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 , 7 Years ago
Grade 12
2 Answers
Shoubhik
Last Activity: 7 Years ago
We know that Electric field due to a ring is
:
Let the charge for convenience,
so force acting on the charge is
For small displacement:
Restoring force acting on charge:
So by both equations we get,
But
So,
Yash Shisode
Last Activity: 5 Years ago
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£
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