A solid sphere of radius R, and dielectric constant ‘k’ has spherical cavity of radius R/4. A point charge q₁ is placed in the cavity. Another charge q₂ is placed outside the sphere at a distance of r from q. Then Coulombic force of interaction between them is found to be ‘F₁’. When the same charges are separated by same distance in vacuum then the force of interaction between them is found to be F₂ then

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$F_{1} = F_{2}$

$F_{1} = F_{2} / k$

$F_{2} = F_{1} / k$

$F_{1} . F_{2} = \frac{1}{k}$

answer is D.

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Detailed Solution

Coulombic force between them remains same as the field at the location of $q_{2}$ remains unchanged.

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