Правка разделов «Statement», «Solution», «Answer»

jzmicer правка от 27 июн. 2026 г., 19:30

правка #19307 предыдущая #19306 ← раньше позже →

en/6.6.22.md +33 −5

		@@ -1,11 +1,39 @@
			### 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]
		+	$$
		+	\boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}.
		+	$$

	@@ -1,11 +1,39 @@
	### Statement		### 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		### 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		#### Answer
			$$
	[Insert a concise answer or boxed result]		\boxed{h = \frac{(\varepsilon - 1) Q^2}{2\varepsilon_0 \varepsilon \rho g S^2}}.
			$$