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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 , 8 Years ago
Grade 12
anser 2 Answers
Shoubhik

Last Activity: 8 Years ago

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}}}

Yash Shisode

Last Activity: 6 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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