Physics Solution: Equilibrium of Pistons with Ideal Gas and Electric Field
Physics Solution: Stability and Dynamics of Pistons
Problem Analysis
We have an ideal gas with relative permittivity $\varepsilon_r$ trapped between two metallic pistons A and B in a tube. The pistons act as a parallel plate capacitor connected to a voltage source $V$.
Forces acting on the pistons:
- Gas Pressure Force ($F_{gas}$): Pushes the pistons outward (increasing separation $x$). For an ideal gas undergoing isothermal changes (assuming slow process), $PV = \text{constant}$. Since Area $A$ is constant, pressure $P \propto 1/x$. Thus, $F_{gas} \propto 1/x$.
Let $F_{gas} = \frac{C_1}{x}$ where $C_1$ is a constant.
- Electrostatic Force ($F_{elec}$): The capacitor plates attract each other, pulling the pistons inward (decreasing $x$).
The force between capacitor plates is given by $F_{elec} = \frac{1}{2} \frac{Q^2}{\varepsilon A} = \frac{1}{2} \varepsilon A E^2$.
With voltage $V$, $E = V/x$.
So, $F_{elec} = \frac{1}{2} (\varepsilon_0 \varepsilon_r) A \left(\frac{V}{x}\right)^2$.
Thus, $F_{elec} \propto 1/x^2$.
Let $F_{elec} = \frac{C_2 V^2}{x^2}$ where $C_2 = \frac{1}{2} \varepsilon_0 \varepsilon_r A$.
Equilibrium Condition: At separation $x_0$ and voltage $V_0$, the forces balance:
$$ F_{gas} = F_{elec} \implies \frac{C_1}{x_0} = \frac{C_2 V_0^2}{x_0^2} $$
Part (a): Stability Analysis
To check stability, we analyze the net outward force $F_{net}$ as a function of a small displacement from $x_0$.
$$ F_{net}(x) = F_{gas} – F_{elec} = \frac{C_1}{x} – \frac{C_2 V_0^2}{x^2} $$
At equilibrium ($x=x_0$), $F_{net}(x_0) = 0$. This implies $C_1 = \frac{C_2 V_0^2}{x_0}$.
Now, consider a small perturbation where $x$ increases ($x > x_0$):
- The gas force (pushing out) decreases as $1/x$.
- The electric force (pulling in) decreases as $1/x^2$.
Since $1/x^2$ drops faster than $1/x$, for $x > x_0$, the attractive electric force becomes smaller than the repulsive gas force.
$$ \frac{1}{x} > \frac{1}{x^2} \text{ (scaling relative to equilibrium)} $$
Therefore, $F_{net} > 0$ (positive outward). The net force pushes the pistons further apart.
Conversely, if $x$ decreases ($x < x_0$):
- The electric force (pulling in) increases faster ($1/x^2$) than the gas force ($1/x$).
- Therefore, $F_{net} < 0$ (net inward force). The pistons are pulled further together.
Since a displacement away from equilibrium results in a force that pushes the system further away from equilibrium, the system is unstable.
Answer (a): The equilibrium is Unstable.
Part (b): Response to Slowly Increasing Voltage
We start at the equilibrium separation $x_0$ with voltage $V_0$. The voltage $V$ is then slowly increased ($V > V_0$).
Looking at the force balance equation:
$$ F_{net} = F_{gas} – F_{elec} = \frac{C_1}{x} – \frac{C_2 V^2}{x^2} $$
At the initial state, these terms were equal.
If $V$ increases, the attractive electric force term $\frac{C_2 V^2}{x^2}$ increases.
Consequently, $F_{net}$ becomes negative (inward).
$$ F_{net} < 0 $$
This net inward force causes the pistons to move towards each other, decreasing $x$.
As $x$ decreases:
- The gas pressure increases as $1/x$.
- The electric force increases as $1/x^2$.
The electric force grows faster than the gas pressure. The dominance of the inward electric force increases as $x$ gets smaller. There is no new equilibrium point established where forces would balance again (until the pistons crash).
Therefore, the pistons will collapse together.
Answer (b): The distance x decreases (ultimately tending to zero/collapse).
(a) Unstable
(b) $x \to 0$ (Distance decreases)
