Two identical point charges, each of 5 C , are fixed at points (-3, 0, 0) m and (3, 0, 0) m . A particle of mass 8 g and charge -3 C is released from rest at point (0, 4, 0) m . The speed of the particle when it reaches the origin (0, 0, 0) is ( 1 4 _ o = 9 10⁹ in SI units )

Step-by-step solution

Initial distance of the particle from each of the fixed charges: r_i = ( 3 - 0)^2 + (0 - 4)^2 + 0^2 = 9 + 16 = 5 m Initial electric potential at (0, 4, 0) m due to both charges: V_i = 2 1 4 ₀ Q r_i = 2 9 10^9 5 10⁻⁶ 5 = 18000 V Final distance of the particle from each of the fixed charges at the origin: r_f = ( 3 - 0)^2 + 0^2 + 0^2 = 3 m Final electric potential at the origin due to both charges: V_f = 2 1 4 ₀ Q r_f = 2 9 10^9 5 10⁻⁶ 3 = 30000 V Change in electrostatic potential energy: U = q(V_f - V_i) = -3 10⁻⁶ (