A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm^

step 1 We have charge Q equals C delta V. C is epsilon A over D times delta V. Putting the numbers in,

A parallel-plate capacitor in air has a plate separation of...

A parallel-plate capacitor in air has a plate separation of 1.50 $\mathrm{cm}$ and a plate area of $25.0 \mathrm{cm}^{2} .$ The plates are charged to a potential difference of 250 $\mathrm{V}$ and disconnected from the source. The capacitor is then immersed in distilled water. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor. Assume the liquid is an insulator.

Key Concepts

Energy Stored in a Capacitor

The energy stored in a capacitor can be expressed as U = ½ CV² or alternatively as U = Q²/(2C). Changes in capacitance and voltage, such as those resulting from the insertion of a dielectric, lead to a corresponding change in the energy stored. Understanding this energy transformation is crucial for analyzing how dielectric insertion impacts the system energetically.

Charge Conservation

In an isolated system where the capacitor is disconnected from any voltage source, the total charge on the plates remains constant. This principle is essential when analyzing the effect of inserting a dielectric material because, despite changes in capacitance and voltage, the stored charge does not change, which in turn affects the energy stored in the capacitor.

Dielectric Materials

A dielectric is an insulating substance that, when inserted between the plates of a capacitor, increases its capacitance by reducing the effective electric field. This is quantified by the dielectric constant (k), which multiplies the original capacitance. However, if the capacitor is disconnected from a voltage source, the charge remains constant even though the capacitance increases and the voltage decreases.

Capacitance Calculation

Capacitance is a measure of a capacitor’s ability to store charge per unit potential difference. For a parallel-plate capacitor, it is calculated using the formula C = ??A/d for vacuum or air, where ?? is the permittivity of free space, A is the area of the plates, and d is the separation distance. This formula highlights the direct proportionality of capacitance with plate area and the inverse relationship with separation distance.

Parallel-Plate Capacitor

This concept refers to a type of capacitor made of two conductive plates separated by a small distance. The geometry determines its capacitance, where the area of the plates and the distance between them play crucial roles. The electric field is considered uniform between the plates, and this configuration is widely used to illustrate basic capacitor principles in electrostatics.

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