Answer:
[tex]\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }[/tex]
Explanation:
Additional information:
The ball has charge [tex]-q_0[/tex], and the ring has positive charge [tex]+Q[/tex] distributed uniformly along its circumference.
The electric field at distance [tex]z[/tex] along the z-axis due to the charged ring is
[tex]E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.[/tex]
Therefore, the force on the ball with charge [tex]-q_0[/tex] is
[tex]F=-q_oE_z[/tex]
[tex]F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}[/tex]
and according to Newton's second law
[tex]F=ma=m\dfrac{d^2z}{dz^2}[/tex]
substituting [tex]F[/tex] we get:
[tex]- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}[/tex]
rearranging we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0[/tex]
Now we use the approximation that
[tex]z^2+a^2\approx a^2[/tex] (we use this approximation instead of the original [tex]d^2+a^2\approx a^2[/tex] since [tex]z<d[/tex], our assumption still holds )
and get
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0[/tex]
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0[/tex]
Now the last equation looks like a Simple Harmonic Equation
[tex]m\dfrac{d^2z}{dz^2}+kz=0[/tex]
where
[tex]\omega=\sqrt{ \dfrac{k}{m} }[/tex]
is the frequency of oscillation. Applying this to our equation we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}[/tex]
[tex]\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}[/tex]
