Two spheres with a conducting wire

AI Thread Summary
Two solid metal spheres, A and B, are connected by a conducting wire, leading to a shared potential after charge redistribution. The initial potentials are +2,346 volts for sphere A and +8,848 volts for sphere B. When connected, both spheres reach an equal final potential of 5,597 volts, allowing for the calculation of the new charge on sphere A using the formula q = v * r. The discussion emphasizes the importance of charge conservation and the relationship between charge, radius, and potential for conducting spheres. The solution process involves setting up equations based on these principles to find the final charges on both spheres.
Abdulayoub
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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 !
 
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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?
 
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