0

Guest

Courses New

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 ?

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 ?

Grade:12

2 Answers

Shoubhik
26 Points
7 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
15 Points
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£
 
 
 
 
 

Think You Can Provide A Better Answer ?

Other Related Questions on Electrostatics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More