Question

when the relative permittivity of the medium is increased the force between two charges placed at a given distance apart​

Answer 1

The force of interaction between two stationary points, such as charges Q1 and Q2, separated by a distance r in space, is determined by Coulomb's law.

$F=(1/4 \pi o) Q1Q2/r^2$………….(1).

Here,eo is the permittivity of space.

The force between the charges will be if they are placed in a dielectric medium with permittivity e.

$F'=(1/4\pi )Q1Q2/r^2$……………….(2)

Here, e is permittivity of the dielectric medium.

The ratio of permittivity of medium to permittivity of space is called relative permittivity $er =(e/eo).$

er is also called the dielectric constant of the medium and is denoted by K.

Now, in equation (2) if we replace e by er.eo, then it can be seen clearly that the force between the charges in medium is $(1/er)$or $(1/K)$ times the force between them in space.

As a result, as the dielectric constant increases, the force between the charges decreases.

The preceding description applies to a linear dielectric medium with polarization P proportional to the externally applied electric field, E. When the electric field is increased, the nonlinear elements in the relationship between P and E become more relevant, the effective dielectric constant rises, and the force between the charges in the medium falls.

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Answer 2

Concept introduction:

The mechanism by which the emf between both the charges is reduced in comparison to space is relative permittivity. Similar to absolute permittivity, relative permittivity is the difference between the capacitance of a capacitor using that substance as its dielectric and that of a similar capacitor using vacuum as its dielectric.

Explanation:

We have to answer the question.

A question regarding relative permittivity has been given to us.

The force of interaction between two stationary points, such as charges $Q1$ and $Q2$, separated by a distance $r$ in space, is determined by Coulomb's law.

$F=(1/4\pi 0)Q1Q2/r^{2}$

The force between the charges will be if they are placed in a dielectric medium with permittivity $e$.

$F^{'} =(1/4\pi )Q1Q2/r^{2}$

The ratio of permittivity of medium to permittivity of space is called relative permittivity $er=(e/eo)$

$er$ is also called the dielectric constant of the medium and is denoted by $K.$

Now, in equation $(2)$ if we replace e by $er.eo$, then it can be seen clearly that the force between the charges in medium is or  times the force between them in space.

As a result, as the dielectric constant increases, the force between the charges decreases.

The preceding description applies to a linear dielectric medium with polarization $P$ proportional to the externally applied electric field, $E.$ When the electric field is increased, the nonlinear elements in the relationship between $P$ and $E$ become more relevant, the effective dielectric constant rises, and the force between the charges in the medium falls.

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