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?
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).
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.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.