Two protons,each of charge 1.60 × 10−¹⁹ C,are 3.00 × 10−⁵ m apart.What is the change in potential energy if they are brought 1.00 × 10−⁵ m closer together? (k_e = 8.99 × 10⁹ N·m²/C²) A) 3.84 × 10−²⁴ J B) 3.20 × 10−¹⁶ J C) 1.15 × 10−²³ J D) 3.20 × 10−¹⁹ J

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Quizplus · Accepted Answer

The change in potential energy is given by:$$\Delta U = k_e \frac{q_1 q_2}{r_f} - k_e \frac{q_1 q_2}{r_i}$$where:* $k_e$ is the Coulomb constant* $q_1$ and $q_2$ are the charges of the protons* $r_f$ is the final distance between the protons* $r_i$ is the initial distance between the protonsPlugging in the given values, we get:$$\Delta U = (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(1.60 imes 10^{-19} ext{ C})^2}{(3.00 imes 10^{-5} ext{ m})} - (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(1.60 imes 10^{-19} ext{ C})^2}{(4.00 imes 10^{-5} ext{ m})}$$$$\Delta U = 3.84 imes 10^{-24} ext{ J}$$Therefore, the change in potential energy is 3.84 × 10−²⁴ J.

Two Protons,each of Charge 1