Two point charges q1 and q2 are at a distance r from each other and interact with the force F.

Two point charges q1 and q2 are at a distance r from each other and interact with the force F. The charge q2 is increased by 3 times, and the distance r is decreased by 2 times. The force of interaction will become equal …

Given:

q1, q2 – point electric charge;

r is the distance between charges;

F is the force of interaction between charges;

q3 = 3 * q2 – charge q2 increased by 3 times;

r1 = r / 2 – the distance between the charges was reduced by 2 times;

k – electrical constant.

It is required to determine F2 – the force of interaction between charges.

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

F1 = k * q1 * q2 / r ^ 2.

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

F2 = k * q1 * q3 / r1 ^ 2 = k * q1 * (3 * q2) / (r / 2) ^ 2 = 12 * k * q1 * q2 / r ^ 2.

Then:

F2 / F1 = (12 * k * q1 * q2 / r ^ 2) / (k * q1 * q2 / r ^ 2) = 12, that is:

F2 = 12 * F1.

Answer: F2 = 12 * F1