A parallel-plate vacuum capacitor has 8.38 J of energy...

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 $\mathrm{mm}$ . If the separation is decreased to $1.15 \mathrm{mm},$ what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conducting plates separated by an insulating medium (or vacuum). Its capacitance is determined by the area of the plates, the separation between them, and the permittivity of the dielectric material between the plates. This type of capacitor is a fundamental example in electromagnetism for studying how geometric factors influence the storage of electric energy.

Capacitance and Plate Separation

The capacitance of a parallel-plate capacitor is inversely proportional to the distance between its plates, as described by the formula C = ??A/d. This means that if the separation (d) is decreased, the capacitance increases. This relationship is key to understanding how changes in a capacitor's geometry affect its ability to store charge and energy.

Energy Storage in Capacitors

The energy stored in a capacitor can be expressed in terms of either the charge and capacitance (U = Q²/2C) or the capacitance and the voltage (U = ½ CV²). The choice of formula depends on whether the capacitor is isolated (keeping the charge constant) or connected to a voltage source (maintaining constant voltage). This distinction is crucial when analyzing how mechanical changes (like varying plate separation) affect the stored energy.

Constant Charge vs Constant Voltage Conditions

When a capacitor is disconnected from a voltage source (isolated), the charge on the plates remains constant even as the geometry—and therefore the capacitance—changes. In this case, a decrease in the separation (and corresponding increase in capacitance) reduces the stored energy, following U = Q²/(2C). Conversely, if the capacitor remains connected to a constant voltage source, the voltage is kept fixed, and an increased capacitance leads to a higher stored energy, according to U = ½ CV². Understanding these two scenarios is essential for predicting how energy storage is influenced by physical modifications to the capacitor.