(c24p56) Consider two widely separated conducting spheres, 1 and 2, the second h
ID: 1874380 • Letter: #
Question
(c24p56) Consider two widely separated conducting spheres, 1 and 2, the second having three times the diameter of the first. The smaller sphere initially has a positive charge q 2.00×10-C, and the larger one is initially uncharged. You now connect the spheres with a long thin wire. How are the final potentials V1 and V2 of the spheres related? Find the final charges ql and q2 q1? Submit Answer Tries 0/99 92? Submit Anwer Tries 0/99 What is the ratio of the final surface charge density of sphere 1 to that of sphere 2? Submit Answer Tries o/99Explanation / Answer
V1_final and V2_final will be the same. This is a necessary condition for electrostatic equilibrium; if it were not so, the potential difference would cause charge to continue to flow. The potential at the surface of the first sphere is given by Q1/(40r1) = Q1/(40r), and the potential at the surface of the second sphere is Q2/(40r2) = Q2/(40(3r)) = Q2/(120r).
Equating these gives:
Q1/(40r) = Q2/(120r). Most of this cancels, leaving
3Q1 = Q2.
Conservation of charge assures us that the initial and final charge is the same: hence,
Q1 + Q2 = 2 C
Solving these equations simultaneously gives:
Q1 + 3Q1 = 2
4Q1 = 2
Q1 = (2/4) C = 0.5 C
Q2 = 3 (2/4) = 1.5 C.
Since the diameter (and thus radius) of sphere 2 is 3 times the radius of sphere 1, its surface area is 3² = 9 times that of sphere 1. You can solve for the charge density explicitly if you like, but even without doing so you can see that sphere 2 has 3 times as much charge, spread out over 9 times the surface area, so its final surface charge density is one-third that of sphere 1.
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