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# Permittivity, Relative Permittivity (or the dielectric constant), practice problems, FAQs

Till now you might know that charge particles exert force on each other but did you know that the participating medium between the charge particles affects this force of interaction? In vacuum the force will be more as compared to some other medium. There is a property of the medium involved on which the force depends. Let’s discuss that property!

Table of content:

• Permittivity
• Relative Permittivity (or the dielectric constant)
• Practice problems
• FAQs

## Permittivity:

The absolute permittivity, or plainly called permittivity (denoted by epsilon, ) is actually measurement of the electric polarizability of a dielectric medium. A material whose permittivity is high polarizes more with an applied electric field than a material with low permittivity. Thus, the dielectric medium with high permittivity stores more energy in the material.

Polarization of medium in an electric field

The magnitude of electric force by a charge q1 on another charge q2 at a distance r is given by ${F}_{e}=k\frac{{q}_{1}{q}_{2}}{{r}^{2}}$ where the proportionality constant k depends on the system of units used.

In SI units,

The numerical value of this constant is defined in terms of speed of light as,

This constant k is written as $k=\frac{1}{4\pi {\epsilon }_{0}}$

Where, 0 is the permittivity of free space.

## Relative Permittivity (or the dielectric constant):

The relative permittivity (or the dielectric constant) is expressed as a ratio of the permittivity of a material to the electric permittivity of a vacuum medium. A dielectric in an insulating material and the dielectric constant of an insulator measures the ability of the insulator to store electric energy in an electrical field. Energy is given by εE22, where E is the electric field.

When the medium is not vacuum,

Permittivity is a material's property that affects the Coulomb force between two point-charges in the material. Relative permittivity is the factor by which the electric field between the charges is decreased relative to vacuum.

r= dielectric constant or the relative permittivity of the medium

## Practice problems:

Q. Deduce the dimension of $k=\frac{1}{4\pi {\epsilon }_{0}}$

A. By Coulomb’s law, ${F}_{e}=\frac{1}{4\pi \epsilon }\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

So, $=\left[\frac{{F}_{e}{r}^{2}}{{q}_{1}{q}_{2}}\right]=\frac{\left[ML{T}^{-2}\right]\left[{L}^{2}\right]}{\left[AT\right]\left[AT\right]}=\left[M{L}^{3}{A}^{-2}{T}^{-4}\right]$

Q. What will be the permittivity of a medium with relative permittivity 5 ?

A. ε=r0

Where =permittivity of a medium

r=relative permittivity

So,

Q. Deduce the dimension of .

A. By Coulomb’s law, ${F}_{e}=\frac{1}{4\pi \epsilon }\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

So, $\left[\epsilon \right]=\frac{\left[{q}_{1}{q}_{2}\right]}{\left[{F}_{e}{r}^{2}\right]}=\frac{\left[AT\right]\left[AT\right]}{\left[ML{T}^{-2}\right]\left[{L}^{2}\right]}=\left[{M}^{-1}{L}^{-3}{A}^{2}{T}^{4}\right]$

Q. Two point-charges are placed in air medium with 1 m distance in between them. The force of electrostatic interaction is 4 N. If they are placed inside glass medium of r=4, what will be the force of interaction?

A. By Coulomb’s law,

So,

## FAQs:

Q. What is negative permittivity?
A.
Negative permittivity means that the electric displacement vector and the electric field vector point in the opposite directions.

Q. What is the dimension of relative permittivity r?
A.
The relative permittivity (or the dielectric constant) is expressed as a ratio of the permittivity of a material to the electric permittivity of a vacuum medium.

ε=r0

Where =permittivity of a medium

r=relative permittivity

So relative permittivity is dimensionless.

Q. What is the relation of electrostatic force with the relative permittivity?
A.
By Coulomb’s law, ${F}_{e}=\frac{1}{4\pi \epsilon }\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

Where q1 and q2 are the point charges, r is the distance between them, =permittivity of a medium

We know, ε=r0

Where =permittivity of a medium

r=relative permittivity

So, ${F}_{e}=\frac{1}{4\pi {{\epsilon }_{r}\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}$

Q. Sort them according to their magnitude of relative permittivity (low to high).

Water, Glass, Air.

A. Air, Glass, Water

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