The force between two point charges placed in a material medium of dielectric constant $\varepsilon_\text{r}$ is $\text{F}.$ If the material is removed, then the force between them becomes

$\frac{\mathrm{F}}{\varepsilon_{\mathrm{r}}}$

$\frac{\varepsilon}{\mathrm{F}}$

JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

1 Medium

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Understanding +4 | -1

The force between two point charges placed in a material medium of dielectric constant

\varepsilon_\text{r}

\text{F}.

If the material is removed, then the force between them becomes

View answer and explanation

Correct option: A

\varepsilon_{\mathrm{r}} \mathrm{F}

Option A (

\varepsilon_{\mathrm{r}} \mathrm{F}

) is correct because removing the dielectric medium increases the force by a factor of the dielectric constant.

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r.

4N

\left( {{\varepsilon _r} = 8} \right)

N

5 \times {10^{ - 2}}