The mutual force of repulsion between two point charges kept a fixed distance apart is 9-10-5N when in vacuum and 4-10-5N when placed in a dielectric medium- What is the value of dielectric constant of the medium-

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Standard XII

Physics

Introducing Epsilon

The mutual fo...

Question

The mutual force of repulsion between two point charges kept a fixed distance apart is $9 \times 10^{- 5} N$ when in vacuum and $4 \times 10^{- 5} N$ when placed in a dielectric medium. What is the value of dielectric constant of the medium?

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Solution

Let the dielectric constant of the medium be $K$ .
The force of interaction between the objects is $\frac{F}{K}$ when placed in
the medium of dielectric constant $K$ and $F$ is the force of interaction
between them in vacuum .
According to question ,
$F = 9 \times 10^{- 5} N$
and $\frac{F}{K} = 4 \times 10^{- 5} N$
Dividing both these equations ,
we get dielectric constant , $K$ = $2.25$

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The mutual force of repulsion between two point charges kept a fixed distance apart is 9×10−5N when in vacuum and 4×10−5N when placed in a dielectric medium. What is the value of dielectric constant of the medium?

The mutual force of repulsion between two point charges kept a fixed distance apart is $9 \times 10^{- 5} N$ when in vacuum and $4 \times 10^{- 5} N$ when placed in a dielectric medium. What is the value of dielectric constant of the medium?