The force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) in a vacuum is given by Coulomb’s law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}, \]
where \(\epsilon_0\) is the permittivity of free space.
In a medium with dielectric constant \(K\), the permittivity changes from \(\epsilon_0\) to \(K \epsilon_0\). This reduces the effective force between the charges by a factor of \(K\). Thus, the force in the medium, if the distance remained \(r\), would be:
\[ F_{\text{medium}} = \frac{1}{4 \pi K \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{F}{K}. \]
For \(K = 5\), this becomes:
\[ F_{\text{medium}} = \frac{F}{5}. \]
Now, since the distance between the charges is reduced to \(\frac{r}{5}\), we need to adjust for this change. Coulomb’s force varies inversely with the square of the distance, so reducing the distance by a factor of \(\frac{1}{5}\) increases the force by a factor of \(\left(\frac{1}{5}\right)^{-2} = 25\).
Combining both effects (the dielectric and the reduced distance), the modified force \(F'\) in the medium is:
\[ F' = \frac{F}{5} \times 25 = 5F. \]
Thus, the force between the charges in the medium, with the distance changed to \(\frac{r}{5}\), is increased by a factor of 5 compared to the original force in a vacuum. Therefore, the answer is: \[ 5F. \]
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: