Question

Consider a capacitor that consists of two parallel plates, one of which can move relative to the other. Air fills the space between the plates, the capacitance is 45.2 pF when the separation between plates is d = 0.360 cm. A battery with potential difference V=12.0 V is connected to the plates to fully charge them and then it is disconnected from the plates. The plates are moved to a separation of 0.180 cm. What is the potential difference V' between the plates now

Answer 1

\[C=\frac{Q_\text{free}}{V}\]
\[E=\frac{\sigma_\text{free}}{\varepsilon_0\kappa}\]
\[V=Ed\]
now since the battery is disconnected the charge is fixed
\(Q_\text{free},\sigma_\text{free},E\) are all constant