What is the charge ratio between two equipotential spheres?

AI Thread Summary
The discussion revolves around calculating the charge ratio between two equipotential spheres, where the first sphere is charged to 20kV and the second sphere's potential drops to 12kV upon connection. The key equations involve the relationship between voltage, charge, and radius, specifically V=kq/r. Participants emphasize the importance of charge conservation during the transfer process and clarify that the second sphere should be considered initially uncharged. The conversation highlights the need to derive the ratio of charges by simplifying the equations while ensuring that both spheres reach the same potential of 12kV after connection. The problem-solving approach focuses on understanding the implications of charge conservation and the relationship between the spheres' radii and charges.
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Homework Statement


A spherical potential conductor of radius R1 is charged to 20kV. When it is connected by a long, fine wire to a second sphere faraway. its potential drops to 12kV. What is the radius of the second sphere.


Homework Equations



V=kq/r


The Attempt at a Solution



So I know that R2= R1* (q2/q1). Thus, I need to find the ratio q2/q1. I am kinda lost in what to use to figure out q2/q1 ratio.

k*q1/R1 = 2x10^4 and k*q2/R2= 4x10^3 (because 2 spheres are equipotential at 12kV and sphere 1 drops 8kV so sphere 2 has to gain 8kV so initially sphere 2 is at 4kV)

k*q1/R1 = k*q2/R2 = 12x10^3.

With these 3 equations, I am still unable to figure out what is q2/q1. Can you guys give me some hints and suggestions ?
 
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nns91 said:
[ (because 2 spheres are equipotential at 12kV and sphere 1 drops 8kV so sphere 2 has to gain 8kV so initially sphere 2 is at 4kV)

Sphere 2 only will gain the same amount of voltage if it has the same radius.
I think you are supposed to assume sphere 2 is initially uncharged. If it has an initial charge, you can't do the problem. Sphere 2 could be gigantic, with a potential slightly below 12 kv, or tiny with a very large negative potential.

the spheres act as capacitors with capacity r/k
 
There is no need to consider them as capacitors.
Your equations are almost there, except for the part on sphere 2 being at 4kV initially of course, but do take care of the variables (why do i see q1 appear twice in different situations?)
Hint: Charge is conserved throughout the transfer process
That should allow you to simplify the expressions.
 
So if I assume that sphere 2 is uncharged initially, then I should make its potential equal to 12kV right ? since 2 spheres are equipotential.

Oh, if charged is conserve then kq1/R1= 20kV = kq1/R1= 12kV ?
 
nns91 said:
So if I assume that sphere 2 is uncharged initially, then I should make its potential equal to 12kV right ? since 2 spheres are equipotential.
Yes. Sphere 1 is at 20kV and Sphere 2 at 0V initially, with both becoming 12kV after charge flows between the two spheres

nns91 said:
Oh, if charged is conserve then kq1/R1= 20kV = kq1/R1= 12kV ?
Er..no...that equation clearly doesn't make sense - 20 = 12??
Charge conservation in the sense that if Sphere 1 possesses Q coulombs of charge initially, and q2 coulombs of charge flow from Sphere 1 to 2, leaving sphere 1 with q1 coulombs remaining, then clearly q1 + q2 = Q.
Then your originally formulated equations can be written as:
k*(q1+q2)/R1 = 2x10^4
k*q1/R1 = k*q2/R2 = 12x10^3
I guess you can carry on?
 
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