A 2.00 -nF parallel-plate capacitor is charged to an initial...

A 2.00 -nF parallel-plate capacitor is charged to an initial potential difference $\Delta V_{i}=100 \mathrm{V}$ and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00 . (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference of the capacitor after the mica is withdrawn?

Key Concepts

Parallel?Plate Capacitor

A parallel?plate capacitor is an electrical component consisting of two conductive plates separated by a small distance. Its capacitance, which indicates the ability to store electric charge, depends on the area of the plates, the distance between them, and the properties of the medium between the plates.

Dielectric Materials

Dielectric materials are insulating substances inserted between the plates of a capacitor to increase its capacitance. This is due to the material's dielectric constant, a factor that quantifies the material’s ability to reduce the effective electric field within the capacitor, thereby enhancing its charge-storing capacity.

Energy Stored in a Capacitor

The energy stored in a capacitor is determined by the capacitor's charge and voltage, or equivalently by its capacitance and the potential difference across its plates. This energy can be calculated using formulas such as (1/2)CV² or (1/2)Q²/C, and changes in any of these parameters directly affect the stored energy.

Isolated Capacitor and Constant Charge Condition

When a capacitor is isolated, it is disconnected from any external circuit, meaning its total charge remains constant even if other parameters, like the capacitance, change. This principle is particularly important when the dielectric material is altered or removed, as the charge on the plates does not change, thereby affecting the voltage and stored energy.

Please give Ace some feedback