The electrical force acting between Sodium and chlorine ions when a piece of salt is put in water can be determined by considering the effect of the dielectric constant of water on the force.

1. Introduction:
When Sodium and chlorine ions of salt (sodium chloride) are put 1 cm apart in air, a force F acts between them. Now, we need to determine the electrical force acting between these ions when they are put 1 cm apart in water.

2. Permittivity of Air and Dielectric Constant of Water:
The permittivity of a medium determines how much the medium can be polarized by an external electric field. In air, the permittivity is denoted by ε0 (epsilon naught), which is approximately equal to 8.85 x 10^-12 F/m. On the other hand, the dielectric constant of a substance describes its ability to store electrical energy in an electric field. For water, the dielectric constant is denoted by K, which is approximately equal to 80.

3. Calculation of Electrical Force in Air:
When the Sodium and chlorine ions are 1 cm apart in air, the electrical force F between them can be calculated using Coulomb's law:

F = (1 / 4πε0) * (q1 * q2) / r^2

Here, q1 and q2 are the charges on the Sodium and chlorine ions, and r is the separation between them.

4. Calculation of Electrical Force in Water:
When the same Sodium and chlorine ions are put 1 cm apart in water, the electrical force between them can be determined by taking into account the effect of the dielectric constant (K) of water.

The electrical force in water can be calculated as:

F' = K * F

This is because the presence of water as a dielectric material increases the effective permittivity of the system, resulting in a stronger electrical force between the ions.

5. Conclusion:
In conclusion, when a piece of salt is put in water and the Sodium and chlorine ions are 1 cm apart, the electrical force acting between them will be K times the force in air. The dielectric constant of water increases the effective permittivity of the system, leading to a stronger electrical force between the ions.