A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 100 cm^2, and a mica die

step 1 For the given problem, we have a parallel plate capacitor. Let's say this is a parallel plate ca

A parallel-plate capacitor has a capacitance of 100 pF, a...

A parallel-plate capacitor has a capacitance of $100 \mathrm{pF},$ a plate area of $100 \mathrm{cm}^{2},$ and a mica dielectric $(\kappa=5.4)$ completely filling the space between the plates. At 50 $\mathrm{V}$ potential difference, calculate (a) the electric field magnitude $E$ in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.

A parallel-plate capacitor has a capacitance of $100 \mathrm{pF},$ a plate area of $100 \mathrm{cm}^{2},$ and a mica dielectric $(\kappa=5.4)$
completely filling the space between the plates. At 50 $\mathrm{V}$ potential
difference, calculate (a) the electric field magnitude $E$ in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.

Key Concepts

Capacitance of a Parallel-Plate Capacitor

This concept revolves around understanding how the geometry of a capacitor (i.e., the plate area and separation) and the dielectric material between the plates determine its capacitance. The capacitance quantifies the ability of the capacitor to store electrical charge for a given potential difference, and with a dielectric, its value is enhanced by the dielectric constant (?).

Electric Field Due to a Potential Difference

This concept involves the relationship between the electric field and the potential difference (voltage) across a given distance in a capacitor. In the context of a parallel-plate capacitor, the electric field in the dielectric is essentially uniform and can be calculated by dividing the voltage by the distance between the plates.

Free Charge on the Capacitor Plates

This idea connects the stored energy and charge in a capacitor with its capacitance and the applied voltage. The free charge on the plates is determined by the relation Q = CV, where C is the capacitance and V is the potential difference, representing the charge that is directly responsible for the established electric field.

Dielectric Polarization and Induced Surface Charge

This concept explains how a dielectric material placed between the capacitor plates becomes polarized in the presence of an electric field. The polarization leads to the development of bound charges at the surfaces of the dielectric, often referred to as induced or polarization charges, which modify the effective field within the capacitor without contributing to the free charge stored on the plates.

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