# Finding the electric potential “on” a plate [closed]

So the actual question is this-

To two plates of parallel plate capacitors, charges $$Q_1$$ and $$Q_2$$ are given. The capacity of the capacitor is C. Then the switch is closed. [assume both $$Q_1$$, $$Q_2$$ to be +ve]

There were options to this problem but that's not the part of my question.

Once the switch is closed, the potential of the second plate will become zero but the trouble is how can I equate the potential to zero when there is no way of calculating the absolute potential on it. I want to calculate the final charges on both sides of both plates.

• Actually, to find final charges on both plate, absolute potentials are not required – maverick Jul 8 at 12:37
• So, how do I do that? – Robin Singh Jul 8 at 12:39
• Does this answer your question? Why are the two outer charge densities on a system of parallel charged plates identical? – Bhavay Jul 8 at 14:26
• This the case before earthing. – Bhavay Jul 8 at 14:32
• I can be wrong (It's better for u to wait for an answer) , but here is what I think : Since the outer side of plate 2 is earthed charge density on outer side of that plate becomes 0. Now again redistribution happen so that net electric field inside plates is 0. and this is only possible if no charge also reside on outer side of plate 1 too. Now on inner side of plate 1 and plate 2 magnitude of Q is same i.e |Q'| therefore when you draw field lines you will see you yourself field inside add up while field outside cancels up. – Bhavay Jul 8 at 14:51

I'll assume this is a standard capacitor with closely spaced plates. When you “ground” something, you are assuming that the ground is the position of zero potential. In this case, when you ground the right hand plate, charge will flow to or from the ground until the charge on the right hand plate is - $$Q_1$$. The charges will attract each other and reside in the “inside” surfaces of the plates. +$$Q_1$$ has no place to go and will not change. The potential of the grounded plate is zero; that of the left hand plate is $$Q_1$$/C.