Calculating the time to reach a potential difference

AI Thread Summary
The discussion revolves around calculating the time required to develop a potential difference of 100 kV between two conducting balls, considering the transfer of electrons at a rate of 10^13 per second. Initial calculations focused on determining the charge needed for the voltage and the charge transfer rate, but discrepancies arose when comparing results with the book. Participants highlighted the importance of accounting for the potential changes on both balls due to electron transfer and the influence of their mutual electric fields. It was noted that the potential difference should consider both spheres' contributions, as the charge distribution is affected by their proximity. The conversation emphasizes the complexity of the problem, suggesting that a simplified approach may not yield accurate results.
ShashVerma
Messages
3
Reaction score
0

Homework Statement


Q: Two conducting balls of radius 0.1 m are situated 3m apart in free space. Electrons are transferred from one ball to another at a rate of 10^13 per second. How long does it take for a p.d of 100 kv to develop?

Homework Equations


8757428f7aeb787795811caa40897baa.png


where ε0 is the electric constant (permittivity of vacuum)
[/B]

The Attempt at a Solution


First I calculated the charge needed for the particular voltage using the potential equation. Then I found C/s by multiplying the number of electrons by charge per electron. Finally I divided the charge needed by the charge per second to get the time to reach the charge. This answer does not match the book.

I think I need to take the increase in potential of one ball and the fall in potential of the other in account. But I have no idea how to do that?

Any help would be appreciated!
 
Last edited by a moderator:
Physics news on Phys.org
ShashVerma said:

Homework Statement


Q: Two conducting balls of radius 0.1 m are situated 3m apart in free space. Electrons are transferred from one ball to another at a rate of 10^13 per second. How long does it take for a p.d of 100 kv to develop?

Homework Equations


8757428f7aeb787795811caa40897baa.png


where ε0 is the electric constant (permittivity of vacuum)
[/B]

The Attempt at a Solution


First I calculated the charge needed for the particular voltage using the potential equation. Then I found C/s by multiplying the number of electrons by charge per electron. Finally I divided the charge needed by the charge per second to get the time to reach the charge. This answer does not match the book.

I think I need to take the increase in potential of one ball and the fall in potential of the other in account. But I have no idea how to do that?

Any help would be appreciated!
Welcome to the PF.

The problem implies that the electrons are transferred at a constant rate, so whatever energy is needed to overcome the growing electrastatic repulsion is apparently available, so I don't think you have to consider that.

Can you post your detailed calculations step-by-step, including units? That will help us to look for any mistakes or oversights. Thanks. :smile:
 
berkeman said:
Welcome to the PF.

The problem implies that the electrons are transferred at a constant rate, so whatever energy is needed to overcome the growing electrastatic repulsion is apparently available, so I don't think you have to consider that.

Can you post your detailed calculations step-by-step, including units? That will help us to look for any mistakes or oversights. Thanks. :smile:

10^13 * 1.602 * 10^-19 = 1.602 * 10^-6 C/s
100*10^3 V= Q/(4*pi*e0), Q = 3.34 * 10^-5 C
t = 3.34 * 10^-5/1.602 * 10^-6
= 20s

But the answer should be .36s
 
ShashVerma said:
100*10^3 V= Q/(4*pi*e0)
The units don't match here.
Always work with units, it helps to find most errors. You forgot one factor.

You also have to consider that this equation is the voltage relative to infinity, not relative to the other ball.
 
I think, as the OP surmised, the question is looking for the potential difference between the spheres, and as electrons are moved from one sphere they leave behind an equivalent net positive charge on the other.

That implies the two spheres will have equal and opposite potentials near their surfaces (assuming negligible induced charge interactions with the given separation between them).
 
mfb said:
The units don't match here.
Always work with units, it helps to find most errors. You forgot one factor.

You also have to consider that this equation is the voltage relative to infinity, not relative to the other ball.
Actually that was a typo.

It was supposed to be 100*10^3 V= Q/(4*pi*e0 * 3). That's what I had done, but I typed it wrong.
 
ShashVerma said:
Actually that was a typo.

It was supposed to be 100*10^3 V= Q/(4*pi*e0 * 3). That's what I had done, but I typed it wrong.
Several problems here.
You seem to have substituted the distance between the spheres, 3m, for r. r should be the radius of one sphere.
Secondly, this only gives you the potential at the surface one one sphere due to its own charge. The potential difference depends on both potentials.
Thirdly, the potential at one sphere is affected by the field of the other sphere.
Fourthly, the charge distribution at one sphere will be affected by the field from the other sphere. The charge distributions on each will not be uniform. However, my understanding is that taking that into account would make it rather an advanced problem. If you pretend that each charge distribution is uniform then the potential at one sphere due its own charge is ok, but if you try to take into account the potential there due to the other sphere it will no longer be a uniform potential. Maybe with a radius one thirtieth of the distance between them it is near enough uniform.
 
Back
Top