. A point charge q_1 is held stationary at the origin. A second point charge q_2 is placed at point a, and the electric potential energy of the pair of charges is +5.4 ×〖10〗^(-8) Joules. When the second charge is moved to point b, the electric force on the charge does -1.9 ×〖10〗^(-8) Joules of work. What is the electric potential energy of a pair of charges when the second charge is at point b?
To find the electric potential energy of the pair of charges when the second charge is at point b, we can use the work-energy theorem. The work done by the electric force is equal to the change in potential energy.
Given that the work done is -1.9 × 10^(-8) Joules, we know that the change in potential energy is also -1.9 × 10^(-8) Joules.
We are given that the electric potential energy of the pair of charges at point a is +5.4 × 10^(-8) Joules. To find the electric potential energy at point b, we need to add the change in potential energy to the initial potential energy.
Hence, the electric potential energy of the pair of charges at point b is:
+5.4 × 10^(-8) Joules + (-1.9 × 10^(-8) Joules) = 3.5 × 10^(-8) Joules
Therefore, the electric potential energy of the pair of charges when the second charge is at point b is 3.5 × 10^(-8) Joules.