# Two spheres with a conducting wire

1. Sep 14, 2015

### Abdulayoub

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
v1=kq1/r1 v2=kq/r2
3. 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

2. Sep 15, 2015

### haruspex

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?

3. Sep 15, 2015

### rude man

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

4. Sep 16, 2015

### Alan I

I'm working on the same problem and this is what I got from following this thread:
- the equality of final potential means: VAf=VBf
⇒ kqAf/rA=kqBf/rB
⇒qAfrB=qBfrA
⇒qAf=qBfrB/rA and qBf=qAfrB/rA (1)

and then from conservation of charge:
qAi+qBi=qAf+qBf
⇒qtotal=(qBfrA/rB) + qBf → solve for qBf
and the same for qAf with substitution from (1).

Is that right?

Last edited: Sep 16, 2015