# Two spheres with a conducting wire

• Abdulayoub
In summary, when the charged spheres are connected by a wire, the potential on each sphere is the same - so the charge on each sphere is 5597 volts.
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

Abdulayoub said:
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?

Abdulayoub said:

## 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: 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.

rude man said:
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: 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:

## 1. How do two spheres with a conducting wire work?

Two spheres with a conducting wire work by creating a closed circuit that allows for the flow of electricity between the two spheres. The conducting wire acts as a pathway for the electrons to travel from one sphere to the other, completing the circuit and allowing for the transfer of charge.

## 2. What is the purpose of the conducting wire in this setup?

The conducting wire serves as a pathway for the flow of electricity between the two spheres. It allows for the transfer of charge from one sphere to the other, completing the circuit and allowing for the movement of electrons.

## 3. How does the distance between the two spheres affect the flow of electricity?

The distance between the two spheres can affect the flow of electricity by creating a difference in potential between the two spheres. The closer the spheres are, the easier it is for electrons to flow between them, resulting in a stronger current. A larger distance may result in a weaker current.

## 4. Can the charge on the spheres be changed?

Yes, the charge on the spheres can be changed by altering the amount of charge on one of the spheres or by changing the distance between the two spheres. This can be done by connecting the spheres to a power source or by bringing them closer together or farther apart.

## 5. What happens if the conducting wire is broken?

If the conducting wire is broken, the circuit will be incomplete and the flow of electricity between the two spheres will cease. This can result in a loss of charge on one or both of the spheres, depending on the amount of time the circuit was broken and the resistance of the wire.

• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
221
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
3K
• Introductory Physics Homework Help
Replies
18
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K