After disconnecting a charged capacitor from a battery, a dielectric material is inserted between the plates. What happens to the energy stored in the capacitor? It decreases. It increases. It stays the same.
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Step 1: The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{1}{2}\frac{Q^2}{C}$, where $C$ is the capacitance, $V$ is the voltage across the capacitor, and $Q$ is the charge on the capacitor. Show more…
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A capacitor is connected to a battery. After being fully charged, the battery is disconnected, and a dielectric is inserted between the plates. How does the charge and the voltage change after the dielectric is inserted? (5 points) A) The charge decreases and the voltage decreases. B) The charge decreases and the voltage increases. C) The charge and the voltage stay the same. D) The charge stays the same and the voltage increases. E) The charge stays the same and the voltage decreases.
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A battery charges an ideal parallel plate capacitor that has nothing between its plates. The battery is then removed so that charges may no longer enter or exit the capacitor plates. As a dielectric material is now inserted between the plates, the voltage across the capacitor ______. a. increases b. remains the same c. decreases A battery remains attached to an ideal parallel plate capacitor while a dielectric material is inserted between its plates. During this process, the charge on the positive capacitor plate ______. a. remains the same b. decreases c. increases
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