Two parallel plates of area 100 cm^2 are given charges of equal magnitudes 8.4 ×10^-7 C but opp

step 1 In this question we've been asked to find out two things. So firstly we have been given two para

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 $8.4 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the dielectric 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

Dielectric Constant

The dielectric constant (or relative permittivity) is a measure of a material's ability to reduce the electric field within it compared to the field in a vacuum. It quantifies how much the material can be polarized by an applied electric field, thereby reducing the effective field strength within the capacitor and increasing its capacitance. This concept is fundamental when analyzing the behavior of capacitors with dielectric materials between their plates.

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a certain distance, with an electric field established between them when they are charged with opposite signs. The capacitance of such a system is determined by the area of the plates, the distance between them, and the permittivity of the dielectric material filling the space. This idealized model forms the basis for understanding more complex capacitor systems.

Polarization and Induced Surface Charges

When a dielectric is placed in an electric field, its molecules become polarized, meaning that they develop induced dipoles with surface charges accumulating at the boundaries of the material. These induced surface charges create an additional field that opposes the applied field, reducing the net field within the dielectric. The concept of polarization is key in determining both the effective dielectric constant and the behavior of charges at the interface between the dielectric material and the conductors.

Transcript

00:01 In this question we've been asked to find out two things.

00:03 So firstly we have been given two parallel plates and this makes a capacitor and the area of the plates is given to us as 100 centimetre square which we can write it as 100 into 10 raised to the power minus 4 meters squared.

00:18 And we've also been given that the magnitude of charge which is given to the plates is 8 .4 into 10 raised to the power minus 7 columns.

00:27 And now they've been now we've also been given another important you know quantity which is the electric field between the plates which is 1 .4 into 10 raised to the power 6 volts per meter firstly we have to find out so number one in the first part we have to find out the electric constant the dielectric constant that is present so this electric field which is the value that is given to us 1 .4 has been produced when an dielectric is present between the plates.

01:02 So they're assuming that there is a dielectric of constant, let's say, k, this is what we have to find out.

01:07 So let's see how we can do this.

01:09 So one important concept that you need to know is that whenever a dielectric is introduced between two plates or in a capacitor, the electric field reduces by the factor of k.

01:23 So this is a new electric field.

01:24 So this value has been given to us.

01:26 And from here we have to find out the value of k.

01:29 So let's see how we can solve this.

01:31 So we can write k as epsilon nor, sorry, this is e0, which is the initial electric field when there is no dielectric upon when there is dielectric.

01:41 So this can be written as q upon a epsilon knot e.

01:47 So this is how we can find out, you know, make an equation for k and find out k.

01:53 Now let's move on.

01:55 So q has been given to us as 8 .4 into 10 raised to the power minus 7...

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