# What will be the steady state Charge distribution of given configuration- battery vs capacitor

I was try to figure it out what should be the charge distribution over the spherical conductor of radius R (in figure below) in following two configurations- Assumptions

length of wire (l) connecting battery (capacitor) to spherical conductor is very very long , wire and battery are ideal and initially charge on spherical conductor is zero(before closing of switch)

1.in first case ,only one end of charged capacitor is connected to the spherical conductor

2.in second case, only one plate of battery is connected to spherical conductor

In first case(capacitor)

as soon as switch closes distribution of charges started until (sphere + plate of capacitor ) becomes equipotential and this distribution causes a New potential difference between the plates!

But when we apply same logic for second case (battery) - similar to above there would be a new potential difference between the plates but it contradict the fact that potential difference between the plates of an ideal battery is constant .

On the other hand if we keep voltage difference between the plates of battery constant then it implies that there would be no charge distribution even if we closed the switch , but isn't it again contradict the fact that conductor connected to same wire should be at constant potential (if current is Zero)?

Can anyone suggest how would distribution takes place in both cases at steady State?

• Are you assuming these devices are in outer space somewhere? Or, close to a big planet like the earth? Mar 10, 2021 at 16:02
• @user69795 for simplicity let's assume this system doesn't intract with anything else (Earth or any other conductor). Mar 10, 2021 at 16:08
• At the moment the switch is connected, a small current will flow in order to bring the now connected parts to the same potential, assuming they were not already at that potential to begin with. How the charge would be distributed is above my skill level, but two things I see are that if you disconnected the switch again after bringing the object up to potential, provided it was well enough insulated, it would store some amount of charge to stay at that potential because of it's parasitic capacitance(it's not a capacitor, so all of it's capacitance is parasitic).
– K H
Mar 11, 2021 at 4:53
• This may indicate negligible distribution of charge towards the surface of the object and in the direction of any connection back to the + of the source, something your example does not take into account.
– K H
Mar 11, 2021 at 4:58