# Inserting metal into parallel plate capacitor

The plates of an isolated parallel plate capacitor with a capacitance C carry a charge Q. The plate separation is d. Initially, the space between the plates contains only air. Then, an isolated metal sheet of thickness 0.5d is inserted between, but not touching, the plates. How does the potential difference between the plates change as a result of inserting the metal sheet?

The answer is: The potential difference will decrease. How do I get this result, it seems right but how do I reason it out?

$$V = Ed$$

There was mention about "E is the same outside the conductor. But the field inside is 0.". But in the formula, E is the field outside conductor right? If so it remains the same? Then $d = 0.5d$?

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If there is a charge $Q$ and $-Q$ on each plate of the capacitor , when you insert a perfect conductor between the plates (parallel), you simply will have a charge $+Q$ on one side(facing negative plate of capacitor) and $-Q$ on other side of the inserted plate. (Given that the area of plates is large enough to assume constant electric field between them)
Now , because the field between two (effectively) infinite plates doesn't depend on their distance, you will have the same field as before (without the new conductor).So , as you mentioned, the new potential difference is $E(d_0-d_{metal})$ .
Yes d=0.5d, Basically start thinking in terms of work per unit charge. So work you do on $1$ coulomb is $E$x$0.5d$. as field in conductor is $0$ in electrostatic condition , you don't have to do any work at all in that region . so $\Delta V$ is 0 in that region . Hence , $\Delta V_{net}= E_{outside}$x$0.5d$. = V for the capacitor . And calculate the new field in the two new air regions divided by the metal plate , you will find that by principle of superposition , it remains same . So it is because of this calculation that Net field also comes out to be same as before, that you're able to use E , basically think in terms of work and everything will be resolved . And then you will know , you will have to evaluate E everywhere again because now the metal plate has been polarised .