In the figure, three identical conducting spheres initially have the following c

Question

In the figure, three identical conducting spheres initially have the following charges: sphere A, 2Q; sphere B, -3Q, and sphere C, 0. Spheres A and B are fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere C is touched to sphere A and then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere C is touched to sphere B and then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between A and B at then end of experiment 2 to that at the end of experiment 1? F_2/F_1 = .5 When two identical conducting spheres are connected, each ends up with half of the initial sum of charge (signs included in the summation). Did you use the Coulomb equation for the force, initially and finally?

Explanation / Answer

Experiment 1:
after first touch, A = C = ½(2 - 0)Q = Q
and after second touch, C = B= ½(1 - 3)Q = -Q
so A*B = Q * -Q = -Q²

Experiment 2:
after first touch, C = B = ½(0 - 3)Q = -1.5*Q
and after second touch, C = A = ½(-1.5 + 2)Q = 0.25Q
so A*B = -1.5Q*0.25Q = -0.375Q²

ratio = -0.375 / -1 = 0.375

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.