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<meta property="og:description" content="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.">
<title>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.</title>
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.
</p>
<center>
<figure>
<img src="statement.png"
loading="lazy" alt="6.5.2" width="200" />
<figcaption>
For problem 6.5.2
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
Let's consider the following figure
</p>
<center>
<figure>
<img src="draw.png"
loading="lazy" alt="6.5.2" width="450" />
<figcaption>
Forces over pistons
</figcaption>
</figure>
</center>
<p>
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
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master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
<meta property="og:title" content="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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<meta property="og:description" content="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.">
<title>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.</title>
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.
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.
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" alt="6.5.2" width="200" />
loading="lazy" alt="6.5.2" width="200" />
<figcaption>
<figcaption>
For problem 6.5.2
For problem 6.5.2
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Let's consider the following figure
Let's consider the following figure
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="draw.png"
<img src="draw.png"
loading="lazy" alt="6.5.2" width="450" />
loading="lazy" alt="6.5.2" width="450" />
<figcaption>
<figcaption>
Forces over pistons
Forces over pistons
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
Applying Newton Second Law over one of the piston, considering the pistons are massless,
Applying Newton Second Law over one of the piston, considering the pistons are massless,
$$F_p = F_{p0} + F_e$$
$$F_p = F_{p0} + F_e$$
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.
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.
$$P = P_0 + \frac{F_e}{S} \;(1)$$
$$P = P_0 + \frac{F_e}{S} \;(1)$$
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
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
$$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$$
$$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$$
Putting $(2)$ into $(1)$
Putting $(2)$ into $(1)$
$$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$$
$$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$$
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
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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
