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.
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<h3> Statement </h3>
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$6.5.2$
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.
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For problem 6.5.2
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<h3>Solution</h3>
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Let's consider the following figure
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Forces over pistons
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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 $(4)$
$$\boxed{\frac{L}{l} = 1 + \frac{Q^2}{2P_0\varepsilon_0 S^2}}$$
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<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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