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I asked this question on electrical engineering stack exchange but I didn't get any reasonable answer ,so I think this question is better suited here.

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

enter image description here

Assumption- 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?

Here is link of my question -

https://electronics.stackexchange.com/questions/552300/what-will-be-the-steady-state-charge-distribution-of-given-configuration-batter

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If the battery and sphere system is isolated with the zero of potential put at infinity, when the connection is made, electrons will flow to the sphere, and the chemical action in the battery will move some electrons to the negative plate. There will now be more positive charge on the positive plate than negative charge on the negative plate but the potential difference will be the same as before the connection was made.

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  • $\begingroup$ @ R.W. Bird, +1 for answer !what I understand from your explanation is -1.charge Transfer takes place which causes imbalance of charge on plates 2.but potential difference between battery will be maintained 3. Sphere and negative plate of battery are at a same potential ,but here (below link) is an answer which says that if one end is not connected to anything charge Transfer will be zero ,can you please check this answer because it is somewhat different from your answer! physics.stackexchange.com/questions/313478/… $\endgroup$
    – user215805
    Mar 10 at 22:19
  • $\begingroup$ It could be argued that the charge on the + plate of the capacitor will attract the electrons and keep them from leaving the negative plate. In fact, if the plates are very large and close, there would be no field outside to move electrons toward the sphere. $\endgroup$
    – R.W. Bird
    Mar 11 at 19:16
  • $\begingroup$ But if charge don't move towards the sphere then isn't sphere and plate develop some potential difference between them , which intuitvily seems wrong since both connected by a wire? $\endgroup$
    – user215805
    Mar 11 at 19:56
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In both cases the excess of charges will spread to the conductor so that there is no E-field inside it.

In the case of the capacitor, as the number of negative charges at the right side doesn't change, the flow of the E-field also is the same. As the area around the charges increased, the E-Field just between the plates is smaller.

In the case of the battery, we can not say that the number of charges are the same. More negative charges are supplied to the sphere so that the E-field inside the battery keeps the same.

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  • $\begingroup$ So is it possible to charge a conductor using battery using same setup that i mentioned ? $\endgroup$
    – user215805
    Mar 11 at 6:29
  • $\begingroup$ Yes. Charges would spread to the conductor, otherwise an E-field would exist inside it. $\endgroup$ Mar 11 at 14:19

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