What happens to the force acting between the charged particles, if the distance between these charged particles is halved?

The correct option is "It increases by four times".

To understand why this is the case, we can use Coulomb's Law, which describes the electrostatic force between two charged particles. Coulomb's Law is mathematically expressed as:

$$F=k·\frac{q_1{q}_2}{r^2}$$

where:
- _F is the force between the charges,
- _k is Coulomb's constant,
- _q1 and _q2 are the magnitudes of the charges, and
- _r is the distance between the two charges.

From this formula, we can see that the force _F is inversely proportional to the square of the distance _r between the charges:

$$F\propto \frac{1}{r^2}$$

Now, let the initial distance be _r and the initial force be _F. If the distance between the charged particles is halved, the new distance _r' becomes:

$$r^{\text{'}}=\frac{r}{2}$$

Substituting this new distance into the proportionality relation, the new force _F' is:

$$F^{\text{'}}\propto \frac{1}{{\left(r^{\text{'}}\right)}^2}=\frac{1}{{\left(\frac{r}{2}\right)}^2}=\frac{1}{\left(\frac{r^2}{4}\right)}=4\left(\frac{1}{r^2}\right)$$

Since the initial force _F is proportional to 1/r², we find:

$$F^{\text{'}}=4F$$

Thus, halving the distance reduces the denominator in Coulomb's Law by a factor of four, which in turn increases the electrostatic force acting between the charged particles by four times.