JEE Class main Answered

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/(4πε_o ) and y is distance of charge q3 from origin O

Resultant force F_R = ( 2 × K × 8 × cosθ) / (y²+9) = ( 16 K y ) / (y²+9)^3/2

acceleration , a = dv/dt = F_R / 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