Force between two point charges \( q_1 \) and \( q_2 \) placed in vacuum at a distance \( r \, \text{cm} \) apart is \( F \). Force between them when placed in a medium having dielectric \( K = 5 \) at \( r/5 \, \text{cm} \) apart will be:
Show Hint
When charges are placed in a medium, always account for the dielectric constant \( K \) and any change in the distance between the charges. Both factors influence the resultant force.
Coulomb's Law defines the electrostatic force between two charges in a vacuum as:\[F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}.\]When these charges are in a medium with a dielectric constant \( K \), the force is modified to:\[F' = \frac{1}{4\pi K \epsilon_0} \cdot \frac{q_1 q_2}{r'^2}.\]Given \( K = 5 \) and \( r' = \frac{r}{5} \), the force becomes:\[F' = \frac{1}{4\pi (5\epsilon_0)} \cdot \frac{q_1 q_2}{\left(\frac{r}{5}\right)^2}.\]Simplifying this expression yields:\[F' = \frac{25}{5} \cdot \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}.\]Which simplifies further to:\[F' = 5F.\] Conclusion: The force between the charges in the specified medium is \( 5F \).