What would be the electrostatic force between two charges of equal magnitude if the distance between them is halved

the forces between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them

The force becomes 4 times as much when the distance is halved.

Explanation:

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

F2 = kq^2/1 / (r/2)^2

Set up a proportion using F1/F2 as you reference.

Notice that you don't know any number that can be used in this equation. However k q^2 does not change as you move the two charges closer together. That being so, you can still show what happens to F1 and F2.

F1/F2 = 1/r^2 // 1/(r^2/4)

Take the denominator, invert the fraction and multiply. That creates another cancellation.

F1/F2 = 1/r^2 * r^2 /4 1

F1/F2 = r^2 // r^2/4          which allows you to cancel r^2

F1/F2 = 1/4                       Now you need to interpret F1 and F2. Cross Multiply

4F1 = F2

d

explanation: