Ask Question
3 May, 08:10

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

+4
Answers (1)
  1. 3 May, 08:28
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers