In figure, a particle has mass m = 5g and charge q' = 2 × 10⁻⁹C starts from rest at point a and moves in a straight line to point b. What is its speed v at point b?

Potential at a point p due to the charge q, Vp = Kq/r
Where, K = 1/4πε₀ = 9 × 10⁹
Total potential at point a, Va = 9 × 10⁹ × (3 × 10⁻⁹)/(10⁻²) + 9 × 10⁹ × (−3 × 10⁻⁹)/(2 × 10⁻²) = 13.5 × 10² V
Total potential at point b, Vb = 9 × 10⁹ × (3 × 10⁻⁹)/(2 × 10⁻²) + 9 × 10⁹ × (−3 × 10⁻⁹)/(10⁻²) = −13.5 × 10² V
Potential difference between point a and b, Vab = Va − Vb = 27 × 10² V
Work done in moving the charge q' from a to b, Wab = q'Vab
Wab = 2 × 10⁻⁹ × 2700 = 5.4 × 10⁻⁶ J
From work-energy theorem : Wab = ΔK.E ⇒ 5.4 × 10⁻⁶ = 1/2 × 0.005 × v² ⇒ v = 0.0465 ms⁻¹ = 4.65 ms⁻¹