A parallel-plate capacitor C is charged up to a potential V0...

A parallel-plate capacitor $C$ is charged up to a potential $V_{0}$ with a charge of magnitude $Q_{0}$ on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation. (a) What is the new capacitance in terms of $C ?$ (b) How much charge is now on the plates in terms of $Q_{0} ?(\mathrm{c})$ What is the potential difference across the plates in terms of $V_{0} ?$

Key Concepts

Relationship Between Voltage, Charge, and Capacitance

The fundamental connection between voltage, charge, and capacitance is given by the equation V = Q/C. This equation implies that for a constant charge, any change in capacitance will inversely affect the voltage across the capacitor's plates. This relationship is essential for understanding how geometric or dielectric changes modify the operating characteristics of a capacitor.

Capacitance of a Parallel-Plate Capacitor

The capacitance of a parallel-plate capacitor is determined by its geometry, specifically being directly proportional to the plate area and inversely proportional to the separation between the plates. This relation is expressed in the formula C = ?A/d, where ? is the permittivity of the medium between the plates, A is the area, and d is the distance between them.

Conservation of Charge in an Isolated Capacitor

When a capacitor is disconnected from its charging source (making it isolated), the total charge stored on its plates remains constant. This conservation of charge is crucial in analyzing circuits where physical parameters such as plate separation are altered, as it affects the subsequent voltage and energy stored in the capacitor.

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