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?

Question

It increases.
It decreases.
It stays the same.

Answer 1

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.