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

jzmicer edited Jun 28, 2026, 6:29 PM

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revision #19347 parent #19346 ← older

en/6.6.26.md +54 −5

		@@ -1,11 +1,60 @@
			### Statement
		+	$6.6.26.$ A capacitor is filled with a dielectric and charged to a potential difference $V$. The plates are connected to each other for a very short time. When the potential difference has decreased by a factor of three, the plates are disconnected. After that, the potential difference slowly increases to $2/3$ of its initial value. How can this effect be explained? Find the dielectric permittivity of the substance filling the capacitor.

		−	$6.6.26.$ [Insert the problem statement]
		−
			### Solution

		−	1
		+	The operation of the capacitor is based on two effects. In an "empty" capacitor (without a dielectric), the field inside is created by the accumulation of charge on the plates. The capacitance of such a capacitor is
		+	$$
		+	C_1 = \frac{\varepsilon_0 S}{d} = C_0.
		+	$$
		+	If a dielectric is inserted between the plates, part of the field is due to its polarisation – the aligned orientation of elementary dipoles in the medium. The capacitance of such a capacitor is
		+	$$
		+	C_2 = \varepsilon C_0.
		+	$$
		+	The idea of the problem is that when the capacitor is short‑circuited, the free charges flow almost instantaneously, while the reorientation of the dipoles in the dielectric requires more time.

		−	#### Answer
		+	The initial field strength in the capacitor:
		+	$$
		+	E_0 = \frac{V}{d} = \frac{\sigma_0}{\varepsilon \varepsilon_0}.
		+	$$
		+	The initial free charge density on the plates:
		+	$$
		+	\sigma_0 = \varepsilon \varepsilon_0 E_0.
		+	$$
		+	The density of bound (polarisation) charge on the surface of the dielectric at the initial moment:
		+	$$
		+	\sigma_{p} = P = (\varepsilon - 1)\varepsilon_0 E_0.
		+	$$

		−	[In~~sert a concise answer or boxed result]~~
		+	When short‑circuited, the free charges redistribute almost instantaneously, while the bound charges do not have time to change. Let the free charge density after this be $\sigma_1$. According to the condition, the potential difference has decreased by a factor of three, so the field strength becomes
		+	$$
		+	E_1 = \frac{V/3}{d} = \frac{E_0}{3}.
		+	$$
		+	The field inside the dielectric is determined as the sum of the fields of the free and bound charges. Remembering that these fields are oppositely directed:
		+	$$
		+	E_1 = \frac{\sigma_1 - \sigma_{p}}{\varepsilon_0}.
		+	$$
		+	From this, the free charge remaining on the plates is
		+	$$
		+	\sigma_1 = \varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right).
		+	$$
		+
		+	After the plates are disconnected, the free charge $\sigma_1$ remains constant. However, the dipoles of the dielectric gradually begin to disorient, the bound charge $\sigma_{p}$ decreases, and the field inside the capacitor increases. According to the condition, when the polarisation has fully adjusted to the new field, the potential difference reaches $2V/3$. Then the final field strength is
		+	$$
		+	E_2 = \frac{2E_0}{3}.
		+	$$
		+	In the equilibrium state, the following holds:
		+	$$
		+	\sigma_1 = \varepsilon \varepsilon_0 E_2 = \varepsilon \varepsilon_0 \frac{2E_0}{3}.
		+	$$
		+
		+	Equating the two expressions for $\sigma_1$:
		+	$$
		+	\varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right) = \varepsilon \varepsilon_0 \frac{2E_0}{3},
		+	$$
		+	$$
		+	\varepsilon - \frac{2}{3} = \frac{2}{3}\varepsilon \quad\Rightarrow\quad \varepsilon = 2.
		+	$$
		+
		+	#### Answer:
		+	The dipole moments in the dielectric align in the electric field with a delay; $\varepsilon = 2$.

	@@ -1,11 +1,60 @@
	### Statement		### Statement
			$6.6.26.$ A capacitor is filled with a dielectric and charged to a potential difference $V$. The plates are connected to each other for a very short time. When the potential difference has decreased by a factor of three, the plates are disconnected. After that, the potential difference slowly increases to $2/3$ of its initial value. How can this effect be explained? Find the dielectric permittivity of the substance filling the capacitor.

	$6.6.26.$ [Insert the problem statement]

	### Solution		### Solution

	1		The operation of the capacitor is based on two effects. In an "empty" capacitor (without a dielectric), the field inside is created by the accumulation of charge on the plates. The capacitance of such a capacitor is
			$$
			C_1 = \frac{\varepsilon_0 S}{d} = C_0.
			$$
			If a dielectric is inserted between the plates, part of the field is due to its polarisation – the aligned orientation of elementary dipoles in the medium. The capacitance of such a capacitor is
			$$
			C_2 = \varepsilon C_0.
			$$
			The idea of the problem is that when the capacitor is short‑circuited, the free charges flow almost instantaneously, while the reorientation of the dipoles in the dielectric requires more time.

	#### Answer		The initial field strength in the capacitor:
			$$
			E_0 = \frac{V}{d} = \frac{\sigma_0}{\varepsilon \varepsilon_0}.
			$$
			The initial free charge density on the plates:
			$$
			\sigma_0 = \varepsilon \varepsilon_0 E_0.
			$$
			The density of bound (polarisation) charge on the surface of the dielectric at the initial moment:
			$$
			\sigma_{p} = P = (\varepsilon - 1)\varepsilon_0 E_0.
			$$

	[In~~sert a concise answer or boxed result]~~		When short‑circuited, the free charges redistribute almost instantaneously, while the bound charges do not have time to change. Let the free charge density after this be $\sigma_1$. According to the condition, the potential difference has decreased by a factor of three, so the field strength becomes
			$$
			E_1 = \frac{V/3}{d} = \frac{E_0}{3}.
			$$
			The field inside the dielectric is determined as the sum of the fields of the free and bound charges. Remembering that these fields are oppositely directed:
			$$
			E_1 = \frac{\sigma_1 - \sigma_{p}}{\varepsilon_0}.
			$$
			From this, the free charge remaining on the plates is
			$$
			\sigma_1 = \varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right).
			$$

			After the plates are disconnected, the free charge $\sigma_1$ remains constant. However, the dipoles of the dielectric gradually begin to disorient, the bound charge $\sigma_{p}$ decreases, and the field inside the capacitor increases. According to the condition, when the polarisation has fully adjusted to the new field, the potential difference reaches $2V/3$. Then the final field strength is
			$$
			E_2 = \frac{2E_0}{3}.
			$$
			In the equilibrium state, the following holds:
			$$
			\sigma_1 = \varepsilon \varepsilon_0 E_2 = \varepsilon \varepsilon_0 \frac{2E_0}{3}.
			$$

			Equating the two expressions for $\sigma_1$:
			$$
			\varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right) = \varepsilon \varepsilon_0 \frac{2E_0}{3},
			$$
			$$
			\varepsilon - \frac{2}{3} = \frac{2}{3}\varepsilon \quad\Rightarrow\quad \varepsilon = 2.
			$$

			#### Answer:
			The dipole moments in the dielectric align in the electric field with a delay; $\varepsilon = 2$.