# Will there any modification happen?

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?

• 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 Sep 25 '17 at 19:33
• 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. – thephysicist Sep 25 '17 at 19:37
• @JMac Pls explain what's wrong with my statement? – thephysicist Sep 26 '17 at 5:11