Consider two widely separated conducting spheres, 1 and 2, the second having fiv

Question

Consider two widely separated conducting spheres, 1 and 2, the second having five times the diameter of the first. The smaller sphere initially has a positive charge q = 9.00×10-6 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 q1 and q2. q1? Submit Answer Tries 0/99 q2? Submit Answer Tries 0/99 What is the ratio of the final surface charge density of sphere 1 to that of sphere 2?Consider two widely separated conducting spheres, 1 and 2, the second having five times the diameter of the first. The smaller sphere initially has a positive charge q = 9.00×10-6 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 q1 and q2.

q1?

q2?

What is the ratio of the final surface charge density of sphere 1 to that of sphere 2?

Explanation / 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(6r)) = Q2/(240r).

Equating these gives:
Q1/(40r) = Q2/(200r). Most of this cancels, leaving
5Q1 = Q2.

Conservation of charge assures us that the initial and final charge is the same: hence,

Q1 + Q2 = 9 C

Solving these equations simultaneously gives:

Q1 + 5Q1 = 9
6Q1 = 9
Q1 = (9/6) C
Q2 = 5(9/6) = 45/6 C.

Since the diameter (and thus radius) of sphere 2 is 5 times the radius of sphere 1, its surface area is 5² = 25 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 five times as much charge, spread out over 25 times the surface area, so its final surface charge density is one-fifth that of sphere 1.

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