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A particle of mass m and positive charge q is projected towards an infinity long line of charge from a distance r the velocity v makes 30 degree with normal to the line charge the minimum distance of approach of charge particle with line of charge

Asked by ayushsahuopp 21st January 2020, 10:03 AM
Answered by Expert
Answer:
Figure shows the trajectory of charged particle moving in the electric field due to infinite line charge.
Initially the charge particle started with velocity v in the direction that makes angle 30° to the normal drawn to the infinite line charge.
 
If we resolve the velocity as ( v cos30 ) along the normal and ( v sin30 ) along the perpendiculardirection  to normal,
 
then ( v cos30 ) component is affected due to force F of repulsion due to electric field of infinite line charge.
 
The vertical componenet ( v sin30 ) is unaffected.
 
Hence the charged particle move in a parabolic path like projectile motion in gravitational field.
 
If vi = ( v cos30 ) is the initial velocity at A along the direction of normal and the velocity becomes zero at B,
 
then by work-energy theorem , begin mathsize 14px style 1 half m space v subscript i superscript 2 space equals space minus integral subscript r subscript i end subscript superscript r subscript f end superscript F space d r end style   ...............................(1)
where F is the force due to electric field begin mathsize 14px style equals space q E space equals space q fraction numerator lambda over denominator 2 pi epsilon subscript o r end fraction end style  .....................(2)
where λ is the charge per unit length in the infinite line charge
hence eqn.(1) is written as, begin mathsize 14px style 1 half m space v subscript i superscript 2 space equals space fraction numerator q lambda over denominator 2 pi epsilon subscript o end fraction integral subscript r subscript f end subscript superscript r subscript i end superscript fraction numerator d r over denominator r end fraction space equals space fraction numerator q lambda over denominator 2 pi epsilon subscript o end fraction space ln open parentheses r subscript i over r subscript f close parentheses end style...................(3)
By substituting vi = ( v cos30 ) in above eqn.(3), we get , rf = ri exp[ - ( 3m v2 π εo ) / ( 4 q λ ) ]
Answered by Expert 21st January 2020, 11:04 PM
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