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en/8.4.11.md
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| + | ### Statement | ||
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| + | $8.4.11.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | During a short time $\Delta t$ the additional charge of $I\Delta t$ is delivered to the capacitor. Since the energy stored in the capacitor is $QV/2$, where $Q$ is the charge on the capacitor and $V=q/C$ is the constant voltage drop across the capacitor (and the generator), the change in the energy stored in the capacitor is $I\Delta tV/2$. Thus, the power absorbed by the capacitor is $IV/2=Iq/(2C)$, which is half of the power $IV=Iq/C$ consumed from the generator. The difference arises from the fact that we need to do negative work to bring the plates of the capacitor (which have a tendency to attract each other) closer to increasing its capacitance while keeping the voltage drop across it constant. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $8.4.11.$ [Insert the problem statement] | |||
| ### Solution | |||
| During a short time $\Delta t$ the additional charge of $I\Delta t$ is delivered to the capacitor. Since the energy stored in the capacitor is $QV/2$, where $Q$ is the charge on the capacitor and $V=q/C$ is the constant voltage drop across the capacitor (and the generator), the change in the energy stored in the capacitor is $I\Delta tV/2$. Thus, the power absorbed by the capacitor is $IV/2=Iq/(2C)$, which is half of the power $IV=Iq/C$ consumed from the generator. The difference arises from the fact that we need to do negative work to bring the plates of the capacitor (which have a tendency to attract each other) closer to increasing its capacitance while keeping the voltage drop across it constant. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||