Question regarding parallel plate capacitor

1. Aug 17, 2011

Lets assume that there is a parallel plate capacitor connected across a supply of V volts. Then, there is charge redistribution between the plates such that finally the potential of positive plate wrt the negative plate is V volts. Let us assume air is the dielectric between the two plates. Now what happens if I introduce a plate in between the existing two plates. As per electrical analogy, its as good as two capacitors connected as a capacitor.

My question is that will such an arrangement of 3 plates with two plates connected across the supply and with a plate in between these two plates still behave as a capacitor. If so, please explain the physics behind it.

Thank you.

2. Aug 17, 2011

Staff: Mentor

Why not? It would be like two capacitors in series.

3. Aug 17, 2011

If so what is the charge on the middle plate assuming that the middle plate is equally spaced from the other two plates?

4. Aug 17, 2011

Staff: Mentor

The net charge would be zero of course. The surface charge on the middle plate will be the same as on the outer plates.

To see if the charge stored changes when the middle plate is introduced, calculate the new capacitance in terms of the original.

5. Aug 19, 2011

Ok. I think I got the logic here. The field lines of positive plate end on one side of the middle plate(the one facing the positive plate) and the field lines again start from other side of the middle plate and end on the negative plate of the original capacitor. Am I right?

6. Aug 19, 2011

Staff: Mentor

Sounds good to me.

7. Aug 19, 2011

aman kapoor

In a capacitor it is known that the two plates are oppositely charged , which is in turn due to attraction of negative charge or the electrons of the second plate by the the first plate which is charged with the help of battery and these both oppositely charged plates form a capacitor.
now when a third plate is inserted , the previously charged positive plate attracts the electron of the third plate due to which one side of the "inserted plate" becomes negatively charged and the other side of this plate becomes positively charged .so the left side separately serve as "-ve"counterpart for the previously the first plate and the right side separately serve as the "+ve" plate for the negative plate.So this forms two complete pairs of positively and negatively charged capacitor plates, hence giving rise to the two new capacitors.