Suppose you charges parallel plate capacitor with a dielectric between the plates using a battery and the. Remove the bater, isolating the capacitor and leaving it charged. You remove the dielectric from between the plates. The potential difference will ?
By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:
[tex]C = \frac{Q}{V} (1)[/tex]
At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:
C = ε₀*A / d (2)
If the space between plates, is filled with a dielectric of dielectric constant κ, the above equation becomes:
[tex]C =\frac{\epsilon_{0}*\kappa*A}{d} (3)[/tex]
If the capacitor, once charged, is disconnected from the battery, charge must keep the same.
Now, if we remove the dielectric, as stated in (3) and (2), the capacitance C will decrease when removing the dielectric.
From (1) if C decreases, and Q remains constant, in order to keep both sides of the equation equal each other, V (the potential difference between plates), must increase.
