Two point charges q1 and q2 are fixed at a distance 3.0 cm as shown in the figure. A dust particle with mass m=5.0×10−9 kg and charge q0=2.0nC starts from rest at point 'a' and moves in a straight line to point ‘b’. What is its speed v at point b (in m/s) ? (round off to nearest integer)

Applying conservation of mechanical energy ΔK+ΔU=0or Kf−Ki+Uf−Ui=0Here, Ki=0 and Kf=12mv2 and Ui=q0Vi and Uf=q0VfHence, Kf−Ki+q0Vf−Vi=012mv2−0+q0Vf−q0Vi=0And solving for v, we find, v=2q0Vi−Vfm            …(i) Electric potential at 'a':Vi=14πε0q1r1a+14πε0q2r2aVi=9.0×109×3.0×10−90.010+−3.0×10−90.020=1350 V      …(ii)Electric potential at  'b′: Vf=14πε0q1r1b+14πε0q2r2bVf=9.0×109×3.0×10−90.020+−3.0×10−90.010=−1350 V       …(iii)From (i), (ii) and (iii) we get v=22.0×10−9[(1350)−(−1350)]5.0×10−9=46 m/srounding off to nearest integer