$6.5.2$ Two electroconductor pistons of cross section $S$ form a plane capacitor in a tube of an isolating material which contains air at atmospheric pressure $P_0$. How changes the distance between pistons if charges of different signs $\pm Q$ are applied to them? The system is good heat conductor and there is no friction.
Let's consider the following figure
Applying Newton Second Law over one of the piston, considering the pistons are massless, $$F_p = F_{p0} + F_e$$ where $F_p$, $F_{p0}$ and $F_e$ are forces due to enclosed air between pistons, due to atmospheric pressure and due to electric field, respectively. $$P = P_0 + \frac{F_e}{S} \;(1)$$ The electric force $F_e$ is defined from the electric field generated by a charged plate that is $E = \frac{\sigma}{2\varepsilon_0}=\frac{Q}{2\varepsilon_0 S}$, so $$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$$ Putting $(2)$ into $(1)$ $$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$$ Taking account that the system is good heat conductor, we can suppose that thermal equilibrium is instantaneously established or that the container acts as a thermos, keeping the temperature unchanged.Then, applying Boyle-Mariotte law $$P_0 S L = P S l$$ $$\frac{L}{l} = \frac{P}{P_0} \;(4)$$ Substituting $(3)$ into $(4)$ $$\boxed{\frac{L}{l} = 1 + \frac{Q^2}{2P_0\varepsilon_0 S^2}}$$
BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba