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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-enter image description here

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?

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  • \$\begingroup\$ Are you assuming these devices are in outer space somewhere? Or, close to a big planet like the earth? \$\endgroup\$
    – user69795
    Mar 10, 2021 at 16:02
  • \$\begingroup\$ @user69795 for simplicity let's assume this system doesn't intract with anything else (Earth or any other conductor). \$\endgroup\$
    – user215805
    Mar 10, 2021 at 16:08
  • \$\begingroup\$ 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). \$\endgroup\$
    – K H
    Mar 11, 2021 at 4:53
  • \$\begingroup\$ 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. \$\endgroup\$
    – K H
    Mar 11, 2021 at 4:58

1 Answer 1

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The battery makes as much charge as it needs, using chemistry, to keep the plate potential the same: you may say “voltage is conserved”. In the capacitor situation charge is conserved, voltage is not.

So when the switch is closed, the charge stays the same, the voltage is less, in the capacitor case. In the battery case, more charge is made and the voltage stays the same.

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  • \$\begingroup\$ But how final distribution on spherical conductor looks like in both cases ? \$\endgroup\$
    – user215805
    Mar 10, 2021 at 16:48
  • \$\begingroup\$ Imagine the electric field in the final state. Most of the negative charge will be on the parallel plate. Very little, but some, of the negative charge will be on the sphere ( due to the fringing field that goes around the plate) . The distribution will be the same in both cases. There will be more charge, and a higher voltage between the positive and negative, in the battery case. To find an exact distribution of charge, you need to use em simulation software to find the exact electric field shape of the very funny shaped capacitor you have constructed with the plates and the sphere. \$\endgroup\$
    – user69795
    Mar 11, 2021 at 5:09
  • \$\begingroup\$ thanks for reply, can we charge a conductor using battery by connecting conductor to only one plate or this method is not feasible for charging ? \$\endgroup\$
    – user215805
    Mar 11, 2021 at 7:09
  • \$\begingroup\$ Make a capacitor with the sphere as one electrode . For example, make a 10pf capacitor. Connect a 1 volt battery to the capacitor. Now, remove the battery and the other electrode of the capacitor. You are left with a sphere with 10pico coulombs of charge on it. \$\endgroup\$
    – user69795
    Mar 13, 2021 at 20:37

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