Introduction:
We have two identical small spherical conductors separated by a distance D and carrying equal positive charges. We are given that half of the charge on one conductor is transferred to the other. We need to determine the new force between them and explain the reasoning behind it.

Understanding the scenario:
To solve this problem, we need to consider the concept of electric force between charged objects. The force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, this can be expressed as F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

Analysis:
1. Initially, both conductors carry equal positive charges, so let's denote the charge on each conductor as q.
2. The force between them is given as F.
3. When half of the charge is transferred from one conductor to the other, the charge on each conductor becomes q/2.
4. We need to find the new force between the conductors.

Solution:
1. Initially, the force between the conductors is given by F = k * (q * q) / D^2.
2. After transferring half of the charge, the charge on each conductor becomes q/2.
3. Using the formula for electric force, the new force between the conductors can be calculated as F' = k * [(q/2) * (q/2)] / D^2.
4. Simplifying the expression, we get F' = (1/4) * F.
5. Therefore, the new force between the conductors is one-fourth of the initial force.

Explanation:
When half of the charge is transferred from one conductor to the other, the force between them decreases by a factor of four. This is because the force between charged objects is directly proportional to the product of their charges. As the charges on the conductors are halved, the force between them is reduced to one-fourth of its initial value.

Conclusion:
The new force between the identical small spherical conductors, after transferring half of the charge from one conductor to the other, is one-fourth of the initial force. This decrease in force can be attributed to the reduction in the charges on the conductors.