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find velocity of particle
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Asked by ashutosharnold1998 | 21 Mar, 2020, 11:01: PM
answered-by-expert 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,

 Error converting from MathML to accessible text.

Above integration is performed using substitution u = (y2 + 9 )

begin mathsize 14px style v squared over 2 equals fraction numerator 16 K over denominator m end fraction integral subscript 3 superscript 5 fraction numerator d u over denominator u to the power of begin display style bevelled 3 over 2 end style end exponent end fraction space equals space fraction numerator 16 K over denominator m end fraction open square brackets fraction numerator square root of 5 space minus space square root of 3 over denominator square root of 15 end fraction close square brackets end style

  By substituting K = 9 × 109  and m = 1 g in above equation, we get v = 6.12 × 106 m/s

 

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