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Lets Suppose there is Conducting plate X and charge Q is given to which means +Q/2 and +Q/2 on each side . Now lets suppose plate Y (uncharged at first) (exactly similar to plate X (area and material wise) )( is brought near to the plate X . Off course redistribution will happen. Q/2 on both sides of plate X and -Q/2 on the inner side and Q/2 on outer side of plate Y such that E is 0 inside the conductor As Explanation provided in the answer for this question

Now my question is suppose Plate Y's outer side is earthed which means -Q/2 charge will flow from earth to plate y thus 0 charge on outside surface of plate Y. Now the system will behave as capacitor (Parallel plate capacitor ).
But the problem is in the redistribution .The whole charge on plate X's outer side Q/2 will flow inside thus making redistribution something like this +Q on the inner side of Plate x and -Q on the inner side of Plate Y. But now if i take an arbitrary point inside Plate X I'm finding E = 0 inside that conductor(plate X){as Electric field direction will be only from plate x to plate y} but not inside an arbitrary point inside Plate Y?Can anyone explain how will redistribution takes place in an earthed plate

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  • $\begingroup$ Your description is confusing to me. In your first sentence you didn't give a negative sign to either $\frac{Q}{2}$ so I assume the plus and minus charges on plate X are uniformly distributed. The second sentence says plate Y (which is exactly the same) will of course redistribute the charge. Why? Is plate Y charged + or minus? Please clear this up or I will vote to close based on being unclear. $\endgroup$ – Bob D Jun 28 '19 at 17:28
  • $\begingroup$ Plate Y is uncharged at first but to make E = 0 inside conductor redistribution will happen (explained in the answer given in the link) $\endgroup$ – user232991 Jun 28 '19 at 17:32
  • $\begingroup$ Is your plate X the same as the left most plate $Q_1$ in the link, i.e. a plate with a net positive charge on it? Because you didn't describe it that way. Then is your Y plate one of those in the middle, i.e., one with no net charge on it (electrons evenly distributed on the plate) before bringing it close to plate X where plate X's field will influence Y? $\endgroup$ – Bob D Jun 28 '19 at 18:49
  • $\begingroup$ @BobD Yes i just assumed 2 plate system . Plate X is on the left and with Uniform Charge distribution of Q/2A on both Sides. Plate Y is initially uncharged and brought in contact(to the right of Plate X) with Plate X. Now redistribution will happen as explained so that E = 0 inside both conductor $\endgroup$ – user232991 Jun 29 '19 at 2:21
  • $\begingroup$ Ok but You should revise your question to say plate X has $+\frac{Q}{2}$ charge on each side of the plate. $\endgroup$ – Bob D Jun 29 '19 at 2:46
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Your final charge distribution on both plates( Q on inner surface of plate X and -Q on inner surface of plate Y) is correct. The electric field inside both the plate is also zero. Since you you assumed it to be a parallel plate capacitor that means you are treating the plates as infinite surface of charge. In this case the electric field at any point due to a single plate will be constant and proportional to charge per unit area on it. thus electric field would be zero everywhere except for the points between the plates.

Electric field lines will be from X to inner surface of Y but will end at inner surface of Y

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  • $\begingroup$ But if we take an arbitrary point inside Plate Y how will it be 0? Isn't e inside Y conductor = σ/2ε + σ/2ε = σ/ε ? $\endgroup$ – user232991 Jun 29 '19 at 2:28
  • $\begingroup$ It will be sigma/2epsilon - sigma/2epsilon = 0. You are making a mistake while assigning direction of electric field due to individual plates. $\endgroup$ – Param_1729 Jun 29 '19 at 3:06

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