Edits to “Statement”, “Answer”

naz edited Jun 24, 2026, 10:08 AM

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revision #19149 parent #19148 ← older newer →

en/11.2.8.md +14 −2

		@@ -1,7 +1,11 @@
			### Statement

		−	$11.2.8.$ [Insert the problem statement]
		+	$11.2.8.$ а. Two capacitors of capacity $C_1$ and $C_2$ are included in the circuit, which has the form of a circle and is located in a uniform magnetic field. The contour is connected along the diameter by a jumper wire $ab$. Determine the charge on the capacitor plates if the rate of change of the magnetic flux through the circuit is $\varphi$.

		+	b*. What would be the charge on the plates of an additional capacitance capacitor $C_3$, turned on as shown in the figure?
		+
		+	![К задаче $11.2.8$\|1415x731, 50%](../../img/11.2.8/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-23 170237.png)
		+
			### Solution

			a) Well, everything is pretty trivial here, the voltage on each capacitor:

			$$U = \frac{\varphi}{2}$$

			So there is a charge on each capacitor:

			$$q_1 = C_1 \frac{\varphi}{2}$$

			$$q_2 = C_2 \frac{\varphi}{2}$$

			b) Let's assume that the field induction is directed inside the drawing. Then the resulting current is shown in the figure:
			![To task $11.2.8$ \|1657x1223, 31%](../../img/11.2.8/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-23 163031.png)

			The figure also shows the charge signs on the capacitor plates and the direction of the field in them (over /against the current).

			Let's write down the sum of the voltage drops for the left and right semicircles:

			$$U_1 + U_3 = \frac{\varphi}{2}$$

			$$U_3 - U_2 = - \frac{\varphi}{2}$$

			From here:

			$$U_1 = \frac{\varphi}{2} - U_3$$
			$$U_2 = U_3 + \frac{\varphi}{2}$$

			Nodes a and b are neutral, so the algebraic sum of the charges on the plates connected to the nodes is zero.:

			$$q_1 - q_2 - q_3 = 0$$

			$$C_1 U_1 - C_2 U_2 = C_3 U_3$$

			Let's substitute our relations for the stresses:

			$$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$

			$$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$

			$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$

			Savchenko apparently chose the opposite direction of induction, so the answer turned out to be a minus.
		@@ -48,4 +52,12 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	а)
		+
		+	$$q_1 = C_1 \frac{\varphi}{2}$$
		+
		+	$$q_2 = C_2 \frac{\varphi}{2}$$
		+
		+	б)
		+
		+	$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$

	@@ -1,7 +1,11 @@
	### Statement		### Statement

	$11.2.8.$ [Insert the problem statement]		$11.2.8.$ а. Two capacitors of capacity $C_1$ and $C_2$ are included in the circuit, which has the form of a circle and is located in a uniform magnetic field. The contour is connected along the diameter by a jumper wire $ab$. Determine the charge on the capacitor plates if the rate of change of the magnetic flux through the circuit is $\varphi$.

			b*. What would be the charge on the plates of an additional capacitance capacitor $C_3$, turned on as shown in the figure?

			![К задаче $11.2.8$\|1415x731, 50%](../../img/11.2.8/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-23 170237.png)

	### Solution		### Solution

	a) Well, everything is pretty trivial here, the voltage on each capacitor:		a) Well, everything is pretty trivial here, the voltage on each capacitor:

	$$U = \frac{\varphi}{2}$$		$$U = \frac{\varphi}{2}$$

	So there is a charge on each capacitor:		So there is a charge on each capacitor:

	$$q_1 = C_1 \frac{\varphi}{2}$$		$$q_1 = C_1 \frac{\varphi}{2}$$

	$$q_2 = C_2 \frac{\varphi}{2}$$		$$q_2 = C_2 \frac{\varphi}{2}$$

	b) Let's assume that the field induction is directed inside the drawing. Then the resulting current is shown in the figure:		b) Let's assume that the field induction is directed inside the drawing. Then the resulting current is shown in the figure:
	![To task $11.2.8$ \|1657x1223, 31%](../../img/11.2.8/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-23 163031.png)		![To task $11.2.8$ \|1657x1223, 31%](../../img/11.2.8/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-23 163031.png)

	The figure also shows the charge signs on the capacitor plates and the direction of the field in them (over /against the current).		The figure also shows the charge signs on the capacitor plates and the direction of the field in them (over /against the current).

	Let's write down the sum of the voltage drops for the left and right semicircles:		Let's write down the sum of the voltage drops for the left and right semicircles:

	$$U_1 + U_3 = \frac{\varphi}{2}$$		$$U_1 + U_3 = \frac{\varphi}{2}$$

	$$U_3 - U_2 = - \frac{\varphi}{2}$$		$$U_3 - U_2 = - \frac{\varphi}{2}$$

	From here:		From here:

	$$U_1 = \frac{\varphi}{2} - U_3$$		$$U_1 = \frac{\varphi}{2} - U_3$$
	$$U_2 = U_3 + \frac{\varphi}{2}$$		$$U_2 = U_3 + \frac{\varphi}{2}$$

	Nodes a and b are neutral, so the algebraic sum of the charges on the plates connected to the nodes is zero.:		Nodes a and b are neutral, so the algebraic sum of the charges on the plates connected to the nodes is zero.:

	$$q_1 - q_2 - q_3 = 0$$		$$q_1 - q_2 - q_3 = 0$$

	$$C_1 U_1 - C_2 U_2 = C_3 U_3$$		$$C_1 U_1 - C_2 U_2 = C_3 U_3$$

	Let's substitute our relations for the stresses:		Let's substitute our relations for the stresses:

	$$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$		$$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$

	$$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$		$$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$

	$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$		$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$

	Savchenko apparently chose the opposite direction of induction, so the answer turned out to be a minus.		Savchenko apparently chose the opposite direction of induction, so the answer turned out to be a minus.
	@@ -48,4 +52,12 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		а)

			$$q_1 = C_1 \frac{\varphi}{2}$$

			$$q_2 = C_2 \frac{\varphi}{2}$$

			б)

			$$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$