The electric force between two point charges separated by a certain distance in air is $F$ the distance at which they should be placed in a medium of relative permittivity $k$ so that the force remain the same is:

d

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\frac{d}{k}

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k d

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\frac{d}{\sqrt{k}}

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Solution

The correct option is D $\frac{d}{\sqrt{k}}$
$F_{m} = F_{a}$

$\frac{1}{4 π ε_{0} k} \frac{q_{1} q_{2}}{{d_{m}}^{2}} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{d_{a}^{2}}$

$\Rightarrow d_{a}^{2} = k d_{m}^{2}$

$\Rightarrow d_{m} = \frac{d}{\sqrt{k}}$

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The electric force between two point charges separated by a certain distance in air is F the distance at which they should be placed in a medium of relative permittivity k so that the force remain the same is:

The correct option is D d√kFm=Fa14πε0kq1q2dm2=14πε0q1q2d2a⇒d2a=kd2m⇒dm=d√k

The electric force between two point charges separated by a certain distance in air is $F$ the distance at which they should be placed in a medium of relative permittivity $k$ so that the force remain the same is:

The correct option is D $\frac{d}{\sqrt{k}}$
$F_{m} = F_{a}$

$\frac{1}{4 π ε_{0} k} \frac{q_{1} q_{2}}{{d_{m}}^{2}} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{d_{a}^{2}}$

$\Rightarrow d_{a}^{2} = k d_{m}^{2}$

$\Rightarrow d_{m} = \frac{d}{\sqrt{k}}$