In Fig. 25-59 , two parallel-plate capacitors A and B are...

In Fig. $25-59$ , two parallel-plate capacitors $A$ and $B$ are connected in parallel across a 600 $\mathrm{V}$ battery. Each plate has area 80.0 $\mathrm{cm}^{2}$ ; the plate separations are 3.00 $\mathrm{mm}$ ; Capacitor $A$ is filled with air; capacitor $B$ is filled with a dielectric of dielectric constant $\kappa=2.60$ . Find the magnitude of the electric field within (a) the dielectric of capacitor $B$ and (b) the air of capacitor $A .$ What are the free charge densities $\sigma$ on the higher-potential plate of (c) capacitor $A$ and (d) capacitor $B ?$ (e) What is the induced charge density $\sigma^{\prime}$ on the top surface of the dielectric?

In Fig. $25-59$ , two parallel-plate capacitors $A$ and $B$ are connected in parallel across a 600 $\mathrm{V}$ battery. Each plate has area 80.0 $\mathrm{cm}^{2}$ ;
the plate separations are 3.00 $\mathrm{mm}$ ;
Capacitor $A$ is filled with air; capacitor $B$ is filled with a dielectric of dielectric constant $\kappa=2.60$ .
Find the magnitude of the electric field within (a) the dielectric of
capacitor $B$ and (b) the air of capacitor $A .$ What are the free charge densities $\sigma$ on the higher-potential plate of (c) capacitor $A$ and
(d) capacitor $B ?$ (e) What is the induced charge density $\sigma^{\prime}$ on the
top surface of the dielectric?

Key Concepts

Induced Bound Charge in Dielectrics

When a dielectric is subjected to an external electric field, its molecules polarize, resulting in induced or bound charges on its surfaces. This polarization leads to an additional charge density (P), given by P = ??(? - 1)E, where E is the electric field within the dielectric. This induced surface (bound) charge opposes the applied electric field and influences the overall behavior of the capacitor without changing the free charge on the plates.

Electric Field in Capacitors

The electric field in a parallel?plate capacitor is typically uniform and is given by the ratio of the voltage (V) across the plates to the distance (d) between them, E = V/d, when the capacitor is filled with air or vacuum. However, when a dielectric with constant ? is used, the internal electric field is reduced to E = V/(?d) because of the polarization of the dielectric material.

Surface Charge Density and Gauss’s Law

Gauss’s law relates the free surface charge density (?) on the plates of a capacitor to the electric field. In an air-filled capacitor, ? = ??E, where ?? is the permittivity of free space. In a dielectric-filled capacitor, although the measured electric field inside is reduced, the free charge density on the plates remains determined by the displacement field, which is given by D = ???E; hence, ? = D remains consistent with the applied voltage and geometry.

Parallel?Plate Capacitor

A parallel?plate capacitor consists of two conductive plates separated by an insulating region (which can be vacuum, air, or another material). Its capacitance, a measure of its ability to store charge for a given voltage, depends on the plate area, the separation distance, and the permittivity of the medium between the plates, following the relation C = ?A/d for a vacuum or air gap, and C = ??A/d when a dielectric is present.

Dielectrics in Capacitors

Dielectrics are insulating materials placed between the plates of a capacitor to increase its capacitance. They are characterized by their dielectric constant (?), which indicates the factor by which the material reduces the effective electric field within the capacitor as compared to a vacuum. Inserting a dielectric leads to a reduction in the electric field (E = V/(?d)) within the material for the same applied voltage, while the free charges on the plates remain determined by the battery.

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