| ### Statement | | ### Statement |
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| $11.2.8.$ 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 $\varphi$. | | $11.2.8.$ 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 $\varphi$. |
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| b*. What would be the charge on the plates of an additional capacitance capacitor $C_3$, turned on as shown in the figure? | | b*. What would be the charge on the plates of an additional capacitance capacitor $C_3$, turned on as shown in the figure? |
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| ### Solution | | ### Solution |
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| a) Well, everything is pretty trivial here, the voltage on each capacitor: | | a) Well, everything is pretty trivial here, the voltage on each capacitor: |
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| $$U = \frac{\varphi}{2}$$ | | $$U = \frac{\varphi}{2}$$ |
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| So there is a charge on each capacitor: | | So there is a charge on each capacitor: |
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| $$q_1 = C_1 \frac{\varphi}{2}$$ | | $$q_1 = C_1 \frac{\varphi}{2}$$ |
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| $$q_2 = C_2 \frac{\varphi}{2}$$ | | $$q_2 = C_2 \frac{\varphi}{2}$$ |
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| 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: |
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| 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). |
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| 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: |
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| $$U_1 + U_3 = \frac{\varphi}{2}$$ | | $$U_1 + U_3 = \frac{\varphi}{2}$$ |
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| $$U_3 - U_2 = - \frac{\varphi}{2}$$ | | $$U_3 - U_2 = - \frac{\varphi}{2}$$ |
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| From here: | | From here: |
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| $$U_1 = \frac{\varphi}{2} - U_3$$ | | $$U_1 = \frac{\varphi}{2} - U_3$$ |
| $$U_2 = U_3 + \frac{\varphi}{2}$$ | | $$U_2 = U_3 + \frac{\varphi}{2}$$ |
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| 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.: |
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| $$q_1 - q_2 - q_3 = 0$$ | | $$q_1 - q_2 - q_3 = 0$$ |
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| $$C_1 U_1 - C_2 U_2 = C_3 U_3$$ | | $$C_1 U_1 - C_2 U_2 = C_3 U_3$$ |
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| Let's substitute our relations for the stresses: | | Let's substitute our relations for the stresses: |
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| $$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$ | | $$C_1\frac{\varphi}{2} - C_1 U_3 - C_2 U_3 - C_2 \frac{\varphi}{2} = C_3 U_3$$ |
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| $$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$ | | $$U_3 = \frac{C_1 - C_2}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$ |
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| $$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$ | | $$q_3 = \frac{C_3(C_1 - C_2)}{C_1 + C_2 + C_3}\frac{\varphi}{2}$$ |
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| 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. |
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| #### Answer | | #### Answer |
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| а) | | а) |
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| $$q_1 = C_1 \frac{\varphi}{2}$$ | | $$q_1 = C_1 \frac{\varphi}{2}$$ |
