Sphere 1 with radius r1 has positive charge q. sphere 2 with radius 8.0r1 is far from sphere 1 and initially uncharged. after the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential v1 of sphere 1 greater than, less than, or equal to potential v2 of sphere 2? what fraction of q ends up on (b) sphere 1 and (c) sphere 2? (d) what is the ratio of the final surface charge density of sphere 1 to that of sphere 2? (answer with 2 significant figures.)

(a) equal

Explanation: when the two spheres are connected through the wire, the charge on the spheres redistribitute such that the two spheres reach the same potential V. This occurs because if the potential on the two spheres was different, then there would be a net electric field that keeps the charges moving from one sphere to another, until an equilibrium condition is reached (and this equilibrium condition is exactly when the potentials on the two spheres are equal).

(b)

As stated in the previous exercise, the potential on the two spheres is equal:

(1)

The capacity of a charged sphere is given by:

(2)

where R is the radius of the sphere. Let's also remind the relationship between charge Q, capacity C and potential V:

So we can rewrite (1) as

and by re-writing the capacity using (2):

(3)

The problem tells us that sphere 2 has a radius 8 times the radius of sphere 1:

Also, the total charge is , which can be rewritten as

By using these last 2 equations into (3), we have:

(c)

Since the first sphere has a charge of , and the total charge must be Q, the second sphere will have a charge of

(d) 8

The surface of a sphere is given by

So, the surface charge density of sphere 1 is

the surface charge density of sphere 2 is

So, the ratio of the two surface charges is

more than, less

less than, more

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this