (ii) as given:

$k=qe/4\pi {\epsilon }_o{R}^3.$

The electron moves in simple harmonic motion with an angular frequency given by substituting the above value in equation (iii) as follows:

$\omega =\sqrt{\frac{eq}{4\pi {ϵ}_{o}m{R}^3}}$

where m is the mass of the electron.

Hence, the value of the angular frequency is.$\sqrt{\frac{eq}{4\pi {ϵ}_{o}m{R}^3}}$