Updated greek laters @ latex compiling

astrosander правка от
правка #10188 предыдущая #9112 GitHub dbb31ba ← раньше
@@ -80,14 +80,14 @@
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.
− $$P = P_0 + \frac{F_e}{S} \;(1)$$
+ $$P = P_0 + \frac{F_e}{S} \quad(1)$$
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
− $$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$$
+ $$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \quad(2)$$
Putting $(2)$ into $(1)$
− $$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$$
+ $$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \quad(3)$$
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
$$P_0 S L = P S l$$
− $$\frac{L}{l} = \frac{P}{P_0} \;(4)$$
+ $$\frac{L}{l} = \frac{P}{P_0} \quad(4)$$
Substituting $(3)$ into $(4)$
$$\boxed{\frac{L}{l} = 1 + \frac{Q^2}{2P_0\varepsilon_0 S^2}}$$
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