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Home / Doubts and Solutions/NEET/Physics

A charged ring having charge Q is held fixed. A point charge q is released on the axis of ring at a distance R from centre of ring. The maximum speed acquired by small charge q is

[Given only coulomb force acts, Q and q are of opposite nature and mass of small charge is m]

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Asked by dipanshusingla029 23rd May 2018, 4:49 PM
Answered by Expert
Answer:
 
The radius of charged ring is R. (This is not given in the posted question).
 
Let P be the point on the axis of charged ring at a distance x from centre O, as shown in figure.
 
let us consider a small charged element λdl in the ring.
 
Electric field at point P along the axial direction due to this charged element λdl  is given by
 
begin mathsize 12px style d E space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript 0 end fraction fraction numerator 1 over denominator open parentheses R squared plus x squared close parentheses end fraction cos theta space equals fraction numerator lambda d l over denominator 4 pi epsilon subscript 0 end fraction fraction numerator 1 over denominator open parentheses R squared plus x squared close parentheses end fraction fraction numerator x over denominator square root of R squared plus x squared end root end fraction space equals space fraction numerator lambda d l over denominator 4 pi epsilon subscript 0 end fraction x over open parentheses R squared plus x squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style

Electric field due to the whole charged ring is simply above relation summed over the circumference.
begin mathsize 12px style E space equals space space fraction numerator Q over denominator 4 pi epsilon subscript 0 end fraction x over open parentheses R squared plus x squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style

This field is zero at centre of ring (x=0) and maximum at a distance R/√2 and afterwards decreasing
and becomes zero at long distance as shown in figure.
 
If a charge q is released at P, it experiences a force F which is given by
begin mathsize 12px style F space equals space q E space equals space space fraction numerator q Q over denominator 4 pi epsilon subscript 0 end fraction x over open parentheses R squared plus x squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end style
due to this force, the charge q is attracted towards O ( Q and q are opposite as per the question).
 
Then the workdone W by the field on charge q to move from a distance R from centre to the centre point O of ring  is given by
 
begin mathsize 12px style W space equals space integral subscript R superscript 0 F d x space equals space fraction numerator q Q over denominator 4 pi epsilon subscript 0 end fraction integral subscript R superscript 0 fraction numerator x space d x over denominator open parentheses R squared plus x squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end fraction end style
The above integration is solved using the substitution u = (R2 + x2) and we get
 
begin mathsize 12px style W space equals space fraction numerator q space Q over denominator 4 pi epsilon subscript 0 space R end fraction open curly brackets fraction numerator 1 over denominator square root of 2 end fraction minus 1 close curly brackets end style
The above workdone by the field is utilized by the charge to get kinetic energy (1/2)mv2 ,
where m is the mass of charge and v is the speed of charge
 
hence we have
 
begin mathsize 12px style W space equals space 1 half m v squared space equals space fraction numerator q space Q over denominator 4 pi epsilon subscript 0 R end fraction open curly brackets fraction numerator 1 over denominator square root of 2 end fraction minus 1 close curly brackets v squared equals fraction numerator q space Q over denominator 2 pi epsilon subscript 0 space m space R end fraction open curly brackets fraction numerator 1 over denominator square root of 2 end fraction minus 1 close curly brackets end style
 
 
 
Answered by Expert 10th June 2018, 5:04 PM
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