Two particles,each of charge 2.5 × 10^-19 C,are 4.8 × 10^-5 m apart.What is the change in potential energy if they are brought 2.4 × 10^-5 m closer together? (k_e = 8.99 × 10⁹ N·m²/C²)

Question

Quizplus · Accepted Answer

The potential energy between two charges is given by U=k(q1q2)/r, where k is Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.
So, the initial potential energy is U=k(q1q2)/r1, where r1=4.8×10^-5 m.
The final potential energy is U=k(q1q2)/r2, where r2=2.4×10^-5 m.
Subtracting these two, we get the change in potential energy:
ΔU=Uf-Ui=k(q1q2)(1/r2-1/r1)=8.99×10^9(2.5×10^-19)^2(1/2.4×10^-5-1/4.8×10^-5)=1.2×10^-23 J.

Therefore, the answer is (A) 1.2E-23 J.

Two Particles,each of Charge 2