Edits to “Statement”, “Solution”, “Answer”

jzmicer edited
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en/6.6.22.md +33 −5
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### Statement
+$6.6.22.$ One plate of an uncharged capacitor is made of a fine grid and lies on the surface of a liquid of density $\rho$ and dielectric permittivity $\varepsilon$. The area of each plate is $S$. To what height will the liquid level rise in the capacitor if a charge $Q$ is given to it?
−$6.6.22.$ [Insert the problem statement]
+![|736x304, 60%](../../img/6.6.22/6.6.22.png)
### Solution
+The capacitance of the capacitor (which can be viewed as two capacitors connected in series) is:
+$$
+C = \left( \frac{d-h}{\varepsilon_0 S} + \frac{h}{\varepsilon \varepsilon_0 S} \right)^{-1}
+ = \frac{\varepsilon_0 S}{d - h + h/\varepsilon}.
+$$
+The field energy at fixed charge is:
+$$
+W_c = \frac{Q^2}{2C} = \frac{Q^2}{2\varepsilon_0 S} \left( d - h + \frac{h}{\varepsilon} \right).
+$$
+The potential energy of the liquid is:
+$$
+W_p = \frac{h}{2} m g = \frac{1}{2} S h^2 \rho g.
+$$
+Equilibrium corresponds to a minimum of the total energy; we find this position using the derivative:
+$$
+\frac{d}{dh} \bigl[ W_p - W_c \bigr] = 0
+\quad\Rightarrow\quad
+S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) = 0.
+$$
+The second derivative:
+$$
+\frac{d}{dh} \left[ S \rho g h - \frac{Q^2}{2\varepsilon_0 S} \left( 1 - \frac{1}{\varepsilon} \right) \right] > 0
+\quad\Rightarrow\quad h > 0 \text{ – obviously.}
+$$
+Hence the only equilibrium height of rise is:
+$$
+h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}.
+$$
−1
−
#### Answer
−
−[Insert a concise answer or boxed result]
+$$
+\boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}.
+$$