Asked by HappyCat:)
A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. What happens to the value of the energy stored in the capacitor?
It increases.
It decreases.
It stays the same.
It increases.
It decreases.
It stays the same.
Answers
Answered by
GPT-4o mini
When a dielectric is inserted between the plates of a parallel-plate capacitor that remains connected to a battery, the following happens:
- The capacitance of the capacitor increases due to the presence of the dielectric.
- Since the capacitor is connected to a battery, the voltage across the capacitor stays constant (equal to the battery voltage).
- The energy stored in the capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
Where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage.
As you insert the dielectric, the capacitance \( C \) increases, while \( V \) remains constant. Since the formula indicates that energy depends on capacitance when voltage is constant, and because the capacitance is increasing, the stored energy \( U \) also increases.
Therefore, the correct answer is: **It increases.**
- The capacitance of the capacitor increases due to the presence of the dielectric.
- Since the capacitor is connected to a battery, the voltage across the capacitor stays constant (equal to the battery voltage).
- The energy stored in the capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
Where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage.
As you insert the dielectric, the capacitance \( C \) increases, while \( V \) remains constant. Since the formula indicates that energy depends on capacitance when voltage is constant, and because the capacitance is increasing, the stored energy \( U \) also increases.
Therefore, the correct answer is: **It increases.**
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