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Initially capacitor 2 is uncharged . Later it is connected with pre-charged-capacitor 1 in an open circuit . So,yet in this arrangement capacitor 1 cannot charge capacitor 2 .

My question is- as no current is flowing , the top plate & bottom plate of capacitor 2 & bottom plate of capacitor 1 are in the same potential.Always true for this arrangement regardless how many charges there are on capacitor 1.So, why does the potential of bottom plate of cap. 2 & bottom plate of cap. 1 become same magnitude always?Will there any modification happen on the plates at the moment of connecting these 2 capacitors in an open circuit?

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  • $\begingroup$ What makes you think " the top plate & bottom plate of capacitor 2 & bottom plate of capacitor 1 are in the same potential"? Some of that is true; but not all of it. $\endgroup$ – JMac Sep 25 '17 at 19:33
  • $\begingroup$ Capacitor 2 is uncharged ,so the PD between the plate is 0. and if there is a PD between the bottom plate of cap 1 & cap 2 ,then current will flow.-thats my thought. $\endgroup$ – thephysicist Sep 25 '17 at 19:37
  • $\begingroup$ @JMac Pls explain what's wrong with my statement? $\endgroup$ – thephysicist Sep 26 '17 at 5:11
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Potential is always measured relative to something. When you consider a capacitor in isolation (not connected to a circuit), then the difference between the two plates gives an obvious reference ("the other plate"). When you connect the two plates in a circuit, they now share a common reference (the wire that connects them) and it makes sense to express all potentials relative to that point (which you might ground).

The potential difference across each of the two capacitors was not affected by connecting just one side; only if you close the switch (so current will flow to equalize the potential difference across the two capacitors) will you affect a change.

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  • $\begingroup$ then,what is the PD between the bottom plate of cap 1 & cap 2,when switch is open? $\endgroup$ – thephysicist Sep 25 '17 at 19:45
  • $\begingroup$ The potential difference is zero when there is a conductor between them. Before there is a conductor, the value is undefined. In principle, given that the capacitor has wires sticking out, it is not a perfect capacitor (there is an external electric field) - and as such it would have some small potential. And as you move towards another conductor, the field would induce a charge, which would cause attraction... but all that is negligible compared to the fields / charges in the "working part" of the capacitor. $\endgroup$ – Floris Sep 25 '17 at 19:49
  • $\begingroup$ What makes you think " the top plate & bottom plate of capacitor 2 & bottom plate of capacitor 1 are in the same potential"? Some of that is true; but not all of it. – JMac . What's wrong with my statement? $\endgroup$ – thephysicist Sep 25 '17 at 19:54
  • $\begingroup$ I don't know why JMac said that, sorry. $\endgroup$ – Floris Sep 25 '17 at 21:28

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