Two protons,each of charge 2.2 × 10^-19 C,are 4.6 × 10^-5 m apart.What is the change in potential energy if they are brought 2.3 × 10^-5 m closer together? (k_e = 8.99 × 10⁹ N·m²/C²)

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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 protonsSubstituting the given values into the equation, we get:$$\Delta U = (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(2.2 imes 10^{-19} ext{ C})^2}{(2.3 imes 10^{-5} ext{ m})} - (8.99 imes 10^9 ext{ N}\cdot ext{m}^2/ ext{C}^2) \frac{(2.2 imes 10^{-19} ext{ C})^2}{(4.6 imes 10^{-5} ext{ m})}$$$$\Delta U = 0.95 imes 10^{-23} ext{ J}$$Therefore, the change in potential energy is 0.95 × 10^-23 J.

Two Protons,each of Charge 2