Two conducting spheres with diameters...?

Question

... of 0.33 m and 0.6 m are seperated by a distance that is large compared to the diameters. The spheres are connected by a thin wire and are charged to 7.8 micro coulombs. The permittivitity of a vacuum is 8.85419*10^-12 C^2/Nm^2. What is the potential of the system of spheres when the reference potential is taken to be 0 at infinity. Answer in units of kV.

The thought I had for this problem was this: Initially, all the charge is in one sphere. After a while, the charge distributes itself, leaving the potentials on each sphere to be equal. After setting V1 equal to V2, you can solve for one of the charges which resides on one sphere. Following this, you can plug that number into the potential equation V = kq/r, with r being the radius of the sphere holding the charge previously found. Unfortunately, this isn't giving me the correct answer. Any help would be appreciated.

Answer 1

The wire should be at the same potential as the spheres, so some of the charge will be on the wire.

Answer 2

To find the distribution of charge: Q1/Q2 = r1/r2

To find the potential, add the potential of the 2 point charges (the spheres act like point charges). V=kQ1/R1 + KQ1/R2, where R1 & R2 are the distances from the point where you measure the potential (it depends on position) and each sphere. At a large enough distance, R1 = R2, approximately.

(My last physics class was in 1971, so you might want a second opinion.)

Answer 3

I was working on the same problem and as far as I can tell, your method is correct (using that similar method I came up with the answer in the back of the book).

V = V_1 = V_2

Q = q_1 + q_2 = 7.8 X 10^-6 C (1)

V = Ke q_1/r_1 = Ke q_2/r_2 (2)

==> q_1/r_1 = q_2/r_2 (3)

Use (1) and (3) to solve for q_1 OR q_2, then plug that back into (2); that should be the solution (like I said, it works here).