Question

A ring of radius 8 cm that lies in the ya plane carries positive charge of 2 μC uniformly distributed over its length. A particle of mass m that carries a charge of-2 oscillates about the center of the ring with an angular frequency of 22 rad/s Find the angular frequency of oscillation of the mass if the radius of the ring is doubled

Answer 1

Answer:

The angular frequency of oscillation of the mass is 11 rad/s.

Explanation:

Given that,

Charge = 2μC

Radius R₁= 8 cm

Radius R₂ = 16 cm

Angular frequency = 22 rad/s

We need to calculate the angular frequency of oscillation of the mass

The electric field produced along x axis

$E=\dfrac{kqx}{\sqrt{R^2+x^2}}$

$E=\dfrac{kqx}{\sqrt{R^2(1+\dfrac{x^2}{R^2})^2}}$

$E=\dfrac{kqx}{R^3}$

The force on the mass is

$F=Eq$

$F=\dfrac{kQqx}{R^3}$....(I)

For,x<<R

Now, using centripetal force

$F = \dfrac{mv^2}{r}$

Put the value of F in equation (I)

$\dfrac{mv^2}{r}=\dfrac{kQq}{R^3}$

We know that,

$v=r\omega$

$m\omega^2r=\dfrac{kQq}{R^3}$

$\omega^2=\dfrac{kQq}{mrR^3}$

For, r<<R

$\omega^2=\dfrac{kQq}{mR^3}$

$\omega=\sqrt{\dfrac{kQq}{mR^3}}$

Here,

$\omega\propto\sqrt{\dfrac{q}{R^3}}$

The ratio of angular frequency

$\dfrac{\omega}{\omega_{1}}=\sqrt{\dfrac{\dfrac{q}{R^3}}{\dfrac{q_{1}}{R_{1}^3}}}$

$\dfrac{\omega}{\omega'}=\sqrt{\dfrac{R^3\times2q}{2R^3\times q}}$

$\omega=\sqrt{\dfrac{8^3\times2\times2\times10^{-6}}{8\times8^3\times2\times10^{-6}}}\times\omega'$

$\omega=0.5\omega'$

Put the value of

$\omega=\dfrac{1}{2}\times22$

$\omega=11\ rad/s$

Hence, The angular frequency of oscillation of the mass is 11 rad/s.