Two similar point charges q1 and q2 are placed at a separation ‘r’ in air. When a dielectric slab of thickness t=r2 is placed between the charges, the Coulomb's repulsion force now reduces to 4/9th of the previous value. The dielectric constant of the slab is …….

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Two similar point charges $q_{1} and q_{2}$ are placed at a separation ‘r’ in air. When a dielectric slab of thickness $(t = \frac{r}{2})$ is placed between the charges, the Coulomb's repulsion force now reduces to 4/9^th of the previous value. The dielectric constant of the slab is …….

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answer is 4.

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Detailed Solution

Initially, $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

If a dielectric slab of thickness ‘t’ and dielectric constant k is introduced, the equivalent separation between charges becomes $\sqrt{k} t + r - t$

$\Rightarrow F' = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{{[\sqrt{k} t + (r - t)]}^{2}}$

Given $F' = \frac{4}{9} F \Rightarrow \frac{1}{{(\sqrt{k} t + r - t)}^{2}} = \frac{4}{9} \frac{1}{r^{2}}$

$= \frac{\sqrt{k}}{2} + \frac{1}{2} = \frac{3}{2} \Rightarrow \frac{\sqrt{k}}{2} = 1 \Rightarrow \sqrt{k} = 2 \Rightarrow k = 4$

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