Solution:

The force between two charges varies inversely with the square of the distance between them, according to Coulomb's law:

\[
F \propto \frac{1}{r^2}
\]

Solution Steps:

1. Initial and Final Distances:
- Initial distance \( r_1 = 0.06 \, \text{cm} \)
- Final distance \( r_2 = 0.06 - 0.01 = 0.05 \, \text{cm} \)

2. Force Ratio:
\[
\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}
\]

3. Substitute Values:
\[
\frac{F_2}{5} = \frac{(0.06)^2}{(0.05)^2}
\]
\[
F_2 = 5 \cdot \frac{0.0036}{0.0025} = 5 \cdot 1.44 = 7.2 \, \text{N}
\]

Conclusion:
The new force \( F_2 \) is \( 7.2 \, \text{N} \).