Question

1. In lecture, we derived the electric field a height z above the center of a thin ring of charge with constant charge per unit length λ. Let's assumie here that λ > 0. Suppose a negative point charge q with mass m s placed a very small distance above the center of the ring. Show that the point charge undergoes simple harmonic motion and find the frequency of small oscillations. Hint: show that near the center of the ring the force on the point charge resembles Hooke's law.

Answer 1

Electric field due to a charged ring at a distance 'z' from the center of the ring is given by

                                                         $E = \frac{Qz}{4 \pi \epsilon_{0} (a^{2} + z^{2})^{3/2}}$

Where, Q (=$2 \pi a \lambda$) is total charge on the ring, a is radius of the ring and

                                                           $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \,\, Nm^{2}/C^{2}$

Given, a negative charge -q is placed at a very small distance from the center of the ring. i.e. z<<a .

Therefore, we can neglect 'z' in the denominator,

                                                       $E = \frac{Qz}{4 \pi \epsilon_{0} a^{3} }$

So, the force on the negative charge is

                                                        $F = -q E = -\left ( \frac{Qq}{4 \pi \epsilon_{0} a^{3} } \right ) z$

                                                 or, $m \, \frac{d^{2}z}{dt^{2}}= -\left ( \frac{Qq}{4 \pi \epsilon_{0} a^{3} } \right ) z$

                                                 or, $\frac{d^{2}z}{dt^{2}}= -(k/m) z$                     .........................................(1)

Where,                               $k = \left ( \frac{Qq}{4 \pi \epsilon_{0} a^{3} } \right )$ , m is mass of the particle.

Equation 1 is the equation of SHM (Simple harmonic Motion).

Hence, Hence, the particle will do SHM with frequency,

                                           $\omega = \sqrt{k/m} = \left ( \frac{Qq}{4 \pi \epsilon_{0} a^{3} m} \right )^{1/2}$

For any doubt please comment.