Two parallel plates of area 100 cm^2 are given charges of equal magnitudes 8.9 ×10^-7 C but oppo

step 1 Alright, so in this problem we have a capacitor which is filled in by some materials and it is c

Two parallel plates of area 100 cm^2 are given charges of...

Two parallel plates of area 100 $\mathrm{cm}^{2}$ are given charges of equal magnitudes $8.9 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the diclectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{V} / \mathrm{m} .$ (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Two parallel plates of area 100 $\mathrm{cm}^{2}$ are given charges of equal magnitudes $8.9 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric
field within the diclectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{V} / \mathrm{m} .$ (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conducting plates separated by a specific distance with a certain area. The device stores electrical energy in the electric field created between the plates, and its behavior can be characterized by its capacitance, which is influenced by both the geometry (area and separation distance) and the dielectric material between the plates.

Capacitance with Dielectric

Capacitance in a parallel-plate capacitor with a dielectric is given by the formula C = (???A)/d, where ? is the dielectric constant, ?? is the permittivity of free space, A is the plate area, and d is the separation between the plates. This relation shows that introducing a dielectric increases the capacitance by a factor equal to its dielectric constant as compared to a vacuum.

Dielectric Constant

The dielectric constant is a fundamental material property defined as the ratio of the permittivity of a dielectric material to the permittivity of free space. It quantifies how much the material reduces the electric field relative to vacuum, effectively increasing the capacitance of a capacitor when the dielectric is inserted between its plates.

Induced Polarization and Bound Charges

When a dielectric is placed in an electric field, its molecules become polarized, leading to the formation of induced charges on its surfaces. These bound charges, although not free to move like conduction electrons, affect the overall field distribution within the capacitor. The magnitude of the induced charge is determined by the extent of polarization, which in turn depends on the strength of the electric field and the intrinsic properties of the dielectric material.

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