Two spheres with a conducting wire

[Abdulayoub]

Homework Statement

1. Two solid metal spheres are very far apart: sphere A (radius 44.1 cm) is charged to potential +2,346 Volts; sphere B (radius 63.6 cm) is charged to potential +8,848. Now, a long conducting wire connects the two spheres. When the charge has stopped moving, find the new charge on sphere A, in μC.

Homework Equations

v1=kq1/r1 v2=kq/r2

The Attempt at a Solution

since there are connected by a conducting wire both of the spheres will have the same potential thus :
v1+v2 / 2 = 5597 V
then:
5597= kq1/r1
q1= .275 μc
im not sure about this so any help will be appreciated !

 

[haruspex]

v1+v2 / 2 = 5597 V

How do you justify that? Is potential some conserved quantity?
What equation do you know relating the charge and radius of a conducting sphere to its surface potential?

 

[rude man]

Homework Statement

1. Two solid metal spheres are very far apart: sphere A (radius 44.1 cm) is charged to potential +2,346 Volts; sphere B (radius 63.6 cm) is charged to potential +8,848. Now, a long conducting wire connects the two spheres. When the charge has stopped moving, find the new charge on sphere A, in μC.

Homework Equations

v1=kq1/r1 v2=kq/r2

You have 4 equations and 4 unknowns: q_Ai, q_Bi, q_Af and q_Bf.
where i = initial and f = final. You can solve for q_Ai and q_Bi by your 'relevant' equation'. Now invoke equality of potential for the final state, and charge conservation.

 

[Alan I]

Now invoke equality of potential for the final state, and charge conservation.

I'm working on the same problem and this is what I got from following this thread:
- the equality of final potential means: V_Af=V_Bf
⇒ kq_Af/r_A=kq_Bf/r_B
⇒q_Afr_B=q_Bfr_A
⇒q_Af=q_Bfr_B/r_A and q_Bf=q_Afr_B/r_A (1)

and then from conservation of charge:
q_Ai+q_Bi=q_Af+q_Bf
⇒q_total=(q_Bfr_A/r_B) + q_Bf → solve for q_Bf
and the same for q_Af with substitution from (1).

Is that right?