JEE Class main Answered
find velocity of particle
Asked by ashutosharnold1998 | 21 Mar, 2020, 11:01: PM
Expert Answer
Figure shows the forces F1 and F2 acting on charge q3 due to the presence of charge q1 and q2
F1 = F2 = ( K × 8 ) / (y2+9) ............... (1)
where K = 1/(4πεo ) and y is distance of charge q3 from origin O
Resultant force FR = ( 2 × K × 8 × cosθ) / (y2+9) = ( 16 K y ) / (y2+9)3/2
acceleration , a = dv/dt = FR / m ;
acceleration a = (dv/dy)(dy/dt) = v(dv/dy) = [ ( 16 K / m ) ] [ y / (y2+9)3/2 ]
above equation is written as, vdv = [ ( 16 K / m ) ] [ ( y dy ) / (y2+9)3/2 ]
By integrating bothsides of equation we get,
Above integration is performed using substitution u = (y2 + 9 )
By substituting K = 9 × 109 and m = 1 g in above equation, we get v = 6.12 × 106 m/s
Answered by Thiyagarajan K | 22 Mar, 2020, 08:33: AM
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