Determine the change in electric potential energy of a system of two charged objects when a - 2.1-C charged object and a - 5.0-C charged object move from an initial separation of 420 km to a final separation of 160 km

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ... 1

k = coulomb's constant = 9.0*10^9 N m^2/C^2

q1 = charge 1 = - 2.1C

q2 = charge 2 = - 5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 - 1/r1) ... 2

Substituting the given values

∆E = 9.0*10^9 * - 2.1 * -5.0 (1/160000 - 1/420000)

∆E = 94.5 * 10^9 (3.87 * 10^-6) J

∆E = 365.72 * 10^3 J

∆E = 365.72 kJ