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A point charge is projected along the axis of circular ring of  charge Q and radius10(2)^1/2 cm.the distance of the point charge from the centre of the ring,where the acceleration of the charged particle is maximum,will be
Asked by jmorakhia | 26 Apr, 2019, 10:55: PM
answered-by-expert Expert Answer
acceleration is maximum where the electric field is maximum.
 
Figure shows electric field dE due to charge dq=λdr on small element dr of circular ring of radius R.
λ is linear charge density which is given by λ = Q/(2πR)
 
Electric field dE can be resolved so that one component is parallel to axis i.e dE cosθ and other one is perpendicular to the axis dE sinθ .
perpendicular component will be cancelled if we consider the electric field contribution of all the points in the circumference.
 
Total electric field E = begin mathsize 12px style integral d E space cos theta space equals space fraction numerator lambda space 2 πR over denominator 4 pi epsilon subscript 0 end fraction space 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 space equals space fraction numerator Q over denominator 4 pi epsilon subscript 0 end fraction space 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 ..........................(1)
 
location of Maximum value of electric field is obtained from ( dE/dx ) = 0
 
Hence by diffrentiating the electric field function as given by eqn.(1) with respect to x, and equating it to zero.
  we get maximum electric field is at x = R/√2
 
Hence acceleration of charged particle is maximum when x = 10√2 / √2  = 10 cm from centre of ring
Answered by Thiyagarajan K | 27 Apr, 2019, 03:55: PM

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