# How will the force of interaction between two charged bodies with charges Q and

**How will the force of interaction between two charged bodies with charges Q and q change, if at q = const, the charge Q is doubled and the distance between the charges is also doubled?**

Given:

q is the first point charge;

Q1 = 2 * Q – the second point charge was doubled;

r1 = 2 * r – the distance between charges was doubled.

It is required to determine F2 / F1 – how the force of interaction between charges will change after the changes made.

The force of interaction between charges in the first case will be equal to:

F1 = k * q * Q / r ^ 2, where k is an electrical constant.

In the second case, the force of the Coulomb interaction will be equal to:

F2 = k * q * Q1 / r1 ^ 2 = k * q * 2 * Q / (2 * r) ^ 2 = k * q * 2 * Q / (4 * r ^ 2) = k * q * Q / (2 * r ^ 2).

Then:

F2 / F1 = (k * q * Q / (2 * r ^ 2)) / (k * q * Q / r ^ 2) = 1/2, that is, will decrease by 2 times.

Answer: after the changes made, the force of the Coulomb interaction will decrease by 2 times.