Question

A 15.0-nC charged object and a -40.0-nC charged object move from an initial separation of 50.0 cm to a final separation of 12.0 cm. Determine the change in the electric potential energy of the system of two charged objects Express your answer with the appropriate units alue Units Take the potential energy to be zero at infinite separation. Calculate the potential energy of the initial state Express your answer with the appropriate units alue Units Take the potential energy to be zero at infinite separation. Calculate the potential energy of the final state Express your answer with the appropriate units atValue Units

Answer 1

The electric potential energy is: U = kq_1 q_2/r where k is constant. q_1 and q_2 are charges. r is distance between two charges. Potential energy at initial separation: U_i = kq_1 q_2/r_i = (9 times 10^9 N middot m^2 middot C^-2) (15 times 10^-9 C) (40 times 10^-9 C)/(50 times 10^-2 m) = 1.08 times 10^-5 N middot m = 1.08 times 10^-5 J Potential energy at final separation: U_f = kq_1 q_2/r_f = (9 times 10^9 N middot m^2 middot C^-2) (15 times 10^-9 C) (40 times 10^-9 C)/(12 times 10^-2 m) = 4.5 times 10^-5 N middot m = 4.5 times 10^-5 J Change in potential energy: U = U_f - U_i = (4.5 times 10^-5 J) - (1.08 times 10^-5 J) = 3.42 times 10^-5 J Potential energy of the initial state: U_i = 1.08 times 10^-5 J Potential energy of the final state: U_f = 4.5 times 10^-5 J