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# find velocity of particle Asked by ashutosharnold1998 21st March 2020, 11:01 PM 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 Expert 22nd March 2020, 8:33 AM
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