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A small particle of mass m and charge -q is placed at point P on the axis of uniformly charged ring and released. If R>>x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to ?

Md Saif Akhtar , 9 Years ago

Grade 12

2 Answers

Shoubhik

We know that Electric field due to a ring is

: $E=\frac{kqx}{2(R^{2}+x^{2})^{3/2}}$

Let the charge $-q=$ $q_{0}$ for convenience,

so force acting on the charge is

$F=q_{0}E$

For small displacement:

$E=\frac{kqx}{2R^{3}}$

Restoring force acting on charge:

$F=Kx$

So by both equations we get,

$Kx=\frac{kqx}{2R^{3}}$

$=>K=\frac{kq}{2R^{3}}$

But

$K=m\omega ^{2}$

So,

$\omega=\sqrt{\frac{kq}{2mR^{3}}}$

Last Activity: 8 Years ago

Yash Shisode

E=(KQx)/(R^2 + x^2) ^3/2

As R>>>x, x^2

Therefore, E=(KQx)/R^3 -----------(1)

F=m(omega) ^2 x -----------------------(2)

F=qE -----------------------(3)

From 1,2&3

m(omega) ^2 x =( kqQx) /R^3

Omega = √(KqQ) /√(mR^3)

= √(qQ) /√(4π£mR^3) as K=1/4π£

Last Activity: 7 Years ago

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