Force between two point charges \( q_1 \) and \( q_2 \) placed in vacuum at \( r \) cm apart is \( F \). Force between them when placed in a medium having dielectric \( K = 5 \) at \( r/5 \) cm apart will be:
Coulomb's Law describes the electrostatic force \(F\) between two point charges \( q_1 \) and \( q_2 \) in a vacuum separated by a distance \( r \), given by \(F = \frac{{k \cdot q_1 \cdot q_2}}{{r^2}}\), where \( k \) is Coulomb's constant.
In a medium with dielectric constant \( K \), the force is reduced by a factor of \( K \), resulting in a modified force \( F_{\text{medium}} = \frac{{F}}{{K}} \).
We need to determine the force when charges are in a medium with \( K = 5 \) and at a new distance \( r' = \frac{r}{5} \).
Adjust for the dielectric medium: The force becomes \(F_{\text{medium}} = \frac{F}{5}\).
Adjust for the new distance: The force is inversely proportional to the square of the distance. With \( r' = \frac{r}{5} \), the force is \(F_{\text{new}} = \frac{k \cdot q_1 \cdot q_2}{\left(\frac{r}{5}\right)^2} = \frac{k \cdot q_1 \cdot q_2}{\frac{r^2}{25}} = 25 \cdot F\).
Combine both effects: The final force is \(F_{\text{final}} = \frac{F_{\text{new}}}{K} = \frac{25F}{5} = 5F\).
Therefore, when the charges are in a medium with \( K = 5 \) and at a distance of \( \frac{r}{5} \), the force between them is \( 5F \).
