Question

Two charges q₁ = ?3.20 nC and q₂ = +8.31 nC are at a distance of 1.62

Answer 1

a) distance travelled by charge q2 from 1.62 micro metre to 0.340 micro metre is = 1.62-0.340 = 1.28 *10^{-6} m

    the force on q2 at 0.340 micro metre is, F = k*|q1*q2|/r^2

                                                                 = 9*10^9 * 3.2*10^{-9}*8.31*10^{-9}/(0.34*10^{-6})^2

                                                                 = 2.07*10^{6} N

    from kinematic equation,

                                         v^2 - u^2 = 2*a*S

   here   u = 0 , S = 1.28*10^{-6} m then

                                             a = v^2/(2*S) = v^2/( 2*1.28*10^{-6} )

Since F= m*a = m*v^2/( 2*1.28*10^{-6} )

       (1/2)*m*v^2 = F*1.28*10^{-6} = 2.6496 J

the kinetic energy is 2.6496 J

b)        if m = 7.85*10^{-6} Kg then

        acceleration,a = F/m = 0.2636*10^{12} m/s^2

           then   speed, v = sqrt(2*a*S) = sqrt(0.674*10^{6}) = 0.82*10^3 m/s